Electrical Engineering Foundations

LESSON 1.1

Charge, Current, Voltage, Energy, and Power

Separate electrical quantities and use a reference direction to make energy calculations meaningful.

Learning objectives

• Distinguish charge, current, voltage, energy, and power by definition and unit.
• Calculate transferred charge and supplied energy for a constant operating point.
• Interpret the sign of power using stated terminal references.

Start with quantities, not component names

• Charge counts the transferred electrical quantity. Charge is measured in coulombs, written C. Current describes how quickly charge crosses a chosen surface: I = delta Q / delta t for an average over an interval. One ampere is one coulomb per second. A current value therefore needs a time interval before it can determine transferred charge. Current is not a quantity that a resistor permanently uses up.
• Voltage compares two locations. Voltage is electrical potential difference, measured in volts. A statement such as “the voltage is five” is incomplete until the two reference locations and unit are identified. In a schematic, label the positive and negative terminals or state that a node voltage is measured relative to the selected reference node. Reversing these references reverses the numerical sign without changing the physical arrangement.
• Energy and power answer different questions. Energy measures a transferred amount in joules; power measures its rate in watts. For constant voltage and current, P = V I and E = P t. A device drawing less power can still transfer more energy if it operates for sufficiently longer. Always convert minutes or hours when an answer is requested in joules, because a watt is one joule per second.

Worked example: a fictional indicator

• Compute charge and energy separately. An ideal source supplies a constant 5 V while a modeled load draws 0.020 A for 30 s. Transferred charge is Q = I t = 0.020 x 30 = 0.60 C. Power is P = V I = 5 x 0.020 = 0.100 W. Energy is E = P t = 0.100 x 30 = 3.00 J. Charge, power, and energy have different units and cannot be exchanged as though they described the same thing.
• Check the power convention. Under the passive sign convention, current entering the labeled positive terminal gives positive absorbed power. A source delivering energy has negative absorbed power when the same convention is applied to it. In the ideal two-element model, the source delivers 0.100 W and the load absorbs 0.100 W. This bookkeeping is an energy check, not evidence that every real source can supply any requested current.
• State the model limits. The calculation assumes constant values and ignores startup behavior, internal resistance, and conversion losses. A real load may vary its current with time. Then energy depends on the time history of power rather than one convenient snapshot. Before comparing designs, state whether your values are instantaneous, average, or measured over a specified interval.

Independent exercise and model reasoning

• Try a changed duration. A simulated load draws 0.015 A from 4 V for 20 s. Predict charge, power, and energy before calculating. Model reasoning: Q = 0.30 C, P = 0.060 W, and E = 1.20 J. Doubling only the duration doubles charge and energy while leaving constant operating power unchanged. Explain that distinction in a short labeled energy-flow sketch.
• Avoid the common shortcut. A voltage rating alone does not specify current or energy use. Record the load model, reference direction, and operating interval before reporting an electrical result.

Applied scenario

A team compares two simulated status indicators and must explain why operating time matters as much as power.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

A constant 3 V load draws 0.010 A for 10 s. How much energy does it absorb?

1. 0.030 J
2. 0.300 J
3. 3.00 J
4. 30.0 J

LESSON 1.2

Ohm’s Law and Resistor Power

Use a linear resistor model and distinguish a calculation from a component operating limit.

Learning objectives

• Solve V = I R with consistent SI prefixes.
• Calculate resistor power using equivalent formulas.
• Explain the constant-resistance assumption and its limits.

A resistor is a relationship in a model

• Begin with the model equation. For an ideal linear resistor, V = I R relates the voltage across the element to current through it. Resistance R is measured in ohms. This equation does not mean current causes voltage in every possible circuit arrangement; it describes the allowed pairs of voltage and current for that element. The rest of the network determines the actual operating point.
• Choose references consistently. Label current entering the terminal marked positive for voltage. Positive resistance then produces V and I with the same sign, so absorbed power is nonnegative: P = V I = I squared R = V squared / R. These are equivalent expressions for this model, not independent laws that can be applied with mismatched component voltages and currents.
• Convert prefixes before judging the answer. A kilo-ohm is 1000 ohms, while a milliampere is 0.001 ampere. Dividing volts by kilo-ohms gives milliamperes numerically, but write the conversion explicitly until the unit reasoning is familiar. A misplaced factor of 1000 can make a plausible-looking calculation inconsistent with the circuit it claims to describe.

Worked example: compare two modeled loads

• Solve the first operating point. Place an ideal 1.0 kilo-ohm resistor across a hypothetical 5.0 V source in a simulation. I = V / R = 5.0 / 1000 = 0.0050 A, or 5.0 mA. Absorbed power is 5.0 x 0.0050 = 0.025 W. Cross-check with V squared / R = 25 / 1000 = 0.025 W. Both methods use the voltage across that same resistor.
• Change one parameter and predict first. If the voltage doubles to 10 V while resistance stays fixed, current doubles to 10 mA. Power becomes 100 / 1000 = 0.100 W, four times the initial value. The square relationship matters: increasing voltage does not merely increase heating-related power at the same linear rate. A parameter sweep makes this distinction visible without building physical hardware.
• Separate ideal behavior from component suitability. An ideal resistor has no temperature coefficient, voltage limit, or manufacturing variation. Real resistance can change with temperature, and a real part has specified operating limits. This lesson calculates ideal power only; it does not approve a physical part or installation. In engineering documentation, distinguish the modeled demand from the separate process of checking component specifications and environment.

Independent exercise and model reasoning

• Use two calculations to detect an error. For 6 V across 2.0 kilo-ohms, predict current and power. Model reasoning: I = 6 / 2000 = 0.003 A = 3 mA; P = 6 x 0.003 = 0.018 W. The alternative expression gives 36 / 2000 = 0.018 W. If the two answers differ, check prefixes and whether the voltage belongs to the selected component.
• Explain a misconception. A larger resistance draws less current only when the voltage across it is held fixed. If a circuit instead fixes current, increasing resistance increases the required voltage and absorbed power. Always name the controlled quantity before describing what “more resistance” will do.

Applied scenario

A simulated indicator design is changed from a 5 V model to a 10 V model, and the team must explain the resulting resistor-power change.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

At fixed voltage, doubling a positive ideal resistance changes absorbed power by what factor?

1. It doubles.
2. It becomes four times as large.
3. It becomes one half.
4. It stays the same.

LESSON 1.3

Series and Parallel Resistor Networks

Identify connectivity from nodes and compare equivalent resistance, branch current, and voltage drop.

Learning objectives

• Identify series and parallel relationships using node connections.
• Calculate equivalent resistance for simple resistor networks.
• Check branch results with total current and power.

Connectivity decides the relationship

• Use nodes instead of drawing position. A node joins points treated as the same electrical potential in an ideal wire model. Two resistors are parallel when both ends connect to the same pair of nodes. They carry the same voltage even if the schematic places them far apart. Two elements are in series when their shared node has no other current path, so their currents must be equal.
• Derive the series rule from voltage addition. With current I through each of two series resistors, the total drop is I R1 + I R2 = I (R1 + R2). The equivalent resistance is therefore R1 + R2. A series element increases the resistance seen by the source in this simple arrangement. Equal current does not imply equal voltage drop unless the resistances are also equal.
• Derive the parallel rule from current addition. Parallel branches have common voltage V, while their currents add: Itotal = V / R1 + V / R2. Thus 1 / Req = 1 / R1 + 1 / R2. For two positive resistors, Req = R1 R2 / (R1 + R2), which is less than either branch resistance. An extra parallel path increases source current at fixed voltage.

