AP Physics 2 Quiz: Simple Circuits

What this quiz covers

This quiz focuses on Simple Circuits, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The current is $0.50\,\text{A}$ everywhere in the loop. (correct answer)
2. The current is $2.0\,\text{A}$ everywhere in the loop.
3. The current is $3.0\,\text{A}$ because the battery sets the current.
4. The current is larger on the side of the loop closer to the $+$ terminal.

Explanation: This question tests understanding of simple circuits. In a single-loop circuit, the current is determined by Ohm's law: I = V/R = 3.0 V / 6.0 Ω = 0.50 A. This current is the same everywhere in the loop because charge cannot accumulate at any point. Choice D reflects the misconception that current varies with position relative to battery terminals, ignoring the fundamental principle of current conservation in circuits. For any single-loop circuit, calculate current using total voltage divided by total resistance.

Question 2

1. It is unchanged because the resistor determines the current.
2. It decreases because a higher battery voltage uses up charge faster.
3. It doubles because the voltage doubles while $R$ stays the same. (correct answer)
4. It splits into two currents because the battery voltage increased.

Explanation: This question tests understanding of simple circuits. In a single-loop circuit, the current is determined by Ohm's law: I = V/R. Initially, with a 1.5 V battery and 3.0 Ω resistor, the current is 0.5 A. When the battery is replaced with a 3.0 V battery (doubling the voltage) while keeping the same 3.0 Ω resistor, the current doubles to 1.0 A. This follows directly from I = V/R: doubling the numerator doubles the result. Choice B incorrectly suggests that higher voltage "uses up" charge faster, misunderstanding that current is charge flow rate, not charge depletion. In simple circuits, current is directly proportional to voltage when resistance is constant.

Question 3

1. It increases because the larger resistor pulls more current.
2. It decreases because the total series resistance increases. (correct answer)
3. It stays the same because the battery voltage is fixed.
4. It splits so less current goes through $R_2$ than $R_1$.

Explanation: This question tests understanding of simple circuits. In a single-loop circuit with resistors in series, the total resistance is the sum of individual resistances: R_total = R₁ + R₂. When R₂ increases while R₁ and the battery voltage remain constant, the total resistance increases. By Ohm's law (I = V/R), increasing the denominator while keeping the numerator constant causes the current to decrease. Choice A incorrectly suggests larger resistors "pull" more current, when actually they oppose current flow. Remember that in series circuits, increasing any resistance decreases the current through all components.

Question 4

1. It is larger than the voltage across the $1.0\,\Omega$ resistor. (correct answer)
2. It is equal to the voltage across the $1.0\,\Omega$ resistor.
3. It is $0\,\text{V}$ because most voltage is used up before it.
4. It is $9.0\,\text{V}$ because each resistor gets the full battery voltage.

Explanation: This question tests understanding of simple circuits. In series circuits, voltage divides proportionally to resistance values. The total resistance is 9.0 Ω, giving a current of I = 9.0 V / 9.0 Ω = 1.0 A. The voltage across each resistor is V = IR, so the 8.0 Ω resistor has V = 1.0 A × 8.0 Ω = 8.0 V, while the 1.0 Ω resistor has only 1.0 V. Choice D reflects the misconception that each series component gets full battery voltage, violating energy conservation. In series circuits, larger resistors always have proportionally larger voltage drops.

Question 5

1. It is $4.5\ \text{V}$ because the only resistor has the full battery voltage. (correct answer)
2. It is $0\ \text{V}$ because the voltage is used up by the current.
3. It is less than $4.5\ \text{V}$ because some voltage is lost in the wires.
4. It is greater than $4.5\ \text{V}$ because the resistor opposes the battery.

Explanation: This question tests understanding of simple circuits. In a single-loop circuit with an ideal battery and one resistor, Kirchhoff's voltage law requires that the voltage across the resistor equals the battery voltage. Since there's only one resistor in the loop, it must have the full 4.5 V across it to complete the circuit. This can be verified using Ohm's law: V = IR = 0.30 A × R, and since the battery provides 4.5 V, the resistor must have 4.5 V across it. Choice B incorrectly suggests voltage is "used up" by current, confusing voltage (potential difference) with energy. In simple circuits, the sum of voltage drops must equal the source voltage.

