Exercise set

Circuit Analysis and Electrical Protection Exercises

Solved circuit analysis and protection exercises for Ohm law, Kirchhoff laws, Thevenin equivalents, transients, insulation, leakage and fault screening.

Branch: Electrical Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.1 · reviewed

These exercises practise circuit analysis and electrical protection as one engineering workflow. The goal is to calculate voltages, currents, power, equivalent sources, stored energy, transient response, insulation quality, leakage and first-pass protection margins before a circuit is accepted for operation.

Assume simplified lumped-circuit models unless an exercise states otherwise. Real protection and commissioning work should also check wiring, grounding, source impedance, temperature, tolerances, measurement uncertainty, device curves, short-circuit duty, arc-flash exposure, enclosure ratings, inspection records and local electrical codes.

How to Use These Exercises

For each problem, state:

whether the values are DC, RMS AC, peak, transient or worst case;
the source, load, cable, instrument and protection boundary;
the circuit law or equivalent model being applied;
whether the result affects normal operation, measurement validity, thermal stress, insulation, fault detection or interrupting duty;
the field evidence needed before release.

The common mistake is separating normal circuit analysis from protection. A circuit that carries normal load can still be unsafe if a minimum fault is not detected, a breaker cannot interrupt maximum fault current, a meter distorts the measurement, or insulation leakage is already close to a trip threshold.

Release Evidence Notes

Circuit release evidence should connect nominal operation, abnormal operation and measurement method. The record should state source voltage, source impedance, load state, cable route, grounding path, protective device, measurement instrument, tolerance basis and whether values are DC, RMS AC, peak, transient or worst case.

Protection evidence should not stop at a normal-load calculation. Useful evidence includes minimum and maximum fault current, pickup setting, interrupting rating, residual-current behavior, insulation resistance, leakage current, voltage drop, thermal energy, device curve or datasheet, inspection state and the field test that proves the installed configuration.

Engineering Boundary Notes

These exercises are first-pass lumped-circuit and protection screens. They do not replace code-compliant design, arc-flash study, selectivity study, short-circuit model, insulation test procedure, power-quality review or commissioned field test. A calculation that passes on nominal values can fail after tolerance, temperature, source impedance or installation effects are included.

Common Release Mistakes

mixing RMS, peak, DC and transient values without stating the basis;
using one-line resistance while ignoring return path, grounding and source impedance;
approving normal load current while minimum fault pickup is not proven;
checking breaker interrupting rating without maximum available fault current;
ignoring instrument loading, insulation trend and leakage accumulation;
treating a first-pass protection window as full coordination evidence.

Scenario Map

Scenario	Exercises	Primary check	Engineering decision
Basic circuit quantities	1, 2, 3, 4, 5	Ohm law, dividers, KCL and Thevenin equivalents	Confirm load current, voltage and source behavior before sizing.
Measurement and dynamic behavior	6, 7, 8, 9, 10	Bridge balance, meter loading, RC/RL time constants and impedance	Decide whether the measured or transient value is trustworthy.
Thermal, insulation and leakage checks	11, 12, 13, 14	Voltage drop, Joule energy, insulation resistance and residual leakage	Identify overheating, excessive drop or insulation weakness.
Protection screening	15, 16, 17, 18	pickup window, interrupting margin, ground-fault detection and release gates	Decide whether the installation can be energized or must be held.

Validation Package Checklist

voltage, current, power and impedance bases are explicit;
source, load, cable, return path and protective device are identified;
measurement instrument and loading effect are reviewed;
transient, thermal, insulation and leakage limits are stated;
minimum fault pickup and maximum interrupting duty are both checked;
installation inspection, device data and field test evidence match the calculation;
final release decision states energize, restrict, retest or redesign.

Exercise 1: Load Current and Resistive Power

A $24\ \text{V DC}$ supply feeds a $12\ \Omega$ heater. Calculate the load current and heater power.