Worked example: the same two values, different wiring

• Analyze series first. Connect ideal 1000-ohm and 2000-ohm resistors in series across 6 V. Req = 3000 ohms and I = 6 / 3000 = 0.002 A. The drops are 2 V and 4 V, adding to the 6 V source. Total absorbed power is V I = 0.012 W. Individual powers, 0.004 W and 0.008 W, sum to the same result.
• Analyze parallel separately. Connect the same resistor values across the same two 6 V nodes. Branch currents are 6 mA and 3 mA, giving a total of 9 mA. Req = 6 / 0.009 = approximately 666.7 ohms. Branch powers are 0.036 W and 0.018 W, totaling 0.054 W. Reusing the series current here would be a connectivity error, not a rounding issue.
• Explain what the ideal source hides. This comparison assumes the source maintains exactly 6 V regardless of demanded current and that wires have no resistance. Real source behavior can differ under load. A network equivalent reproduces the terminal voltage-current relationship; it does not preserve every internal voltage or make all internal components interchangeable for other purposes.

Independent exercise and model reasoning

• Test an intuitive prediction. Two 3000-ohm resistors are placed in parallel across 9 V. Find Req, each branch current, and total power. Model reasoning: Req = 1500 ohms, each branch carries 3 mA, total current is 6 mA, and power is 0.054 W. The equivalent resistance is smaller than 3000 ohms because two current paths are available.
• Avoid a visual trap. Crossing lines are not automatically connected; a schematic’s junction convention matters. Redraw the circuit with labeled nodes, identify each element’s two endpoints, and only then choose a reduction rule. Preserve that node-labeled drawing as evidence of your reasoning.

Applied scenario

Two drawings use identical resistor values but different connections; an analyst must explain why their source-current predictions differ.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

Two 2 kilo-ohm resistors are connected in parallel. What is their equivalent resistance?

1. 4 kilo-ohms
2. 2 kilo-ohms
3. 1 kilo-ohm
4. 0.5 kilo-ohms

LESSON 2.1

Kirchhoff’s Laws and Node Analysis

Use conservation with explicit signs to solve a node connected to several branches.

Learning objectives

• Apply current conservation at a node.
• Write a consistent voltage-loop equation.
• Solve a one-node resistor network and verify its result.

Conservation makes a circuit solvable

• Current conservation is a statement about a node. Kirchhoff’s current law, KCL, says the algebraic sum of currents at an ideal lumped-circuit node is zero. Choose a consistent sign convention, such as positive current leaving the node. A negative calculated value then means the actual current flows opposite your assumed direction. It does not mean the mathematics has failed or charge is disappearing.
• Voltage conservation is a statement about a closed path. Kirchhoff’s voltage law, KVL, says the signed voltage changes around a closed lumped-circuit loop sum to zero under the model’s assumptions. Choose a traversal direction, record a rise when moving from a source’s negative to positive terminal, and record drops consistently. Do not add magnitudes without regard to polarity or count one element twice.
• Treat node voltage as the unknown. Node analysis expresses branch currents using voltage differences divided by resistance. For a branch between node V and a known node Vs, the current leaving V through that resistor is (V - Vs) / R. The same form works even when the answer is negative. Ground is the chosen reference potential in this model, not an automatic claim about a physical earth connection.

Worked example: one unknown node

• Write the network before simplifying. A node V connects to a 6 V node through 1000 ohms and to the 0 V reference through two separate 2000-ohm branches. KCL with all currents leaving V gives (V - 6) / 1000 + V / 2000 + V / 2000 = 0. Multiplying by 2000 gives 2V - 12 + V + V = 0, so V = 3 V.
• Interpret the branch signs. The first branch current is (3 - 6) / 1000 = -3 mA, meaning 3 mA enters V from the source side. Each branch to the reference carries 3 / 2000 = 1.5 mA out. Incoming 3 mA equals outgoing 1.5 + 1.5 mA. A loop through the source, input resistor, and either output resistor has a 6 V rise and two 3 V drops.
• Check what the model assumes. Ideal wires have negligible voltage drop and the lumped model treats each element with localized terminal quantities. It is not a complete electromagnetic description at all frequencies or geometries. At this stage, the purpose is to make reference directions and conservation visible, not to claim every real interconnect is an ideal node.

Independent exercise and model reasoning

• Change the source and keep the topology. Replace 6 V with 8 V in the same network. Model reasoning: the two output resistors are equivalent to 1000 ohms, matching the input resistor, so V = 4 V. Input current is 4 mA and each output branch carries 2 mA. Verify using the full KCL equation as well as the equivalent-resistance shortcut.
• Use a failed check diagnostically. If currents do not balance, inspect reference directions, node labels, and whether a branch was omitted. Do not repair the result by changing signs after the fact without explaining the chosen convention. Preserve the original equation and mark the corrected reasoning.

Applied scenario

A simulated sensor input feeds two identical loads; a reviewer asks for proof that the predicted node voltage satisfies conservation.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

A node receives 7 mA and has two outgoing branches. One carries 2 mA. What must the other carry in the steady ideal model?

1. 9 mA
2. 7 mA
3. 5 mA
4. 2 mA

LESSON 2.2

Capacitance, Stored Energy, and Initial Conditions

Explain charge storage and why a capacitor requires a time-dependent description.

Learning objectives

• Relate capacitance, voltage, and stored charge.
• Calculate ideal capacitor energy and interpret its voltage dependence.
• Use initial conditions to distinguish transient from steady behavior.

Storage changes what a circuit remembers

• Capacitance relates stored charge to voltage. In an ideal linear capacitor, Q = C V, where capacitance C is measured in farads. This C symbol names capacitance, while the unit C used for coulombs names charge; context and units keep them separate. A capacitor stores separated charge and energy in an electric field. It is not simply a resistor whose value happens to be large.
• Current depends on voltage change. The ideal relationship is i = C dv/dt under the passive sign convention. A constant capacitor voltage gives zero ideal capacitor current. A rapidly changing voltage can require substantial current even if its instantaneous value is small. Conversely, finite current changes voltage progressively. An instantaneous voltage jump would require an impulse of current in the ideal model, so ordinary finite-current switching preserves the initial capacitor voltage.
• Energy depends on voltage squared. Stored energy is E = one half C V squared. Doubling the voltage at fixed capacitance quadruples the stored energy. This is a model calculation, not an instruction to charge hardware. Real capacitors have specified voltage limits, leakage, series resistance, and other behaviors that the ideal element does not capture. All course experiments remain on paper or in simulation.

Worked example: compare charge and energy

• Convert capacitance carefully. A 100 microfarad capacitor has C = 100 x 10^-6 F = 0.000100 F. At 5 V, Q = 0.000100 x 5 = 0.000500 coulomb, or 0.500 millicoulomb. Stored energy is 0.5 x 0.000100 x 25 = 0.00125 J, or 1.25 millijoules. Microfarads and millijoules involve different prefixes; carry units through each line.
• Compare another operating point. At 10 V on the same ideal capacitance, charge becomes 0.00100 coulomb, twice the first result. Energy becomes 0.00500 J, four times the first result. The different scaling follows the area under the voltage-versus-charge relationship. It explains why a statement about “twice the charge” does not automatically imply twice the energy.
• Explain an initial condition. Suppose the modeled capacitor is initially at 2 V when it is connected through a positive resistor to a 5 V ideal source. Its voltage begins at 2 V and changes toward 5 V, while the resistor initially sees 3 V. Treating the capacitor as an immediate short circuit would discard the stored state and give the wrong initial current for this case.

