---
title: "AP Physics C: E&M Unit 11 Practice Test"
description: "A timed AP Physics C: E&M Unit 11 (Electric Circuits) practice test in the AP format, drawn from 1,637 multiple-choice questions and 38 FRQs written for the unit. The standard version takes about 88 minutes."
canonical: "https://fiveable.me/ap-physics-c-e-m/unit-test/11"
type: "unit-test"
subject: "AP Physics C: E&M"
unit: "Unit 11"
---

# AP Physics C: E&M Unit 11 Practice Test

## Overview

A timed AP Physics C: E&M Unit 11 (Electric Circuits) practice test in the AP format, drawn from 1,637 multiple-choice questions and 38 FRQs written for the unit. The standard version takes about 88 minutes.

## Test Format

- 20 Multiple Choice Questions (40 minutes)
- 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes)
- 1 FRQ – FRQ 2 – Translation Between Representations (24 minutes)

## Unit FRQs by Type

- [FRQ 1 – Mathematical Routines](/ap-physics-c-e-m/frqs/mathematical-routines?unit=11): 10 questions
- [FRQ 2 – Translation Between Representations](/ap-physics-c-e-m/frqs/translation-between-representations?unit=11): 4 questions
- [FRQ 3 – Experimental Design](/ap-physics-c-e-m/frqs/experimental-design?unit=11): 6 questions
- [FRQ 4 – Qualitative/Quantitative Translation](/ap-physics-c-e-m/frqs/qualitative-quantitative-translation?unit=11): 18 questions

## Practice Questions

### Question 1

A battery with emf E and internal resistance r is connected to a load resistor with resistance R = r. The terminal potential difference across the battery is measured as V_1. If the load resistance is increased to R = 3r, what is the new terminal potential difference V_2 in terms of V_1?

- (A) The new potential difference is 1.5 times the original value
- (B) The new potential difference is 2.0 times the original value
- (C) The new potential difference is 0.75 times the original value
- (D) The new potential difference is 3.0 times the original value

**Answer: (A)** Analyzing the circuit using Kirchhoff's laws establishes the functional dependence of terminal voltage on load resistance. Initially, with R = r, the current is I = E/(2r) and terminal voltage is V_1 = E - Ir = E/2. With R = 3r, the current becomes I = E/(4r) and terminal voltage V_2 = E - Ir = 3E/4. Comparing the two states shows V_2 = 1.5 V_1.

### Question 2

A nonideal battery with internal resistance r is connected to a variable resistor. When the current in the circuit is I, the rate of thermal energy dissipation within the battery is P. If the variable resistor is adjusted so the current decreases to I/2, what is the new rate of energy dissipation within the battery?

- (A) The rate of dissipation decreases to 0.25P
- (B) The rate of dissipation decreases to 0.50P
- (C) The rate of dissipation increases to 2.00P
- (D) The rate of dissipation increases to 4.00P

**Answer: (A)** Applying the functional relationship for power dissipation P = I^2r reveals the quadratic dependence on current. When the current is reduced by a factor of 2 (I becomes I/2), the square of the current is reduced by a factor of 4, resulting in a new power of 0.25P.

### Question 3

A battery with internal resistance r is connected to a load resistor R. When R = r, the fraction of the total power generated by the battery that is delivered to the load is 0.50. If the load resistance is increased to R = 3r, what is the new fraction of power delivered to the load?

- (A) The fraction increases from 0.50 to 0.75
- (B) The fraction increases from 0.50 to 1.00
- (C) The fraction decreases from 0.50 to 0.25
- (D) The fraction decreases from 0.50 to 0.33

**Answer: (A)** Deriving the efficiency function n = P_load / P_total = (I^2 R) / (I^2 (R+r)) = R/(R+r) allows for prediction of the new value. Substituting R = 3r yields n = 3r/(3r+r) = 3/4 = 0.75. This demonstrates that efficiency increases as load resistance increases relative to internal resistance.

### Question 4

Two identical batteries, each with emf E and internal resistance r, are connected in parallel to a circuit. If the batteries are reconnected in series to the same circuit, how do the equivalent emf and equivalent internal resistance of the battery combination change compared to the parallel arrangement?

- (A) Equivalent emf doubles and equivalent resistance quadruples
- (B) Equivalent emf doubles and equivalent resistance doubles
- (C) Equivalent emf stays constant and equivalent resistance doubles
- (D) Equivalent emf stays constant and equivalent resistance quadruples

**Answer: (A)** Comparing the functional combinations of the two arrangements reveals the scaling factors. In parallel, E_eq = E and r_eq = r/2. In series, E_eq = 2E and r_eq = 2r. Comparing series to parallel: EMF scales by 2E/E = 2, and resistance scales by 2r/(r/2) = 4.

### Question 5

A circuit contains three resistors with resistances of 20 Ω, 30 Ω, and 60 Ω. If the 20 Ω and 30 Ω resistors are connected in parallel, and this combination is connected in series with the 60 Ω resistor, what is the equivalent resistance of the entire configuration?

- (A) 10 Ω
- (B) 45 Ω
- (C) 72 Ω
- (D) 110 Ω

**Answer: (C)** Calculating the parallel combination first gives (20×30)/(20+30) = 12 Ω. Adding this in series with the 60 Ω resistor results in a total equivalent resistance of 72 Ω.

