A circuit’s connections tell you which quantities are shared. Learn to trace those connections before using a formula, starting with a parallel circuit that makes the difference between voltage and current clear.
What you’ll learn
Learn the method
An ideal 12-volt source connects across two parallel resistors, one 6 ohms and one 3 ohms. What current flows through each resistor? Ignore wire and source resistance.
Both resistors connect between the same top and bottom wires. Each branch therefore has the full 12-volt potential difference across it.
Rearrange V = IR as I = V ÷ R. For the 6-ohm resistor, I = 12 ÷ 6 = 2 amperes. For the 3-ohm resistor, I = 12 ÷ 3 = 4 amperes.
At the same voltage, the smaller resistance allows the larger current. The source supplies the sum of the branch currents: 2 + 4 = 6 amperes.
2 A through the 6 Ω resistor and 4 A through the 3 Ω resistor.
Adding 6 Ω and 3 Ω to get 9 Ω would treat the resistors as a series connection. In this diagram, current has two paths; it does not pass through both resistors in sequence.
If the 3-ohm branch is opened, does the current in the 6-ohm branch change in this ideal model?
No. The ideal source still places 12 V across the 6 Ω resistor, so its current stays 2 A. Total source current falls to 2 A because the other path is interrupted.
The example teaches one method. Open a concept below when you need a rule or a different type of example during practice.
Voltage is an electric potential difference, measured in volts. Current is the rate of charge flow, measured in amperes. Resistance opposes current and is measured in ohms. For a resistor modeled by Ohm’s law, V = IR. A 12-volt potential difference across a 4-ohm resistor produces a current of 3 amperes.
A series circuit has one path through its components, so the same current passes through each. Parallel branches share the same voltage between their common connection points, while branch currents can differ. Follow wires and junctions rather than judging which components look close together on the page.
An open circuit interrupts a conducting path. An ideal short circuit provides a path with negligible resistance that may bypass another component. They are different changes: a break stops current along that path, while a short can greatly increase current. Interpret these as circuit models for the question.
A resistor limits current in a circuit, a capacitor stores energy in an electric field, and a diode conducts much more readily in one direction than the other. A switch opens or closes a path. Start with the function being asked about, then match the component or circuit symbol.
Now apply what you know
Try a few questions without opening the solutions, then explain why each answer works. Use scratch paper for calculations and work without a calculator. These original, untimed questions sample the subject; the result is not an official score.
Work through one question at a time. Explain your approach, check the worked answer, then move on. Use scratch paper for calculations.
0 of 12 checked · 0 correct.
Progress is saved in this browser. These original study questions do not produce an official AFQT score.
Answer: C. A capacitor
Two separated parallel plates represent a capacitor. An inductor uses a coil symbol, and a transformer includes two coupled coils rather than this single pair of plates.
Answer: C. C
Only the point marked C lies below the zero-voltage axis. Negative relative voltage means P is at a lower potential than Q at that instant; points above zero have the opposite polarity.
Answer: A. 0 A
Wire W directly joins the two terminals of the 8 Ω resistor, making their potential difference zero. Its current is therefore 0 ÷ 8 = 0 A. The supply still sends 12 ÷ 4 = 3 A through the 4 Ω resistor and the bypass wire.
Answer: B. 6 W
The parallel combination is (3 × 6) ÷ (3 + 6) = 2 Ω. With the series 2 Ω resistor, total resistance is 4 Ω and source current is 3 A. The parallel section therefore has 6 V across it, so the 6 Ω resistor dissipates 6² ÷ 6 = 6 W. Using the full 12 V across that branch would incorrectly give 24 W.
Answer: A. A and B
A and B share a junction with no branching path, so the same branch current passes through both. C is in a separate branch across the supply and does not form a series pair with either one individually.
Answer: B. Either switch may be closed, or both may be closed
S1 and S2 form alternative parallel paths. Closing S1 completes one path through the lamp even with S2 open. Opening both breaks every available path.
Answer: A. South; it attracts the retreating north pole
The induced field opposes the change caused by the magnet moving away. The near coil face becomes south, attracting the retreating north pole and opposing separation. Lenz’s law opposes the change in magnetic flux, not every existing field.
Answer: B. About 2.5 V because the output and input resistances form equal halves of a divider
The two 50 Ω resistances divide the ideal 5.0 V pulse equally: Vinput=5×50/(50+50)=2.5 V. A 1 MΩ input draws little current, leaving nearly the full 5 V. A 50 Ω input is a finite load, not a zero-ohm short.
Answer: A. A
A shows the conventional earth symbol, with a vertical connection and successively shorter horizontal bars. B is a resistor, C a capacitor, and D a cell. These symbols describe different circuit functions.
Answer: C. 1.5 A
The incoming currents total 2.8 + 0.6 = 3.4 A. The outgoing currents must have the same total, so I = 3.4 − 1.9 = 1.5 A. Adding every labeled current ignores the directions of flow.
Answer: C. It is forward biased
A diode is forward biased when its anode is sufficiently positive relative to its cathode. Here the stated polarity is forward; the resistor limits current rather than reversing the diode’s bias.
Answer: C. 0.5 A
The turns ratio makes secondary voltage 240 × 200/800 = 60 V rms. The 30 Ω load draws 60/30 = 2 A, so output power is 120 W. In an ideal transformer input power is also 120 W; primary current is 120/240 = 0.5 A. The 2 A choice is secondary current.
Before moving on, choose one error you can explain. Rework that question with the solution closed, then revisit the skill in a later session to check what you remember.
Label the source, the component being asked about and the known units. Rearrange V = IR to isolate the requested quantity before substituting values. For power, P = VI relates watts, volts and amperes; do not confuse power with energy used over time.
Before calculating, predict what happens when resistance rises at a fixed voltage, or when a switch opens a particular branch. Then use the formula or diagram to check the prediction. This exposes an incorrect circuit interpretation early.
Electronics Information does not enter the AFQT. It can contribute to service composites used for technical job qualification. Meeting a composite threshold is only one part of a job’s eligibility requirements.
The computer test includes unscored tryout questions in selected subtests. The longer time shown is the total for a section with tryouts, not extra time to add. These figures describe the proctored CAT-ASVAB, not untimed PiCAT or our practice sets. See the official questions and timing table and the guide to test versions.
Sources and materials. Subtest descriptions and timing were checked on September 29, 2026 against the Department of Defense’s What to Expect guide, sample questions and score guide. Teaching examples, diagrams and practice questions are independently authored Test Ninjas materials. Test Ninjas is not affiliated with or endorsed by the Department of Defense.