Question 6

1. The current increases because the battery supplies constant current
2. The current decreases because $I=\dfrac{V}{R}$ (correct answer)
3. The current stays the same because voltage is used up in the resistor
4. The current splits into two paths as the resistance increases

Explanation: This question tests understanding of simple circuits. In a single-loop circuit with one battery and one resistor, Ohm's law states that current I = V/R, where V is the battery voltage and R is the resistance. Since the battery voltage remains constant at 9.0 V and the resistance increases, the current must decrease according to this inverse relationship. Choice B incorrectly assumes batteries supply constant current rather than constant voltage, which is a common misconception about how batteries work. In a single-loop circuit, apply Ohm's law directly: when resistance increases with constant voltage, current must decrease.

Question 7

1. The current stays the same because the battery keeps pushing charges
2. The current splits into two currents at the break
3. The current becomes zero because there is no complete path (correct answer)
4. The current increases because resistance decreases when the circuit opens

Explanation: This question tests understanding of simple circuits. In a single-loop circuit, current can only flow if there is a complete conducting path from the positive terminal of the battery through the circuit and back to the negative terminal. When the wire is broken at any point, this path is interrupted and current immediately drops to zero throughout the entire circuit. Choice B incorrectly assumes the battery can somehow push charges through an open circuit, not understanding that a complete loop is essential for current flow. In a single-loop circuit, any break in the conducting path stops all current flow - the circuit must be closed for current to exist.

Question 8

1. It doubles because the battery forces the same voltage through more resistance.
2. It is halved because the resistance is doubled while the voltage stays the same. (correct answer)
3. It is unchanged because current is used up in the resistor.
4. It splits into two equal currents because the resistance doubled.

Explanation: This question tests understanding of simple circuits. In a single-loop circuit, the current is determined by Ohm's law: I = V/R. With the battery voltage constant at 9.0 V and the original current at 0.30 A, the original resistance is R = V/I = 30 Ω. When resistance doubles to 60 Ω, the new current becomes I = 9.0 V / 60 Ω = 0.15 A, which is half the original current. Choice D incorrectly suggests current splits into two paths, revealing the misconception that doubling resistance creates multiple current paths in a single loop. In a single-loop circuit, current is inversely proportional to resistance when voltage is constant.

Question 9

1. It decreases because $P=\dfrac{V^2}{R}$ when battery voltage is fixed. (correct answer)
2. It increases because higher resistance always means more power.
3. It stays constant because the battery supplies constant power.
4. It becomes zero because the voltage is used up in the battery first.

Explanation: This question tests understanding of simple circuits. Power dissipated in a resistor can be expressed as P = V²/R when the voltage across it is fixed. In a single-loop circuit with one resistor, the voltage across the resistor equals the battery voltage (8.0 V). When resistance increases while voltage remains constant, the power decreases because it's inversely proportional to resistance. Initially, R = V²/P = 64/2 = 32 Ω, and increasing R decreases P. Choice A incorrectly assumes higher resistance means more power, confusing the relationships in different circuit configurations. For constant voltage sources, power dissipation decreases as resistance increases.

Question 10

1. It is smaller in the $2.0\ \Omega$ resistor because current is used up by the $4.0\ \Omega$ resistor.
2. It splits between the resistors, so each has different current.
3. It is larger in the $2.0\ \Omega$ resistor because smaller resistance means larger current.
4. It is the same in both resistors because they are in series in a single loop. (correct answer)

Explanation: This question tests understanding of simple circuits. In a series circuit, components are connected end-to-end forming a single path for current flow. By the principle of charge conservation, the same amount of charge must flow through each component per unit time, meaning the current is identical through both resistors. The different resistances affect the voltage drop across each resistor (V = IR), not the current through them. Choice A incorrectly applies the relationship between resistance and current as if the resistors were in parallel. In any series connection within a single loop, the current through all components is always the same.

