AP Physics 2 Quiz: Compound Direct Current Dc Circuits

What this quiz covers

This quiz focuses on Compound Direct Current Dc 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 in the $R_2$ branch is less than the current in the $R_3$ branch.
2. The current in the $R_2$ branch equals the current in the $R_3$ branch.
3. The current in the $R_2$ branch is greater than the current in the $R_3$ branch. (correct answer)
4. The current in the $R_2$ branch is zero because $R_1$ limits the current.

Explanation: This problem tests understanding of compound DC circuits. At junction C, current from R₁ splits between parallel branches containing R₂ (10 Ω) and R₃ (30 Ω). Since parallel resistors share the same voltage and follow Ohm's law (I = V/R), the branch with lower resistance carries more current. With R₂ having one-third the resistance of R₃, it carries three times the current: I₂ = V/10 while I₃ = V/30, making I₂ = 3I₃. Choice A incorrectly reverses this relationship, suggesting higher resistance means more current, a common misconception about Ohm's law. When current splits at a junction in compound circuits, always remember that more current flows through the path of least resistance.

Question 2

1. $R_2$ dissipates more power than $R_3$. (correct answer)
2. $R_2$ dissipates the same power as $R_3$.
3. $R_2$ dissipates less power than $R_3$.
4. Both dissipate zero power because the current is consumed in $R_1$.

Explanation: This problem tests understanding of compound DC circuits. R₁ (6Ω) is in series with parallel branches containing R₂ (6Ω) and R₃ (12Ω). In parallel branches, voltage is the same across each resistor, but current divides inversely with resistance—R₂ gets twice the current of R₃ since it has half the resistance. Power dissipation follows P = V²/R for parallel resistors with the same voltage, so lower resistance means higher power. Since R₂ (6Ω) has half the resistance of R₃ (12Ω), it dissipates twice the power. Choice D incorrectly assumes current is "consumed" in R₁, violating conservation of charge. To analyze power in compound circuits, first determine whether resistors are in series (same current) or parallel (same voltage), then apply the appropriate power formula.

Question 3

1. The current through $R_1$ is greater than the current through $R_2$. (correct answer)
2. The current through $R_1$ equals the current through $R_2$.
3. The current through $R_1$ is less than the current through $R_2$.
4. The current through $R_1$ is zero because $R_3$ is in series after junction K.

Explanation: This problem tests understanding of compound DC circuits. The parallel combination of R₁ (5Ω) and R₂ (10Ω) is in series with R₃ (5Ω). In parallel branches, current divides inversely with resistance—since R₁ has half the resistance of R₂, it carries twice the current of R₂. The total current from the parallel section then flows through R₃, which equals the sum of currents through R₁ and R₂. Therefore, the current through R₁ is greater than the current through R₂. Choice D incorrectly assumes that having R₃ in series somehow blocks current through R₁, misunderstanding how series and parallel combinations work. When solving compound circuits, trace the current path and apply junction rules: current entering a junction equals current leaving.

Question 4

1. The current through $R_1$ is greater than the current through $R_2$. (correct answer)
2. The current through $R_1$ equals the current through $R_2$.
3. The current through $R_1$ is less than the current through $R_2$.
4. The current through $R_1$ is zero because the junction blocks current.

Explanation: This question examines current relationships in compound DC circuits where a series resistor precedes identical parallel branches. R₁ is in series with the entire circuit, so it carries the total current before it splits at junction J into two equal branches (R₂ = R₃ = 8 Ω). Since the parallel branches have equal resistance, the current divides equally between them, with each branch carrying half the total current. Therefore, the current through R₁ (total current) is greater than the current through R₂ (half the total current). Choice D incorrectly claims junctions block current, misunderstanding that junctions are simply connection points where current conservation applies. To analyze compound circuits systematically, identify which components carry total current versus partial current based on their position relative to junctions.

Question 5

1. $\Delta V_{R_1}$ is greater than $\Delta V_{R_2}$ because $R_1$ is smaller.
2. $\Delta V_{R_1}$ is less than $\Delta V_{R_2}$ because $R_2$ is larger.
3. $\Delta V_{R_1}$ equals $\Delta V_{R_2}$ because they share junctions $J$ and $K$. (correct answer)
4. $\Delta V_{R_2}$ is zero because the $4\ \Omega$ branch takes all the voltage.

Explanation: This problem tests understanding of voltage in compound DC circuits with a parallel section followed by series resistance. R₁ = 4 Ω and R₂ = 8 Ω are in parallel between junctions J and K, followed by R₃ = 4 Ω in series. The key principle is that all components connected between the same two junctions (J and K) must have the same potential difference, regardless of their individual resistances. Therefore, the voltage across R₁ equals the voltage across R₂. Choice D incorrectly suggests one branch "takes all the voltage," misunderstanding that parallel branches share the same voltage drop. When analyzing compound circuits, identify junction pairs and remember that all paths between the same junctions have equal potential differences.

Question 6

1. The current through $R_2$ is greater than the current through $R_3$.
2. The current through $R_2$ equals the current through $R_3$.
3. The current through $R_2$ is less than the current through $R_3$. (correct answer)
4. The current through $R_2$ is zero because current is used up in $R_1$.