Independent exercise and model reasoning

• Practice the squared relationship. An ideal 200 microfarad capacitor is at 3 V. Find its charge and energy. Model reasoning: Q = 200 x 10^-6 x 3 = 0.000600 C, and E = 0.5 x 200 x 10^-6 x 9 = 0.000900 J. If the voltage becomes 6 V, charge doubles and energy quadruples to 0.00360 J.
• Avoid a steady-state shortcut during switching. An ideal capacitor acts as an open branch only for constant-voltage DC steady state, after relevant transients have settled. Record the initial voltage and the time at which an observation is made before deciding which simplified picture applies.

Applied scenario

A learner compares two simulation screenshots taken just after switching and long after switching, then explains why both can be correct.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

At fixed capacitance, tripling capacitor voltage changes stored energy by which factor?

1. Three
2. Six
3. Nine
4. One third

LESSON 2.3

RC Charging and the Time Constant

Predict a first-order charging curve and use its limiting cases as evidence.

Learning objectives

• Calculate the time constant of a series RC network.
• Predict capacitor voltage during charging from an initial state.
• Check a transient formula at the initial time and long-time limit.

One time constant describes a first-order model

• Connect the physical relationships. In a source-resistor-capacitor series model, the resistor limits current while the capacitor voltage changes. As the capacitor approaches the source voltage, the voltage across the resistor decreases, so current decreases too. The resulting exponential response is not a constant-rate ramp. Its shape follows both Ohm’s law and the capacitor current relation, rather than an arbitrary curve chosen for appearance.
• Define the time constant with units. Tau = R C. Ohms multiplied by farads gives seconds. For a source step to Vs and initial capacitor voltage V0, vc(t) = Vs + (V0 - Vs) exp(-t/tau), for t at or after switching. This expression assumes a single ideal resistor-capacitor first-order circuit and a constant source after the step. More complicated networks can require different equivalent resistance or more than one state.
• Interpret the percentage correctly. After one tau, about 63.2% of the total change from V0 to Vs has occurred, leaving about 36.8% of the initial difference. This does not mean every capacitor reaches 63.2% of its source voltage when it starts from a nonzero voltage. About five tau brings the response within 0.674% of its final difference; the ideal exponential never becomes exactly final at a finite time.

Worked example: a five-volt step

• Calculate before plotting. Let R = 10 kilo-ohms, C = 100 microfarads, Vs = 5 V, and V0 = 0 V. Tau = 10000 x 0.000100 = 1.00 s. At t = 1 s, vc = 5(1 - exp(-1)) = approximately 3.16 V. The resistor voltage is approximately 1.84 V, so current is 1.84 / 10000 = 0.184 mA. Initially, the current is 5 / 10000 = 0.500 mA.
• Verify the endpoints. At t = 0, exp(0) = 1, giving vc = 0 V as required by the initial condition. As time grows, the exponential tends to zero, giving vc approaching 5 V and current approaching zero. At 5 s, vc is approximately 4.966 V. These checks catch many sign mistakes that a plausible-looking mid-curve value might hide.
• Explain the effect of a changed parameter. Doubling R while holding C fixed doubles tau and halves initial current for this case. It does not change the final capacitor voltage in the ideal no-load model. Adding a load or source resistance changes the model and may change both the effective time constant and final value. State the network being analyzed before comparing curves.

Independent exercise and model reasoning

• Use a nonzero starting state. A 2 s time-constant network starts at 1 V and approaches 5 V. Find vc at t = 2 s. Model reasoning: vc = 5 + (1 - 5) exp(-1) = approximately 3.53 V. The rise is 2.53 V, which is 63.2% of the total 4 V change. It is not 63.2% of 5 V.
• Record a model limitation. An ideal charging trace omits tolerances, leakage, and measurement loading. Preserve predicted values and assumptions alongside the curve, then explain which change would require you to revise the model.

Applied scenario

A simulated indicator should delay its response; the designer needs to distinguish a slower transient from a different final voltage.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

For an ideal no-load RC charging circuit, doubling capacitance at fixed resistance changes the time constant how?

1. It doubles.
2. It halves.
3. It stays the same.
4. It becomes zero.

LESSON 3.1

Sinusoidal Signals, Frequency, and RMS

Read a waveform and separate peak, average, RMS, and frequency.

Learning objectives

• Relate sinusoidal frequency and period.
• Calculate RMS for a zero-offset sine wave.
• Use RMS to calculate average power in a resistor under stated assumptions.

A waveform needs more than one number

• Describe amplitude, time scale, and offset. A sinusoidal voltage can be written v(t) = Voffset + Vpeak sin(2 pi f t + phase). Vpeak is the sinusoidal amplitude around the offset, f is frequency in hertz, and phase specifies the position of the cycle relative to a reference. The same amplitude can appear at many frequencies, so a graph without a time axis cannot establish the rate of oscillation.
• Relate period to frequency. Period T is the time for one complete repeating cycle, and f = 1 / T. A period of 0.020 s corresponds to 50 Hz. This relation does not require a particular amplitude. When reading a sampled graph, distinguish the interval between data points from the full period; counting sample spacing as the cycle length can produce a large frequency error.
• Separate average from RMS. A zero-offset sine wave has zero average voltage over complete cycles, but it can transfer positive average power to a resistor. Root-mean-square voltage captures the mean of the squared waveform before taking the square root. For a pure zero-offset sine wave, Vrms = Vpeak / square root of 2. The formula does not apply unchanged to every waveform or to the total RMS of an offset signal.

Worked example: an ideal low-amplitude signal

• Compute the time and voltage measures. Consider v(t) = 4 sin(2 pi x 100 t) volts across an ideal 200-ohm resistor in a simulation. Frequency is 100 Hz and period is 0.010 s, or 10 ms. Peak voltage is 4 V and peak-to-peak voltage is 8 V. Vrms = 4 / 1.4142 = approximately 2.828 V. Average voltage over a complete number of cycles is zero.
• Compute average resistor power. Pavg = Vrms squared / R = 8 / 200 = 0.040 W. Peak current is 4 / 200 = 0.020 A and RMS current is approximately 14.14 mA. Instantaneous power is v(t) squared / R, so it stays nonnegative even during negative-voltage portions of the cycle. The resistor absorbs energy on both halves under the passive sign convention.
• State what is excluded. This calculation uses a pure sine wave, an ideal resistor, and a steady periodic condition. Capacitors and inductors can produce phase differences between current and voltage; multiplying voltage RMS by current RMS then gives apparent power unless the phase relationship is included. Do not extend the resistor formula to every AC network merely because the units look familiar.

Independent exercise and model reasoning

• Change amplitude and frequency independently. A zero-offset 6 V peak sine wave has a 5 ms period across 300 ohms. Model reasoning: f = 200 Hz, Vrms = approximately 4.243 V, and average power is 18 / 300 = 0.060 W. Changing only frequency leaves this ideal resistor power unchanged, though that conclusion would need reconsideration for frequency-dependent components.
• Avoid the common misconception. Zero average voltage does not imply zero heating-related average power. Preserve the waveform definition, observation interval, and component model when explaining your result.

Applied scenario

A waveform viewer reports several voltage measures, and a student must select the correct one for an ideal resistor-power calculation.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

What is the RMS voltage of a pure zero-offset sine wave with a 10 V peak?

1. 10 V
2. 20 V
3. Approximately 7.07 V
4. 0 V

LESSON 3.2

Diode Models and a Simulated Indicator

Use polarity and a piecewise model to explain one-way conduction without treating the model as a universal device law.