Solution

Using Ohm law:

\displaystyle I=\frac{V}{R}

\displaystyle I=\frac{24}{12}=2\ \text{A}

Power:

P=VI

P=24 \times 2=48\ \text{W}

The heater draws $2\ \text{A}$ and dissipates $48\ \text{W}$ .

Engineering Comment

The current checks supply and conductor loading. The power checks thermal behavior. A resistor calculation is not complete unless the component wattage, enclosure temperature and duty cycle are also acceptable.

Plausibility Check

A $12\ \Omega$ load on $24\ \text{V}$ is a low-voltage, moderate-current load. A few amperes and tens of watts are plausible.

Exercise 2: Series Resistance and Voltage Division

A $48\ \text{V}$ source feeds three series resistors: $R_1=6\ \Omega$ , $R_2=10\ \Omega$ and $R_3=8\ \Omega$ . Find the current and the voltage across $R_2$ .

Solution

Total resistance:

R_{tot}=6+10+8=24\ \Omega

Current:

\displaystyle I=\frac{48}{24}=2\ \text{A}

Voltage across $R_2$ :

V_2=IR_2=2 \times 10=20\ \text{V}

Engineering Comment

In a series circuit, all components carry the same current. The largest resistance receives the largest voltage drop. That matters for insulation rating and component power dissipation.

Plausibility Check

$R_2$ is $10/24$ of the total resistance, so it should receive $10/24$ of $48\ \text{V}$ , which is $20\ \text{V}$ .

Exercise 3: Loaded Voltage Divider

A voltage divider uses $R_1=10\ \text{k}\Omega$ from a $12\ \text{V}$ source to the output node and $R_2=5\ \text{k}\Omega$ from the output node to ground. A load of $R_L=5\ \text{k}\Omega$ is connected from output to ground. Find the loaded output voltage.

Solution

The lower leg is:

R_{eq}=R_2 \parallel R_L

\displaystyle R_{eq}=\frac{5\text{k}\Omega \times 5\text{k}\Omega}{5\text{k}\Omega+5\text{k}\Omega}=2.5\text{k}\Omega

Loaded output voltage:

\displaystyle V_o=12\frac{R_{eq}}{R_1+R_{eq}}

\displaystyle V_o=12\frac{2.5}{10+2.5}=2.4\ \text{V}

Engineering Comment

The unloaded divider would produce $4\ \text{V}$ . The load pulls the output down to $2.4\ \text{V}$ . Divider outputs should not be used as stiff supplies unless loading is included.

Plausibility Check

Adding a parallel load lowers the lower-leg resistance, so the output voltage must be less than the unloaded value. $2.4\ \text{V}$ is therefore credible.

Exercise 4: KCL at a Node

A node receives $6\ \text{A}$ from a source. Two branches leave the node: branch A carries $2.2\ \text{A}$ and branch B carries $1.5\ \text{A}$ . Find the current in branch C.

Solution

By Kirchhoff current law:

I_{in}=I_A+I_B+I_C

So:

6=2.2+1.5+I_C

I_C=6-3.7=2.3\ \text{A}

Engineering Comment

KCL is also a diagnostic tool. If measured branch currents do not close within instrument uncertainty, there may be an unmeasured leakage path, wrong CT polarity, meter range issue or wiring error.

Plausibility Check

The two known outgoing currents total less than the incoming current, so the remaining branch current should be positive.

Exercise 5: Thevenin Equivalent and Load Current

A source network is represented by a Thevenin voltage $V_{th}=18\ \text{V}$ and Thevenin resistance $R_{th}=3\ \Omega$ . A load $R_L=6\ \Omega$ is connected. Find the load current and load voltage.

Solution

Load current:

\displaystyle I_L=\frac{V_{th}}{R_{th}+R_L}

\displaystyle I_L=\frac{18}{3+6}=2\ \text{A}

Load voltage:

V_L=I_LR_L=2 \times 6=12\ \text{V}

Engineering Comment

The source resistance causes a $6\ \text{V}$ internal drop. In protection work, Thevenin impedance also limits fault current, which can help interrupting duty but hurt minimum-fault detection.