Question 11

1. It is $3.0\ \Omega$ because $R=IV$.
2. It is $12\ \Omega$ because $R=V/I$. (correct answer)
3. It is $0.083\ \Omega$ because $R=I/V$.
4. It is $6.0\ \Omega$ because the battery supplies $1\ \Omega$ per volt.

Explanation: This question tests understanding of simple circuits. To find resistance when current and voltage are known, we rearrange Ohm's law to R = V/I. With a voltage of 6.0 V and current of 0.50 A, the resistance is R = 6.0 V / 0.50 A = 12 Ω. This is the fundamental relationship between voltage, current, and resistance in any circuit element. Choice A incorrectly multiplies current and voltage (R = IV), which would give units of power (watts), not resistance (ohms). When solving circuit problems, always check that your formula gives the correct units for the quantity you're finding.

Question 12

1. The current is the same through $R_1$ and $R_2$ (correct answer)
2. The current is larger through $R_1$ because it has smaller resistance
3. The current is larger through $R_2$ because it has larger resistance
4. The current splits so each resistor gets half the battery current

Explanation: This question tests understanding of simple circuits. In a single-loop circuit, there is only one path for current to flow, so the same current must pass through every component in the circuit. This is a consequence of charge conservation - charges cannot accumulate or disappear at any point in the circuit. The current through both R₁ and R₂ is I = V/(R₁ + R₂) = 12V/(2Ω + 4Ω) = 2.0 A. Choice A incorrectly assumes that smaller resistance means larger current within the same series circuit, confusing the behavior of individual resistors with parallel circuits. In a single-loop circuit, remember that the same current flows through all components regardless of their individual resistances.

Question 13

1. It becomes $0.30\ \text{A}$ because higher voltage is used up faster.
2. It stays $0.60\ \text{A}$ because the resistor sets the current regardless of voltage.
3. It splits into two currents because the battery voltage increased.
4. It becomes $1.2\ \text{A}$ because doubling the voltage doubles the current. (correct answer)

Explanation: This question tests understanding of simple circuits. In a single-loop circuit, current is proportional to voltage when resistance is constant, following I = V/R. The original resistance is R = 3.0 V / 0.60 A = 5.0 Ω. With the new 6.0 V battery and the same 5.0 Ω resistor, the new current is I = 6.0 V / 5.0 Ω = 1.2 A, exactly double the original. Choice C incorrectly suggests higher voltage is "used up faster," revealing the misconception that voltage is consumed like a substance rather than being a driving force for current. In a single-loop circuit with constant resistance, doubling the voltage doubles the current.

Question 14

1. It decreases because lower voltage forces lower resistance.
2. It stays the same because resistance is a property of the resistor, not the battery. (correct answer)
3. It increases because the battery now supplies less current.
4. It becomes zero because voltage is used up in the circuit.

Explanation: This question tests understanding of simple circuits. Resistance is an intrinsic property of a resistor that depends on its material, dimensions, and temperature, not on the applied voltage or resulting current. When the battery voltage changes from 5.0 V to 2.5 V, the resistor's resistance remains unchanged. The current will decrease proportionally (halving when voltage halves), but the resistance stays constant. Choice A incorrectly suggests voltage determines resistance, revealing the misconception that circuit components change their properties based on applied voltages. In a single-loop circuit, a resistor's resistance value is independent of the battery voltage.

Question 15

1. It is zero because there is no complete path for charge flow. (correct answer)
2. It is nonzero because the battery supplies constant current.
3. It splits into two equal currents on each side of the open switch.
4. It increases until the battery voltage is used up by the resistor.

Explanation: This question tests understanding of simple circuits. In any circuit, current requires a complete conducting path from the battery's positive terminal through the circuit and back to the negative terminal. With the switch open, this path is broken, preventing any charge flow, so the current is zero everywhere in the circuit. The battery voltage still exists across the open switch, but no current flows. Choice B incorrectly assumes batteries supply constant current regardless of circuit conditions, revealing the misconception that batteries are current sources rather than voltage sources. In a single-loop circuit, an open switch always results in zero current throughout the entire loop.