### Question 6

Two cylindrical wires, Wire X and Wire Y, are made of the same material and have the same length. Wire X has a radius of r and a resistance of 24 Ω. Wire Y has a radius of 2r. If the two wires are connected in parallel, what is the equivalent resistance of the combination?

- (A) 4.8 Ω
- (B) 8.0 Ω
- (C) 16.0 Ω
- (D) 30.0 Ω

**Answer: (A)** Since resistance is inversely proportional to cross-sectional area (A = πr²), doubling the radius reduces resistance by a factor of 4, making Wire Y 6 Ω. The parallel combination of 24 Ω and 6 Ω is (24×6)/(24+6) = 4.8 Ω.

### Question 7

A battery with an emf of 12.0 V and internal resistance r is connected to a load resistor R = 5.0 Ω. The terminal voltage across the battery is measured to be 10.0 V. What is the value of the internal resistance r?

- (A) 1.0 Ω
- (B) 2.0 Ω
- (C) 5.0 Ω
- (D) 6.0 Ω

**Answer: (A)** The current is determined by the load resistor and terminal voltage (I = 10 V / 5 Ω = 2 A). Using the terminal voltage equation V = ε - Ir, the internal resistance is r = (12 V - 10 V) / 2 A = 1.0 Ω.

### Question 8

A network consists of three identical resistors, each with resistance R. Two of the resistors are connected in series, and this series combination is connected in parallel with the third resistor. Which expression correctly gives the equivalent resistance of the network?

- (A) R/3
- (B) 2R/3
- (C) 3R/2
- (D) 3R

**Answer: (B)** The series branch has a resistance of R + R = 2R. Placing this in parallel with the third resistor R yields an equivalent resistance of (2R × R) / (2R + R) = 2R² / 3R = 2R/3.

### Question 9

A cylindrical copper wire of radius $$r$$ carries a steady current $$I$$. The charge carriers in the wire have a drift velocity $$v_d$$. If this wire is replaced by another copper wire of radius $$2r$$ carrying the same current $$I$$, what is the new drift velocity of the charge carriers?

- (A) $$v_d / 4$$
- (B) $$v_d / 2$$
- (C) $$2 v_d$$
- (D) $$4 v_d$$

**Answer: (A)** Drift velocity is inversely proportional to the cross-sectional area for a constant current ($$v_d = I / (nqeA)$$). Since area $$A = \pi r^2$$, doubling the radius increases the area by a factor of 4, reducing the drift velocity to one-fourth.

### Question 10

A metal wire of length $$L$$ and resistance $$R$$ is stretched uniformly until its length becomes $$3L$$, while its volume remains constant. What is the new resistance of the wire?

- (A) $$R / 3$$
- (B) $$3 R$$
- (C) $$6 R$$
- (D) $$9 R$$

**Answer: (D)** Since volume $$V = AL$$ is constant, tripling $$L$$ reduces $$A$$ to $$A/3$$. Resistance $$R = \rho L/A$$ becomes $$\rho(3L)/(A/3) = 9(\rho L/A) = 9R$$.

## Review the Unit

- [11.1 Electric Current](/ap-physics-c-e-m/unit-11/1-electric-current/study-guide/9YRMrkv1PVy23BzH)
- [11.2 Electric Circuits](/ap-physics-c-e-m/unit-11/2-electric-circuits/study-guide/17WyJIXaesWwOEX8)
- [11.3 Resistance, Resistivity, and Ohm's Law](/ap-physics-c-e-m/unit-11/3-resistance-resistivity-and-ohms-law/study-guide/TnRPkql9C75GQe0d)
- [11.4 Electric Power](/ap-physics-c-e-m/unit-11/4-electric-power/study-guide/u2cRqQTlthIAJtwp)
- [11.5 Compound Direct Current Circuits](/ap-physics-c-e-m/unit-11/5-compound-direct-current-circuits/study-guide/lvbJLaPd4EqBAUf6)
- [11.6 Kirchhoff's Loop Rule](/ap-physics-c-e-m/unit-11/6-kirchhoffs-loop-rule/study-guide/Nxj12AcsMRzP3ejn)
- [11.7 Kirchhoff's Junction Rule](/ap-physics-c-e-m/unit-11/7-kirchhoffs-junction-rule/study-guide/2tEeZRNyXUYgq63j)
- [11.8 Resistor-Capacitor (RC) Circuits](/ap-physics-c-e-m/unit-11/8-resistor-capacitor-rc-circuits/study-guide/qy6QreLu93jx043L)

## Other Unit Tests

- [Unit 8: Electric Charges & Fields: Gauss's Law](/ap-physics-c-e-m/unit-test/8)
- [Unit 9: Electric Potential](/ap-physics-c-e-m/unit-test/9)
- [Unit 10: Conductors and Capacitors](/ap-physics-c-e-m/unit-test/10)
- [Unit 12: Magnetic Fields & Electromagnetism](/ap-physics-c-e-m/unit-test/12)
- [Unit 13: Electromagnetic Induction](/ap-physics-c-e-m/unit-test/13)

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