Learning objectives

• Determine a diode’s modeled operating state from polarity.
• Calculate current in a simple constant-drop indicator model.
• Explain the limits of ideal and constant-drop diode approximations.

Select the model before calculating

• A diode has an orientation. Its two terminals are the anode and cathode. Forward bias means the anode is at higher potential than the cathode. The ideal diode model permits forward conduction with zero voltage drop and blocks reverse current completely. Those are convenient approximations for selected questions, not exact statements about real semiconductor behavior in every condition.
• A constant-drop model adds a threshold approximation. In a simplified forward-conducting model, a specified voltage drop is assigned to the diode while a resistor determines current. A fictional indicator diode might be modeled with a 2.0 V forward drop for one exercise. This number is an assumption for that example, not a universal LED value. Device current, temperature, and construction affect actual behavior.
• Do not let a voltage source hide the current question. Connecting a diode model to an ideal source without a suitable current-setting element can make the model invalid or underdetermined. In this course, a positive series resistor creates a bounded simulated current. Real component selection also needs specifications, tolerances, and operating limits; the calculation below does not approve physical wiring or a component choice.

Worked example: a fictional five-volt indicator

• Write the loop equation. A 5.0 V ideal source drives a 300-ohm resistor and a forward-oriented indicator modeled with a constant 2.0 V drop. KVL gives 5.0 - I x 300 - 2.0 = 0. Therefore I = (5.0 - 2.0) / 300 = 0.010 A, or 10 mA. The resistor drops 3.0 V and absorbs I squared R = 0.030 W in this model.
• Check the energy allocation. The source delivers 5.0 x 0.010 = 0.050 W. The resistor absorbs 0.030 W and the diode model absorbs 2.0 x 0.010 = 0.020 W. Their sum is 0.050 W. This confirms internal arithmetic consistency but does not tell us optical efficiency or actual emitted brightness, which the constant-drop model does not contain.
• Test the assumed state. If the source is only 1.0 V, forcing the same forward-conducting equation would produce negative current. That contradicts the chosen forward-state approximation. Use the nonconducting state for this simplified model instead. A real diode’s small-current behavior is continuous rather than an exact hard threshold, so distinguish a model-state check from a complete semiconductor curve.

Independent exercise and model reasoning

• Try a changed resistor. Use the same 5.0 V source and 2.0 V drop with 600 ohms. Model reasoning: I = 3 / 600 = 5 mA; resistor power is 0.015 W and modeled diode power is 0.010 W. Current halves because the modeled voltage available across the resistor is unchanged. Document the assumptions before comparing the result with another design.
• Avoid a misleading safety conclusion. “Reverse biased” is not equivalent to unlimited reverse-voltage tolerance. Our ideal reverse-blocking state omits leakage and breakdown. The learning activity uses simulation only and makes no claim about a real device’s permitted operating region.

Applied scenario

A simulated status lamp is too strong in the model, and the learner must reduce predicted current without inventing a universal LED forward voltage.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

In the stated 5 V source, 2 V diode-drop, 300-ohm series model, what is the current?

1. Approximately 16.7 mA
2. 10 mA
3. Approximately 6.67 mA
4. 300 mA

LESSON 3.3

Transistors as Controlled Switches

Separate a control signal from the load path using a simplified transistor-switch model.

Learning objectives

• Identify control and load terminals in a simplified MOSFET model.
• Estimate on-state voltage drop and power with a stated resistance.
• Explain why a control threshold does not guarantee a fully enhanced device.

A control input changes another current path

• Separate signal from load power. A transistor can let a control signal influence a larger current path. In a simplified n-channel MOSFET low-side switch, the load connects between a positive source and the drain, the source terminal connects to the reference, and gate-to-source voltage controls the channel. These words name terminals; the MOSFET source terminal is not the same concept as the external power source.
• Choose a bounded switch approximation. Our model uses an open circuit when off and a specified on-resistance when on. This permits current and conduction-loss calculations without solving a full semiconductor model. The on-resistance is an assumed value for the fictional operating condition. Real devices have gate-voltage, temperature, transient, and operating-region dependencies that require separate specifications and verification.
• Do not confuse threshold with full conduction. A published threshold voltage identifies a specified small-current condition; it does not by itself establish low resistance at a desired load current. Even though a MOSFET gate can draw little steady current in an ideal model, changing gate voltage requires charging capacitance. Thus a switching system cannot be fully evaluated by treating its gate as a cost-free, instantaneous command.

Worked example: a resistive simulated load

• Include the switch in the loop. A hypothetical 5 V supply drives a 100-ohm load and an on-state MOSFET modeled as 0.5 ohm. Total series resistance is 100.5 ohms. Current is 5 / 100.5 = approximately 0.04975 A. The load voltage is approximately 4.975 V, and the switch drop is approximately 0.0249 V. These two drops add to 5 V within rounding.
• Compare conduction power. Switch power is I squared x 0.5 = approximately 0.00124 W. Load power is I squared x 100 = approximately 0.2475 W. The input power, about 0.2488 W, equals their sum. Replacing the switch with a perfect zero-resistance wire would slightly overestimate load voltage and erase the modeled switch loss entirely.
• Keep the experiment within its scope. The off-state approximation predicts zero current in this resistive network. It omits leakage, switching transitions, and parasitic paths. Inductive loads require additional analysis of stored energy and switching behavior and are outside this lab. Do not reinterpret this paper model as instructions for driving a real motor, relay, or other physical load.

Independent exercise and model reasoning

• Explore a poorer modeled switch. Change only on-resistance to 5 ohms. Model reasoning: current is 5 / 105 = approximately 47.62 mA; switch drop is approximately 0.238 V; switch conduction power is approximately 0.01134 W. The load receives less voltage while the switch dissipates more power. Explain why a small current change can accompany a much larger switch-loss change.
• Avoid a binary-performance assumption. “On” does not always mean negligible loss, and “off” does not describe every real transient. State the control condition, load type, model resistance, and omitted behavior whenever you compare switching designs.

Applied scenario

A simulated controller turns a resistive indicator load on and off; the team must explain what a nonideal on-state adds to the energy calculation.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

Why is MOSFET threshold voltage alone insufficient to establish low-loss switching at a specified load current?

1. It describes a specified small-current condition, not necessarily the required low on-resistance.
2. It means the gate has no capacitance.
3. It guarantees zero power loss.
4. It specifies the load resistance.

LESSON 4.1

Digital Logic and Truth-Table Verification

Translate a verbal requirement into Boolean logic and test every input combination.

Learning objectives

• Construct a complete two-input truth table.
• Distinguish AND, OR, NOT, and exclusive-OR behavior.
• Verify a small alarm requirement against all possible input cases.

Define the meaning of each bit

• A logic value is an interpreted state. Digital models use 0 and 1 to represent two classes of conditions. In a physical electronic system, voltage thresholds and timing rules determine how signals are interpreted; zero and one are not automatically exact zero volts and one volt. This lesson uses abstract Boolean values so the decision rule can be tested independently of a particular hardware implementation.
• Translate the requirement precisely. Suppose A means “an abnormal sensor condition is present” and E means “monitoring is enabled.” An alarm should be active only when both are true. The expression is Y = A AND E. If the requirement instead said either event should trigger an output, OR would fit. Exclusive OR, XOR, is true when the two inputs differ, which is a different requirement again.
• Enumerate instead of guessing. Two independent binary inputs have four combinations: 00, 01, 10, and 11. A complete truth table includes all four, even when some feel unlikely. A passing normal case does not prove disabled or abnormal cases are correct. Exhaustive testing is practical for this small input space and exposes errors that a single demonstration can hide.