Plausibility Check

The load resistance is twice the source resistance, so the load should receive two thirds of the source voltage. Two thirds of $18\ \text{V}$ is $12\ \text{V}$ .

Exercise 6: Wheatstone Bridge Imbalance

A Wheatstone bridge has $R_1=100\ \Omega$ , $R_2=200\ \Omega$ , $R_3=150\ \Omega$ and $R_4=300\ \Omega$ . Check whether it is balanced using:

\displaystyle \frac{R_1}{R_2}=\frac{R_3}{R_4}

Solution

Left ratio:

\displaystyle \frac{R_1}{R_2}=\frac{100}{200}=0.5

Right ratio:

\displaystyle \frac{R_3}{R_4}=\frac{150}{300}=0.5

Since the ratios are equal, the bridge is balanced.

Engineering Comment

A balanced bridge has zero ideal detector voltage. In field work, a nonzero reading can come from sensor change, lead resistance, temperature drift, excitation instability or detector loading.

Plausibility Check

Both lower resistors are exactly twice their upper partners, so balance is expected.

Exercise 7: Multimeter Loading Error

A circuit node is fed through $R_s=900\ \text{k}\Omega$ from a $10\ \text{V}$ source. A multimeter with input resistance $R_m=10\ \text{M}\Omega$ measures the node to ground. Find the measured voltage.

Solution

The meter forms a divider:

\displaystyle V_m=10\frac{R_m}{R_s+R_m}

\displaystyle V_m=10\frac{10{,}000{,}000}{900{,}000+10{,}000{,}000}

V_m=9.17\ \text{V}

Engineering Comment

The true open-circuit node would be close to $10\ \text{V}$ , but the meter loads the circuit. High-impedance measurements need instrument loading checks, especially in sensor, insulation, leakage and analog input circuits.

Plausibility Check

The meter resistance is much larger than the source resistance, so the error should be noticeable but not catastrophic. A reading near $9.2\ \text{V}$ is plausible.

Exercise 8: RC Charging Time to a Threshold

A capacitor charges through a resistor toward $24\ \text{V}$ . The circuit has $R=20\ \text{k}\Omega$ and $C=100\ \mu\text{F}$ . Estimate the time for the capacitor voltage to reach $90\%$ of final value using:

V_C(t)=V_s(1-e^{-t/RC})

Solution

At $90\%$ :

0.90=1-e^{-t/RC}

e^{-t/RC}=0.10

t=-RC\ln(0.10)

Time constant:

RC=20{,}000 \times 100\times 10^{-6}=2.0\ \text{s}

Therefore:

t=2.0 \times 2.303=4.61\ \text{s}

Engineering Comment

This delay can affect undervoltage relays, reset circuits, soft-start timing and discharge safety. A device may appear off while stored energy remains significant.

Plausibility Check

A capacitor reaches about $95\%$ after three time constants. Reaching $90\%$ after $2.3$ time constants is consistent.

Exercise 9: RL Current Rise

A relay coil has $R=40\ \Omega$ and $L=0.8\ \text{H}$ . It is energized from a $24\ \text{V}$ DC source. Find the final current and the time constant.

Solution

Final current:

\displaystyle I_{\infty}=\frac{V}{R}

\displaystyle I_{\infty}=\frac{24}{40}=0.60\ \text{A}

Time constant:

\displaystyle \tau=\frac{L}{R}

\displaystyle \tau=\frac{0.8}{40}=0.020\ \text{s}=20\ \text{ms}

Engineering Comment

The coil does not reach final current instantly. Timing matters for relays, contactors, solenoids and trip coils. Suppression components can also slow current decay after de-energization.

Plausibility Check

A $24\ \text{V}$ coil with $40\ \Omega$ resistance should draw less than $1\ \text{A}$ . A $20\ \text{ms}$ electrical time constant is plausible for a relay-scale inductive load.

