Practice a College Board-style free response question on Electric Circuits. Write your response, then reveal the model answer to see exactly what earns each point.
Free Response Question · Unit 3 · Series-Parallel Circuit Analysis
A circuit consists of a 12 V battery (negligible internal resistance) connected to a 4 Ω resistor (R1) in series with a parallel combination of two resistors: R2 = 6 Ω and R3 = 3 Ω.
Component
Value
Battery EMF
12 V
R1 (in series with the parallel combination)
4 Ω
R2 (in parallel with R3)
6 Ω
R3 (in parallel with R2)
3 Ω
A
Calculate the equivalent resistance of the entire circuit.
✓ Model answer (earns the point)
First find the equivalent resistance of R2 and R3 in parallel: 1/R_23 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, so R_23 = 2 Ω. Since R1 is in series with this parallel combination, the total equivalent resistance is R_total = R1 + R_23 = 4 + 2 = 6 Ω.
Why it scores: Correctly identifies that R2 and R3 are in parallel and applies the reciprocal formula, correctly identifies that R1 is in series with that combination and adds directly, AND arrives at the correct final equivalent resistance.
B
Calculate the total current supplied by the battery.
✓ Model answer (earns the point)
Using Ohm's law on the entire circuit: I_total = V/R_total = 12 V / 6 Ω = 2 A. This is the current that flows through R1 and then splits between R2 and R3.
Why it scores: Uses the equivalent resistance found in Part A with the full battery voltage (since R1 + R_23 is the entire circuit as seen by the battery), AND arrives at the correct current value.
C
Calculate the current through R2, and explain how Kirchhoff's junction rule applies to the currents through R2 and R3.
✓ Model answer (earns the point)
First find the voltage across the parallel combination: V_23 = I_total × R_23 = (2 A)(2 Ω) = 4 V. Since R2 and R3 share this same voltage (parallel branches), the current through R2 is I_2 = V_23/R2 = 4/6 = 0.67 A. By Kirchhoff's junction rule, the total current entering the R2–R3 junction (2 A) must equal the sum of the currents leaving through each branch: I_2 + I_3 = I_total, so I_3 = 2 − 0.67 = 1.33 A. This confirms charge is conserved — none is lost or created at the junction, it simply divides between the two available paths.
Why it scores: Correctly finds the voltage across the parallel section, correctly uses Ohm's law to find I_2, AND explicitly states and applies Kirchhoff's junction rule (I_2 + I_3 = I_total) as the justification — not just computing currents without connecting them to the conservation principle.
How to score points on AP Physics 2 FRQs
Simplify the circuit step by step, and show each step. Combine parallel sections into a single equivalent resistor before treating the rest of the circuit as series, and show that intermediate value.
State which rule you're using and why. "Same voltage because they're in parallel" or "same current because they're in series" — explicit reasoning earns points beyond just the right number.
Explicitly invoke Kirchhoff's rules when asked to "explain" or "justify." Writing out I_in = I_out or ΣΔV = 0 shows the grader you understand the underlying principle, not just the shortcut formulas.
Carry exact or appropriately rounded values between parts. Don't round too aggressively in an early part if a later part depends on that value.
Double-check that current and voltage values make sense. Currents in parallel branches should sum to the total; voltages in series should sum to the source voltage.