Worked example: an enabled alarm

• Build the table from the words. For input order A then E, the required Y values are 00 gives 0, 01 gives 0, 10 gives 0, and 11 gives 1. The 10 case matters: an abnormal condition exists, but monitoring is disabled, so the specified output remains off. This is a fictional logic exercise, not a recommendation to disable a real protective system.
• Detect the wrong operator. If someone implements Y = A OR E, the output becomes 1 for 01 and 10 as well as 11. Those two mismatches are concrete evidence against the implementation. If someone implements XOR, the 11 case incorrectly becomes 0. Name the failed input row and expected output rather than reporting only that the design “does not work.”
• Add a separate reset condition. Let R mean “reset is requested,” and change the requirement to Y = A AND E AND NOT R. There are now eight input combinations. For example, A = 1, E = 1, R = 1 produces 0 because NOT R is 0. This combinational rule has no memory: releasing reset while A and E remain 1 immediately makes Y 1 in the ideal model. A latched alarm would need a state model and timing specification.

Independent exercise and model reasoning

• Test a disagreement indicator. A two-sensor diagnostic should activate when exactly one of its binary sensors reads 1. Model reasoning: Y = A XOR B, with output sequence 0, 1, 1, 0 for 00, 01, 10, 11. OR would fail the final row because it also activates when both are 1. Explain how this table follows from the word “exactly.”
• State the abstraction boundary. Truth-table verification establishes the steady Boolean rule, not electrical compatibility, noise margin, propagation delay, or safe operation of physical machinery. Preserve the requirement, bit definitions, and complete table together so another learner can verify the intended decision.

Applied scenario

A fictional monitoring dashboard has an enable setting; a learner must choose an operator that matches the written alarm rule.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

Which operator is true when exactly one of two binary inputs is true?

1. AND
2. OR
3. XOR
4. NOT applied only to the first input

LESSON 4.2

Voltage Dividers, Sensor Loading, and Thresholds

Model a resistive sensor input and explain why an attached measurement load can change its voltage.

Learning objectives

• Derive an unloaded two-resistor divider voltage.
• Calculate the effect of a finite parallel load.
• Evaluate a threshold decision using modeled values and uncertainty.

A divider is a network, not a guaranteed voltage source

• Derive the unloaded expression. Connect Rtop from a known source Vs to output node Vout, and Rbottom from Vout to the reference. With no additional output branch, the same current flows through both resistors: I = Vs / (Rtop + Rbottom). The output is the drop across Rbottom, so Vout = Vs Rbottom / (Rtop + Rbottom). Label which resistor is on the bottom; reversing them generally changes the answer.
• Use the output as a sensor model. A resistive sensing element can be represented by one divider resistance in an introductory model. Its changing resistance changes node voltage, which can feed a later decision. The divider equation alone does not establish the relationship between a real physical quantity and resistance. That relationship needs calibration, operating conditions, and a device model appropriate to the sensor.
• Include the attached load. A measurement input with finite resistance adds a branch from the output to the reference. It is then parallel with Rbottom. Replace those two resistances by their equivalent and solve the updated network. Assuming a meter or input is invisible can produce a substantial error when its resistance is not large compared with the divider’s output resistance.

Worked example: loading changes the decision

• Solve the no-load case. A hypothetical 5 V source feeds Rtop = 10 kilo-ohms and Rbottom = 10 kilo-ohms. With no load, Vout = 5 x 10 / (10 + 10) = 2.50 V. Divider current is 0.250 mA. A fictional decision block that activates for Vout at least 2.0 V would report active in this ideal unloaded model.
• Add a finite input resistance. Attach a 10 kilo-ohm load from Vout to the reference. The effective bottom resistance becomes 10 kilo-ohms parallel with 10 kilo-ohms = 5 kilo-ohms. Therefore Vout = 5 x 5 / (10 + 5) = approximately 1.667 V. The input resistor carries approximately 0.333 mA; bottom and load branches each carry approximately 0.167 mA. The same 2.0 V threshold now reports inactive.
• Distinguish a threshold from confidence. A numerical output of 2.001 V does not automatically imply a robust decision near a 2.0 V threshold. Source variation, resistance tolerance, input behavior, and measurement uncertainty can change the result. Our exercise uses exact ideal values first, then asks what margin is available. Real systems may need filtering, hysteresis, or different interfaces, which require additional design analysis.

Independent exercise and model reasoning

• Try a lighter load. Replace the 10 kilo-ohm output load with 90 kilo-ohms. Model reasoning: 10 parallel 90 gives 9 kilo-ohms, so Vout = 5 x 9 / 19 = approximately 2.368 V. The result is closer to 2.50 V but still below it. Loading has decreased because the added branch draws less current, not because its current is magically zero.
• Avoid a misleading label. A divider output should not be described as a fixed supply independent of load. Record the source, both divider resistances, attached input, and threshold before claiming the sensor decision has been verified.

Applied scenario

A simulated sensor reads above a threshold when disconnected from a decision input but below it after the input is included in the model.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

A 5 V divider has 10 kilo-ohms on top and 10 kilo-ohms on the bottom. Adding another 10 kilo-ohms across the bottom gives approximately what output?

1. 5.00 V
2. 3.33 V
3. 2.50 V
4. 1.67 V

LESSON 4.3

Capstone: Verify a Simulated Sensor Indicator

Combine requirements, circuit models, logic, power accounting, and evidence into one bounded design review.

Learning objectives

• Create a traceable requirement and test matrix for a simulated indicator.
• Combine a loaded divider, threshold rule, enable logic, and modeled output load.
• Document calculation evidence, model limits, and one design improvement.

Start with an observable requirement

• Define the system boundary. Design a browser or paper simulation that reports whether a fictional resistive sensor exceeds a voltage threshold while monitoring is enabled. The input model is a divider; the decision model is an ideal comparator; the enable rule is Boolean AND; the output is an abstract status indicator. Nothing in this project controls physical equipment or claims to implement a protective function.
• Write acceptance tests before choosing values. Use a 5 V ideal source, a 10 kilo-ohm top resistor, and two sensor states: 5 kilo-ohms for state L and 20 kilo-ohms for state H. Include a 100 kilo-ohm decision-input load from the output node to reference. The comparator reports A = 1 when Vout is at least 2.5 V. Final output Y = A AND E, where E is the monitoring-enable bit.
• Keep analog and digital claims separate. The divider calculates a continuous voltage. The comparator converts that voltage into a Boolean condition using the stated threshold. AND then combines the condition with enable. An indicator symbol displaying ON confirms only the simulated Boolean output. It does not verify physical output current, noise immunity, timing, or a real comparator’s electrical operating range.

Worked example: calculate both sensor states

• Solve the lower resistance state. With a 5 kilo-ohm sensor parallel to the 100 kilo-ohm input, the effective bottom resistance is (5 x 100) / (5 + 100) = approximately 4.762 kilo-ohms. Vout = 5 x 4.762 / (10 + 4.762) = approximately 1.613 V. This is below 2.5 V, so A = 0 and Y = 0 for either enable value.
• Solve the higher resistance state. With a 20 kilo-ohm sensor parallel to 100 kilo-ohms, the effective resistance is 16.667 kilo-ohms. Vout = 5 x 16.667 / (10 + 16.667) = 3.125 V. This exceeds the threshold, so A = 1. With E = 0 the final output remains 0; with E = 1 it becomes 1. Record all four sensor-state and enable combinations, including the disabled high-state case.
• Add a separate output-load calculation. If an abstract ON indicator is represented by the earlier 5 V source, 2 V constant diode drop, and 300-ohm series resistor model, its calculated current is 10 mA. This is a separate output-stage assumption. The comparator symbol is not assumed to directly supply that current, and no physical connection is implied. A real interface would require an appropriate driver and verified specifications.