Exercise 10: Series RL Impedance

A $30\ \Omega$ resistor is in series with an inductor of $L=0.12\ \text{H}$ at $50\ \text{Hz}$ . Find the inductive reactance and impedance magnitude.

Solution

Inductive reactance:

X_L=2\pi fL

X_L=2\pi(50)(0.12)=37.7\ \Omega

Impedance magnitude:

|Z|=\sqrt{R^2+X_L^2}

|Z|=\sqrt{30^2+37.7^2}=48.2\ \Omega

Engineering Comment

This is a basic impedance calculation, not a full AC power study. It still matters for inrush, relay coils, solenoids, filters and source loading because the current is limited by both resistance and reactance.

Plausibility Check

Because the reactance is slightly larger than the resistance, the impedance magnitude should be larger than either alone and near $50\ \Omega$ .

Exercise 11: Cable Voltage Drop

A $24\ \text{V DC}$ control circuit supplies $3.5\ \text{A}$ through a two-conductor cable. The round-trip cable resistance is $0.42\ \Omega$ . Find the voltage drop and load voltage.

Solution

Voltage drop:

\Delta V=IR

\Delta V=3.5 \times 0.42=1.47\ \text{V}

Load voltage:

V_L=24-1.47=22.53\ \text{V}

Percent drop:

\displaystyle \frac{1.47}{24}\times 100=6.1\%

Engineering Comment

Voltage drop can cause contactor chatter, sensor error, PLC input misread or undervoltage trips. The calculation should use round-trip resistance for DC two-wire circuits.

Plausibility Check

A few amperes through less than half an ohm gives a drop around one to two volts, which matches the result.

Exercise 12: Joule Heating During an Overload

A cable segment has resistance $0.08\ \Omega$ . During a short overload it carries $80\ \text{A}$ for $4\ \text{s}$ . Estimate the resistive energy deposited in the cable.

Solution

Power:

P=I^2R

P=80^2 \times 0.08=512\ \text{W}

Energy:

E=Pt

E=512 \times 4=2048\ \text{J}

Engineering Comment

This is an energy screen, not a cable damage proof. Real acceptability depends on conductor size, insulation class, ambient temperature, repeated events, heat dissipation and protective-device clearing curves.

Plausibility Check

The current is large, but the event lasts only a few seconds. Energy in the low kilojoule range is plausible.

Exercise 13: Insulation Resistance from Leakage Current

An insulation tester applies $500\ \text{V DC}$ and measures $50\ \mu\text{A}$ leakage. Estimate insulation resistance.

Solution

Using:

\displaystyle R=\frac{V}{I}

with:

I=50\times 10^{-6}\ \text{A}

gives:

\displaystyle R=\frac{500}{50\times 10^{-6}}=10{,}000{,}000\ \Omega

So:

R=10\ \text{M}\Omega

Engineering Comment

Insulation resistance should be interpreted with voltage rating, equipment type, cable length, humidity, temperature and trend history. One passing number can still be weak evidence if it is far below previous measurements.

Plausibility Check

Microampere leakage at hundreds of volts corresponds to megohm-scale resistance, so $10\ \text{M}\Omega$ is reasonable.

Exercise 14: Normal Leakage Versus Ground-Fault Trip

A machine has three EMI filters, each contributing $3.2\ \text{mA}$ of normal leakage to ground. The residual-current device trips at $30\ \text{mA}$ . Find total normal leakage and the margin to trip.

Solution

Total leakage:

I_{leak}=3 \times 3.2=9.6\ \text{mA}

Trip margin:

M=30-9.6=20.4\ \text{mA}

Engineering Comment

The normal leakage is below the trip threshold, but nuisance trips can still occur if additional equipment, moisture, transient filters or harmonics increase residual current. Leakage should be measured in the installed configuration.

Plausibility Check

Three small milliampere leakages sum to less than $10\ \text{mA}$ , so a $30\ \text{mA}$ device still has margin.