Question 16

1. It is unpredictable because voltage is used up by the resistor.
2. It is unchanged because the battery supplies a constant current.
3. It is smaller because higher current at the same voltage implies less resistance. (correct answer)
4. It is larger because higher current requires more resistance.

Explanation: This question tests understanding of simple circuits. In a single-loop circuit, Ohm's law states I = V/R, meaning current and resistance are inversely related when voltage is constant. Since the battery voltage remains at 10 V and the current increases after replacing the resistor, the new resistance must be smaller than the original. This follows from rearranging Ohm's law: R = V/I, where larger current means smaller resistance. Choice A incorrectly suggests higher current requires more resistance, revealing the misconception that current and resistance are directly proportional. In a single-loop circuit with constant voltage, higher current always indicates lower resistance.

Question 17

1. It stays the same because the battery supplies a constant current.
2. It is cut in half because $I$ is proportional to the battery voltage for fixed $R$. (correct answer)
3. It doubles because the resistance forces the same power output.
4. It becomes zero because the smaller battery's voltage is used up in the resistor.

Explanation: This question tests understanding of simple circuits. In a single-loop circuit, Ohm's law states that current I = V/R. When the battery voltage changes from 10 V to 5.0 V while the resistance stays constant, the current becomes I = 5.0V/R instead of I = 10V/R. This means the current is reduced by the same factor as the voltage reduction, which is half. Since voltage and current are directly proportional for a fixed resistance, halving the voltage halves the current. Choice B incorrectly assumes batteries supply constant current regardless of voltage. In simple circuits with fixed resistance, current is directly proportional to the applied voltage.

Question 18

1. $R=4.5\,\Omega$ because $R=VI$
2. $R=0.020\,\Omega$ because $R=\dfrac{I}{V}$
3. $R=15\,\Omega$ because the battery provides $15\,\text{V}$ to each coulomb
4. $R=50\,\Omega$ because $R=\dfrac{V}{I}$ (correct answer)

Explanation: This question tests understanding of simple circuits. In a single-loop circuit, Ohm's law can be rearranged to find resistance when voltage and current are known: R = V/I. With a 15 V battery and 0.30 A of current, the resistance is R = 15V/0.30A = 50 Ω. This calculation shows how much the resistor opposes current flow in the circuit. Choice C incorrectly inverts the formula to R = I/V, which would give units of 1/Ω (conductance) rather than Ω (resistance). In a single-loop circuit, always use R = V/I to calculate resistance from measured voltage and current.

Question 19

1. It is $4.1\ \Omega$ because current is used up in the resistor.
2. It is $5.0\ \Omega$ because $R=V/I$. (correct answer)
3. It is $0.20\ \Omega$ because $R=VI$.
4. It is $9.0\ \Omega$ because the battery supplies constant current.

Explanation: This question tests understanding of simple circuits. In a single-loop circuit, Ohm's law relates voltage, current, and resistance through R = V/I. With the battery providing 4.5 V and the measured current at 0.90 A, the resistance is R = 4.5 V / 0.90 A = 5.0 Ω. The entire battery voltage appears across the single resistor in the loop. Choice A incorrectly multiplies voltage and current (R = VI), revealing the misconception that resistance equals the product rather than the quotient of voltage and current. In a single-loop circuit, always use R = V/I to find resistance from known voltage and current.

Question 20

1. It decreases because the total resistance in the loop increases. (correct answer)
2. It stays the same because the voltage is used up by the resistor.
3. It increases because the battery supplies a constant current.
4. It splits between parts of the resistor, so each part has less current.

Explanation: This question tests understanding of simple circuits. In a single-loop circuit with one battery and one resistor, Ohm's law (V = IR) governs the relationship between voltage, current, and resistance. Since the battery voltage remains constant at 12 V and the new resistor has larger resistance, the current must decrease according to I = V/R. Choice C incorrectly suggests current can split in a single-loop circuit, revealing the misconception that current divides even when there's only one path. In a single-loop circuit, the same current flows through all components, and increasing resistance while keeping voltage constant always decreases current.