Independent exercise and model reasoning

• Challenge the design with loading. Change the decision-input load from 100 kilo-ohms to 10 kilo-ohms while keeping the 20 kilo-ohm sensor. Model reasoning: bottom equivalent becomes 6.667 kilo-ohms and Vout = 2.0 V. The high sensor state now fails to cross 2.5 V. The correct response is to revise the interface model or requirements, not to hide the failed test.
• Finish with evidence and limits. Submit the schematic, equations, four-case truth table, changed-load failure, and one improvement proposal. Identify exact-value assumptions and behavior omitted by the ideal comparator, source, and diode. A strong engineering report shows where the model succeeds, where it fails, and what additional evidence a later design stage would need.

Applied scenario

A reviewer receives a simulated sensor-indicator proposal and asks for a complete calculation trail, disabled-state tests, and a loading failure case.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

In the capstone, the 20 kilo-ohm sensor with a 100 kilo-ohm input load gives 3.125 V. What is Y when monitoring enable E is zero?

1. 1, because the sensor is high
2. 0, because Y = A AND E
3. 3.125, because Y is the analog voltage
4. 2.5, because the threshold sets the output

LESSON 5.1

Calculate a divider with its load connected — worked model

A simulated ideal 5 V source feeds a 2 kΩ upper resistor and a 3 kΩ lower resistor in series. The output is measured across the lower resistor. Start with an ideal meter that draws no current.

Learning objectives

• Calculate a divider with its load connected
• 2.5 V
• Defend the conclusion with scoped and reproducible evidence.

The practice brief

• A simulated ideal 5 V source feeds a 2 kΩ upper resistor and a 3 kΩ lower resistor in series. The output is measured across the lower resistor. Start with an ideal meter that draws no current.

Understand the mechanism

• The unloaded series resistance is 5 kΩ, current is 5 V / 5 kΩ = 1 mA, and output is 1 mA × 3 kΩ = 3 V. This result depends on the unloaded assumption. A device attached across the lower resistor creates another current path and changes the circuit.

• Keep the source, upper branch, lower branch, and output node on a sketch before applying a divider formula. A voltage label does not identify an unlimited source of current. Check the loaded network using equivalent resistance and then verify node currents.

• The units help: volts divided by kilohms yield milliamps. Summing the two series voltage drops should recover the ideal source voltage. A simulator reports the model it was given; an omitted input resistance in the model remains an omitted real-world effect even if the numerical solution converges.

• In a practical design, source resistance, input behavior, tolerances, power ratings, noise, and allowable signal error may matter. This worksheet does not require wiring a physical circuit. Its goal is to expose why a reference value taken from an unloaded schematic may fail once an actual interface is connected, and to make the recalculation and its assumptions reproducible.

Build the initial evidence table

• Write one row for each supplied input or observation. Keep identifiers, units, timestamps, roles, counts, and states distinct where the case uses them. Label information supplied by the fixture separately from a conclusion you calculate. Do not add an unstated business rule to make the answer easier.

Worked reasoning

2.5 V

Check the scope

• Explain what this evidence supports and what it leaves unresolved. A correct calculation, a parsed object, a matching identifier, and an authorized action are different kinds of claims. Use only the claims established by this specific case.

Offline practice

• [Download the BEAT IT engineering and quantum arithmetic checks](/resources/academy/engineering-quantum-checks.py). Run with Python 3 in a practice folder. The program uses only the standard library, makes no network calls, and checks synthetic arithmetic and ideal-model results.

• Passing these checks does not establish real equipment safety, hardware execution, or deployment readiness. Retain your predictions and the observed program output separately.

Further study

• [OpenStax: series and parallel resistors](https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel)

Applied scenario

A simulated ideal 5 V source feeds a 2 kΩ upper resistor and a 3 kΩ lower resistor in series. The output is measured across the lower resistor. Start with an ideal meter that draws no current.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

What is the loaded output voltage?

1. 2.5 V
2. 3 V regardless of load
3. 5 V
4. 1.25 V

LESSON 5.2

Calculate a divider with its load connected — controlled variation

A simulated ideal 5 V source feeds a 2 kΩ upper resistor and a 3 kΩ lower resistor in series. The output is measured across the lower resistor. Start with an ideal meter that draws no current.

Learning objectives

• Calculate a divider with its load connected
• Connect a 6 kΩ load across the 3 kΩ lower resistor. The parallel equivalent is (3×6)/(3+6)=2 kΩ. The total becomes 4 kΩ, source current 1.25 mA, and output 2.5 V. Lower-resistor current is about 0.8333 mA and load current about 0.4167 mA; they sum to 1.25 mA.
• Defend the conclusion with scoped and reproducible evidence.

Keep the original baseline

• A simulated ideal 5 V source feeds a 2 kΩ upper resistor and a 3 kΩ lower resistor in series. The output is measured across the lower resistor. Start with an ideal meter that draws no current.

The controlled change

• Connect a 6 kΩ load across the 3 kΩ lower resistor. The parallel equivalent is (3×6)/(3+6)=2 kΩ. The total becomes 4 kΩ, source current 1.25 mA, and output 2.5 V. Lower-resistor current is about 0.8333 mA and load current about 0.4167 mA; they sum to 1.25 mA.

Compare both paths

• Retain the original result and change only the stated condition. Predict the new result first, then trace the affected relationship, policy decision, state transition, or denominator. Mark the earliest step where the old and new reasoning differ. A paper trace is a legitimate exercise, but it is not an executed test.

Reference reasoning

• The unloaded series resistance is 5 kΩ, current is 5 V / 5 kΩ = 1 mA, and output is 1 mA × 3 kΩ = 3 V. This result depends on the unloaded assumption. A device attached across the lower resistor creates another current path and changes the circuit.

• Keep the source, upper branch, lower branch, and output node on a sketch before applying a divider formula. A voltage label does not identify an unlimited source of current. Check the loaded network using equivalent resistance and then verify node currents.

• The units help: volts divided by kilohms yield milliamps. Summing the two series voltage drops should recover the ideal source voltage. A simulator reports the model it was given; an omitted input resistance in the model remains an omitted real-world effect even if the numerical solution converges.

• In a practical design, source resistance, input behavior, tolerances, power ratings, noise, and allowable signal error may matter. This worksheet does not require wiring a physical circuit. Its goal is to expose why a reference value taken from an unloaded schematic may fail once an actual interface is connected, and to make the recalculation and its assumptions reproducible.

Review the changed conclusion

• Connect a 6 kΩ load across the 3 kΩ lower resistor. The parallel equivalent is (3×6)/(3+6)=2 kΩ. The total becomes 4 kΩ, source current 1.25 mA, and output 2.5 V. Lower-resistor current is about 0.8333 mA and load current about 0.4167 mA; they sum to 1.25 mA.

Investigate disagreement

• If your result differs, inspect the first diverging step. Check whether you changed the population, units, time window, authority, or outcome definition. Record the correction rather than hiding the failed attempt.

Offline practice

• [Download the BEAT IT engineering and quantum arithmetic checks](/resources/academy/engineering-quantum-checks.py). Run with Python 3 in a practice folder. The program uses only the standard library, makes no network calls, and checks synthetic arithmetic and ideal-model results.