Exercise 15: Minimum Fault Current and Pickup Window

A protective device must carry a maximum load of $18\ \text{A}$ and detect a minimum credible fault of $72\ \text{A}$ . A proposed pickup is $40\ \text{A}$ . Check the screening window:

I_{load,max}<I_{pickup}<I_{fault,min}

Solution

Substitute:

18<40<72

The pickup passes the first-pass window.

Load margin:

40-18=22\ \text{A}

Fault sensitivity margin:

72-40=32\ \text{A}

Engineering Comment

This check does not prove coordination. It only shows that the pickup is above normal load and below minimum fault. Time-current curves, tolerance, inrush and upstream/downstream selectivity still need review.

Plausibility Check

The pickup is between load and minimum fault with tens of amperes of margin on both sides, so the first-pass result is credible.

Exercise 16: Interrupting Duty Margin

A panel has maximum available fault current $I_{fault,max}=18\ \text{kA}$ . The breaker interrupting rating is $25\ \text{kA}$ . Calculate the interrupting margin.

Solution

Interrupting margin:

M_{interrupt}=I_{rating}-I_{fault,max}

M_{interrupt}=25-18=7\ \text{kA}

The breaker passes this simple interrupting-duty screen.

Engineering Comment

Interrupting rating must be checked at the applied voltage and system configuration. If a generator, transformer replacement or utility upgrade increases available fault current, the margin can disappear.

Plausibility Check

The rating is higher than the available fault current, so the positive margin is expected.

Exercise 17: Ground-Fault Residual Current

Three phase conductors carry $41.0\ \text{A}$ , $38.5\ \text{A}$ and $39.5\ \text{A}$ out to loads. The neutral returns $116.0\ \text{A}$ . Estimate residual current using:

I_{res}=|I_A+I_B+I_C-I_N|

The ground-fault alarm threshold is $2.0\ \text{A}$ .

Solution

Sum of phase currents:

I_A+I_B+I_C=41.0+38.5+39.5=119.0\ \text{A}

Residual:

I_{res}=|119.0-116.0|=3.0\ \text{A}

Since:

3.0\ \text{A}>2.0\ \text{A}

the alarm threshold is exceeded.

Engineering Comment

Residual current can indicate ground leakage, wiring error, measurement polarity error or harmonic neutral effects depending on the measurement method. The next step is not only tripping; it is verifying CT polarity, conductor routing and normal leakage sources.

Plausibility Check

The phase sum exceeds neutral return by $3\ \text{A}$ , so a $3\ \text{A}$ residual follows directly.

Exercise 18: Electrical Release Decision Gate

A small control panel has the following screening evidence:

Check	Result	Gate
Maximum load current	$18\ \text{A}$	breaker pickup $>25\ \text{A}$
Minimum fault current	$90\ \text{A}$	breaker pickup $<75\ \text{A}$
Maximum fault current	$6.5\ \text{kA}$	interrupting rating $\ge 10\ \text{kA}$
Insulation resistance	$4\ \text{M}\Omega$	$\ge 5\ \text{M}\Omega$
Normal leakage	$11\ \text{mA}$	$<20\ \text{mA}$

The proposed breaker pickup is $32\ \text{A}$ . Decide whether the panel can be released from this evidence alone.

Solution

Load check:

32\ \text{A}>25\ \text{A}

passes the stated load gate.

Minimum fault detection:

32\ \text{A}<75\ \text{A}

passes, and the actual minimum fault is $90\ \text{A}$ , above the pickup.

Interrupting duty:

6.5\ \text{kA}<10\ \text{kA}

passes.

Insulation:

4\ \text{M}\Omega<5\ \text{M}\Omega

fails.

Normal leakage:

11\ \text{mA}<20\ \text{mA}

passes.

The panel should not be released from this evidence alone because the insulation-resistance gate fails.

Engineering Comment

Most electrical checks pass, but insulation failure is a hold point. The correct action is to investigate moisture, contamination, wiring damage, connected filters, test setup and trend history before energization or shipment.

Plausibility Check

The conclusion is consistent with the gate table: a single failed safety-related insulation gate is enough to hold release even when current and interrupting checks pass.

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