• Passing these checks does not establish real equipment safety, hardware execution, or deployment readiness. Retain your predictions and the observed program output separately.

Further study

• [OpenStax: series and parallel resistors](https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel)

Applied scenario

A simulated ideal 5 V source feeds a 2 kΩ upper resistor and a 3 kΩ lower resistor in series. The output is measured across the lower resistor. Start with an ideal meter that draws no current.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

Which conclusion follows from the changed practice case?

1. The changed condition cannot affect the result.
2. The original result proves every future outcome.
3. The evidence gap can be filled with an assumption.
4. Connect a 6 kΩ load across the 3 kΩ lower resistor. The parallel equivalent is (3×6)/(3+6)=2 kΩ. The total becomes 4 kΩ, source current 1.25 mA, and output 2.5 V. Lower-resistor current is about 0.8333 mA and load current about 0.4167 mA; they sum to 1.25 mA.

LESSON 5.3

Calculate a divider with its load connected — evidence and review

A simulated ideal 5 V source feeds a 2 kΩ upper resistor and a 3 kΩ lower resistor in series. The output is measured across the lower resistor. Start with an ideal meter that draws no current.

Learning objectives

• Calculate a divider with its load connected
• Draw both networks, calculate equivalent resistance and output, and verify the loaded node-current balance. Compare the output change with a fictional 2.9–3.1 V requirement.
• Defend the conclusion with scoped and reproducible evidence.

Your deliverable

Draw both networks, calculate equivalent resistance and output, and verify the loaded node-current balance. Compare the output change with a fictional 2.9–3.1 V requirement.

Inputs and constraints

• A simulated ideal 5 V source feeds a 2 kΩ upper resistor and a 3 kΩ lower resistor in series. The output is measured across the lower resistor. Start with an ideal meter that draws no current.

Reference mechanism

• The unloaded series resistance is 5 kΩ, current is 5 V / 5 kΩ = 1 mA, and output is 1 mA × 3 kΩ = 3 V. This result depends on the unloaded assumption. A device attached across the lower resistor creates another current path and changes the circuit.

• Keep the source, upper branch, lower branch, and output node on a sketch before applying a divider formula. A voltage label does not identify an unlimited source of current. Check the loaded network using equivalent resistance and then verify node currents.

• The units help: volts divided by kilohms yield milliamps. Summing the two series voltage drops should recover the ideal source voltage. A simulator reports the model it was given; an omitted input resistance in the model remains an omitted real-world effect even if the numerical solution converges.

• In a practical design, source resistance, input behavior, tolerances, power ratings, noise, and allowable signal error may matter. This worksheet does not require wiring a physical circuit. Its goal is to expose why a reference value taken from an unloaded schematic may fail once an actual interface is connected, and to make the recalculation and its assumptions reproducible.

Acceptance review

2.5 V

Required variation

• Connect a 6 kΩ load across the 3 kΩ lower resistor. The parallel equivalent is (3×6)/(3+6)=2 kΩ. The total becomes 4 kΩ, source current 1.25 mA, and output 2.5 V. Lower-resistor current is about 0.8333 mA and load current about 0.4167 mA; they sum to 1.25 mA.

Review with another engineer

Explain the problem, the selected method, your actual evidence, a plausible alternative, and the remaining limit. A reviewer should be able to reconstruct the case from the retained inputs and steps without relying on a polished screenshot.

Portfolio evidence

• Keep a concise README with the fictional fixture, assumptions, original result, changed-input trace, any executed test command and output, and one limitation. Clearly label designs and predictions that were not executed. These examples do not authorize production changes or prove production readiness.

Offline practice

• [Download the BEAT IT engineering and quantum arithmetic checks](/resources/academy/engineering-quantum-checks.py). Run with Python 3 in a practice folder. The program uses only the standard library, makes no network calls, and checks synthetic arithmetic and ideal-model results.

• Passing these checks does not establish real equipment safety, hardware execution, or deployment readiness. Retain your predictions and the observed program output separately.

Further study

• [OpenStax: series and parallel resistors](https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel)

Applied scenario

A simulated ideal 5 V source feeds a 2 kΩ upper resistor and a 3 kΩ lower resistor in series. The output is measured across the lower resistor. Start with an ideal meter that draws no current.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

Which deliverable addresses this case?

1. A universal success claim without tests.
2. A copy of a result without a method.
3. Draw both networks, calculate equivalent resistance and output, and verify the loaded node-current balance. Compare the output change with a fictional 2.9–3.1 V requirement.
4. A screenshot without its input or scope.

LESSON 6.1

Check power against stated tolerances — worked model

A simulated resistor is nominally 660 Ω across an ideal 12 V source. A fictional specification lists a 0.25 W rating. The source can be up to 10% high and the resistance can be 5% low. Treat these as supplied limits, not a complete real-component qualification.

Learning objectives

• Check power against stated tolerances
• About 0.2779 W exceeds the 0.25 W limit
• Defend the conclusion with scoped and reproducible evidence.

The practice brief

• A simulated resistor is nominally 660 Ω across an ideal 12 V source. A fictional specification lists a 0.25 W rating. The source can be up to 10% high and the resistance can be 5% low. Treat these as supplied limits, not a complete real-component qualification.

Understand the mechanism

• At nominal conditions, current is 12/660 ≈ 0.0181818 A and resistor power is V²/R = 144/660 ≈ 0.218182 W. That is below 0.25 W, but checking only the nominal point omits the stated tolerances. For fixed positive resistance, power rises with the square of voltage; for fixed voltage, it rises as resistance falls.

• Therefore the maximum under the supplied independent bounds is at the highest voltage and lowest resistance. A current calculation and a power calculation answer different questions. Compare both P=V²/R and P=VI using the same chosen operating point as an arithmetic cross-check.

• Do not mix a nominal current with a worst-case voltage. The rating comparison also has a scope limit. A real resistor’s allowable dissipation can depend on ambient temperature, mounting, duty cycle, derating rules, and manufacturer data not supplied here.

• This exercise can reject the proposed choice under its own stated limit without declaring a different physical design safe. Record assumptions, worst-case inputs, calculated power, and the pass/fail rule before recommending a next design review. No powered hardware test or mains connection is part of the worksheet.

Build the initial evidence table

• Write one row for each supplied input or observation. Keep identifiers, units, timestamps, roles, counts, and states distinct where the case uses them. Label information supplied by the fixture separately from a conclusion you calculate. Do not add an unstated business rule to make the answer easier.

Worked reasoning

About 0.2779 W exceeds the 0.25 W limit

Check the scope

• Explain what this evidence supports and what it leaves unresolved. A correct calculation, a parsed object, a matching identifier, and an authorized action are different kinds of claims. Use only the claims established by this specific case.

Offline practice

• [Download the BEAT IT engineering and quantum arithmetic checks](/resources/academy/engineering-quantum-checks.py). Run with Python 3 in a practice folder. The program uses only the standard library, makes no network calls, and checks synthetic arithmetic and ideal-model results.

• Passing these checks does not establish real equipment safety, hardware execution, or deployment readiness. Retain your predictions and the observed program output separately.

Further study

• [OpenStax: series and parallel resistors](https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel)

Applied scenario

A simulated resistor is nominally 660 Ω across an ideal 12 V source. A fictional specification lists a 0.25 W rating. The source can be up to 10% high and the resistance can be 5% low. Treat these as supplied limits, not a complete real-component qualification.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

What does the supplied worst-case check show?

1. The nominal pass proves every operating point passes
2. The current is exactly 0.25 A
3. Lower resistance always lowers power at fixed voltage
4. About 0.2779 W exceeds the 0.25 W limit

LESSON 6.2

Check power against stated tolerances — controlled variation

A simulated resistor is nominally 660 Ω across an ideal 12 V source. A fictional specification lists a 0.25 W rating. The source can be up to 10% high and the resistance can be 5% low. Treat these as supplied limits, not a complete real-component qualification.

Learning objectives

• Check power against stated tolerances
• At 13.2 V and 627 Ω, power is 13.2²/627 ≈ 0.277895 W, exceeding the stated 0.25 W limit. The nominal calculation passes but this supplied worst-case check fails.
• Defend the conclusion with scoped and reproducible evidence.

Keep the original baseline

• A simulated resistor is nominally 660 Ω across an ideal 12 V source. A fictional specification lists a 0.25 W rating. The source can be up to 10% high and the resistance can be 5% low. Treat these as supplied limits, not a complete real-component qualification.

The controlled change

At 13.2 V and 627 Ω, power is 13.2²/627 ≈ 0.277895 W, exceeding the stated 0.25 W limit. The nominal calculation passes but this supplied worst-case check fails.

Compare both paths

• Retain the original result and change only the stated condition. Predict the new result first, then trace the affected relationship, policy decision, state transition, or denominator. Mark the earliest step where the old and new reasoning differ. A paper trace is a legitimate exercise, but it is not an executed test.

Reference reasoning

• At nominal conditions, current is 12/660 ≈ 0.0181818 A and resistor power is V²/R = 144/660 ≈ 0.218182 W. That is below 0.25 W, but checking only the nominal point omits the stated tolerances. For fixed positive resistance, power rises with the square of voltage; for fixed voltage, it rises as resistance falls.

• Therefore the maximum under the supplied independent bounds is at the highest voltage and lowest resistance. A current calculation and a power calculation answer different questions. Compare both P=V²/R and P=VI using the same chosen operating point as an arithmetic cross-check.

• Do not mix a nominal current with a worst-case voltage. The rating comparison also has a scope limit. A real resistor’s allowable dissipation can depend on ambient temperature, mounting, duty cycle, derating rules, and manufacturer data not supplied here.

• This exercise can reject the proposed choice under its own stated limit without declaring a different physical design safe. Record assumptions, worst-case inputs, calculated power, and the pass/fail rule before recommending a next design review. No powered hardware test or mains connection is part of the worksheet.

Review the changed conclusion

At 13.2 V and 627 Ω, power is 13.2²/627 ≈ 0.277895 W, exceeding the stated 0.25 W limit. The nominal calculation passes but this supplied worst-case check fails.

Investigate disagreement

• If your result differs, inspect the first diverging step. Check whether you changed the population, units, time window, authority, or outcome definition. Record the correction rather than hiding the failed attempt.

Offline practice

• [Download the BEAT IT engineering and quantum arithmetic checks](/resources/academy/engineering-quantum-checks.py). Run with Python 3 in a practice folder. The program uses only the standard library, makes no network calls, and checks synthetic arithmetic and ideal-model results.

• Passing these checks does not establish real equipment safety, hardware execution, or deployment readiness. Retain your predictions and the observed program output separately.

Further study

• [OpenStax: series and parallel resistors](https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel)

Applied scenario

A simulated resistor is nominally 660 Ω across an ideal 12 V source. A fictional specification lists a 0.25 W rating. The source can be up to 10% high and the resistance can be 5% low. Treat these as supplied limits, not a complete real-component qualification.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

Which conclusion follows from the changed practice case?

1. The original result proves every future outcome.
2. The evidence gap can be filled with an assumption.
3. At 13.2 V and 627 Ω, power is 13.2²/627 ≈ 0.277895 W, exceeding the stated 0.25 W limit. The nominal calculation passes but this supplied worst-case check fails.
4. The changed condition cannot affect the result.

LESSON 6.3

Check power against stated tolerances — evidence and review

A simulated resistor is nominally 660 Ω across an ideal 12 V source. A fictional specification lists a 0.25 W rating. The source can be up to 10% high and the resistance can be 5% low. Treat these as supplied limits, not a complete real-component qualification.

Learning objectives

• Check power against stated tolerances
• Calculate nominal and worst-case current and power. State the violated requirement and the additional component/environment data needed before selecting a replacement.
• Defend the conclusion with scoped and reproducible evidence.

Your deliverable

Calculate nominal and worst-case current and power. State the violated requirement and the additional component/environment data needed before selecting a replacement.

Inputs and constraints

• A simulated resistor is nominally 660 Ω across an ideal 12 V source. A fictional specification lists a 0.25 W rating. The source can be up to 10% high and the resistance can be 5% low. Treat these as supplied limits, not a complete real-component qualification.

Reference mechanism

• At nominal conditions, current is 12/660 ≈ 0.0181818 A and resistor power is V²/R = 144/660 ≈ 0.218182 W. That is below 0.25 W, but checking only the nominal point omits the stated tolerances. For fixed positive resistance, power rises with the square of voltage; for fixed voltage, it rises as resistance falls.

• Therefore the maximum under the supplied independent bounds is at the highest voltage and lowest resistance. A current calculation and a power calculation answer different questions. Compare both P=V²/R and P=VI using the same chosen operating point as an arithmetic cross-check.

• Do not mix a nominal current with a worst-case voltage. The rating comparison also has a scope limit. A real resistor’s allowable dissipation can depend on ambient temperature, mounting, duty cycle, derating rules, and manufacturer data not supplied here.

• This exercise can reject the proposed choice under its own stated limit without declaring a different physical design safe. Record assumptions, worst-case inputs, calculated power, and the pass/fail rule before recommending a next design review. No powered hardware test or mains connection is part of the worksheet.

Acceptance review

About 0.2779 W exceeds the 0.25 W limit

Required variation

At 13.2 V and 627 Ω, power is 13.2²/627 ≈ 0.277895 W, exceeding the stated 0.25 W limit. The nominal calculation passes but this supplied worst-case check fails.

Review with another engineer

Explain the problem, the selected method, your actual evidence, a plausible alternative, and the remaining limit. A reviewer should be able to reconstruct the case from the retained inputs and steps without relying on a polished screenshot.

Portfolio evidence

• Keep a concise README with the fictional fixture, assumptions, original result, changed-input trace, any executed test command and output, and one limitation. Clearly label designs and predictions that were not executed. These examples do not authorize production changes or prove production readiness.

Offline practice

• [Download the BEAT IT engineering and quantum arithmetic checks](/resources/academy/engineering-quantum-checks.py). Run with Python 3 in a practice folder. The program uses only the standard library, makes no network calls, and checks synthetic arithmetic and ideal-model results.

• Passing these checks does not establish real equipment safety, hardware execution, or deployment readiness. Retain your predictions and the observed program output separately.

Further study

• [OpenStax: series and parallel resistors](https://openstax.org/books/university-physics-volume-2/pages/10-2-resistors-in-series-and-parallel)

Applied scenario

A simulated resistor is nominally 660 Ω across an ideal 12 V source. A fictional specification lists a 0.25 W rating. The source can be up to 10% high and the resistance can be 5% low. Treat these as supplied limits, not a complete real-component qualification.

Your evidence and reasoning

Inputs and assumptions → diagram and equation → observed result → units and limits → one improvement.

Practice check

Which deliverable addresses this case?

1. A copy of a result without a method.
2. Calculate nominal and worst-case current and power. State the violated requirement and the additional component/environment data needed before selecting a replacement.
3. A screenshot without its input or scope.
4. A universal success claim without tests.