Explanation: This problem tests understanding of compound DC circuits. In this circuit, R₁ is in series with a parallel combination of R₂ and R₃, where the parallel branches rejoin before returning to the battery. In parallel branches, the voltage across each branch is the same, but current divides inversely proportional to resistance—more current flows through the smaller resistance. Since R₃ = 3 Ω is smaller than R₂ = 6 Ω, more current flows through R₃ than through R₂. Choice D incorrectly assumes current is "used up" in R₁, reflecting the misconception that current is consumed rather than conserved. To solve compound circuit problems, first identify series and parallel regions, then apply the appropriate rules for current and voltage distribution in each region.

Question 7

1. The current through $R_3$ is twice the current through $R_2$.
2. The current through $R_2$ is twice the current through $R_3$. (correct answer)
3. The currents through $R_2$ and $R_3$ are equal because they are parallel.
4. The current through $R_2$ is zero because $R_1$ uses up the current first.

Explanation: This problem tests understanding of compound DC circuits. In this circuit, R₁ is in series with a parallel combination of R₂ and R₃, meaning all current through R₁ must split between the two parallel branches. Since parallel resistors share the same voltage drop, and current follows Ohm's law (I = V/R), the branch with lower resistance carries more current. With R₂ = 6.0 Ω and R₃ = 12.0 Ω, the current through R₂ is I₂ = V/6 while the current through R₃ is I₃ = V/12, making I₂ = 2I₃. Choice C incorrectly assumes equal currents in parallel branches, reflecting the misconception that parallel elements always have equal currents rather than equal voltages. When analyzing compound circuits, first identify series and parallel regions, then apply the appropriate rules: series elements share current, parallel elements share voltage.

Question 8

1. The current through $R_2$ is half the current through $R_3$.
2. The current through $R_2$ equals the current through $R_3$.
3. The current through $R_2$ is twice the current through $R_3$. (correct answer)
4. The current through $R_3$ is greater because it has the larger resistance.

Explanation: This problem tests understanding of compound DC circuits. Between junctions M and N, resistors R₂ (10 Ω) and R₃ (20 Ω) are in parallel, sharing the same voltage drop. By Ohm's law (I = V/R), current through each branch is inversely proportional to its resistance. Since R₃ has twice the resistance of R₂, it carries half the current: I₂ = V/10 and I₃ = V/20, making I₂ = 2I₃. Choice D incorrectly suggests larger resistance means more current, reversing the actual relationship between resistance and current in parallel circuits. When solving compound circuits, remember that in parallel branches with the same voltage, lower resistance always means higher current.

Question 9

1. More current goes through $R_2$ than through $R_3$. (correct answer)
2. More current goes through $R_3$ than through $R_2$.
3. Equal current goes through $R_2$ and $R_3$ because they are parallel.
4. All current goes through $R_3$ because it has the larger resistance.

Explanation: This problem tests understanding of compound DC circuits. At junction J, the total current from R₁ splits between parallel branches R₂ (4.0 Ω) and R₃ (12.0 Ω). Since parallel resistors share the same voltage and follow Ohm's law (I = V/R), current divides inversely with resistance. With R₂ having one-third the resistance of R₃, it carries three times the current: if I₃ = I, then I₂ = 3I. Choice C incorrectly assumes equal currents in parallel branches, a common misconception that ignores how resistance affects current distribution. To analyze current splitting in compound circuits, apply the principle that lower resistance paths carry proportionally more current when voltage is constant.

Question 10

1. The current in Branch 1 is greater than the current in Branch 2.
2. The current in Branch 1 is less than the current in Branch 2.
3. The current in Branch 1 is zero because $R_1$ and $R_2$ consume the current.
4. The current in Branch 1 equals the current in Branch 2. (correct answer)

Explanation: This problem tests understanding of compound DC circuits. The circuit has two parallel branches between junctions J and K: Branch 1 contains R₁ + R₂ = 3Ω + 3Ω = 6Ω total, while Branch 2 contains R₃ = 6Ω. Since both branches have equal total resistance (6Ω each) and experience the same voltage drop (18V between J and K), they carry equal currents according to Ohm's law. The current in each branch equals 18V/6Ω = 3A. Choice D incorrectly assumes current is "consumed" by resistors, which is a common misconception about current conservation. Strategy: calculate total resistance in each parallel branch before comparing currents.

Question 11

1. The current in Branch 1 is greater than the current in Branch 2. (correct answer)
2. The current in Branch 1 is zero because Branch 2 has more resistors.
3. The current in Branch 1 equals the current in Branch 2.
4. The current in Branch 1 is less than the current in Branch 2.

Explanation: This problem tests understanding of compound DC circuits. Between junctions J and K, Branch 1 has R₁ = 3Ω while Branch 2 has R₂ + R₃ = 3Ω + 3Ω = 6Ω total resistance. Since both branches experience the same 12V across J-K but Branch 1 has half the resistance of Branch 2, Branch 1 carries twice the current according to I = V/R. Specifically, I₁ = 12V/3Ω = 4A while I₂ = 12V/6Ω = 2A, making the current in Branch 1 greater. Choice D incorrectly assumes having more resistors means zero current, misunderstanding that series resistors add resistance but don't block current. Remember: lower total branch resistance means higher branch current in parallel circuits.

Question 12

1. The current through $R_2$ is less than the current through $R_3$.
2. The current through $R_2$ is zero because current is consumed by $R_1$.
3. The current through $R_2$ is greater than the current through $R_3$. (correct answer)
4. The current through $R_2$ equals the current through $R_3$.

Explanation: This problem tests understanding of compound DC circuits. After R₁, the circuit splits at junction J into parallel branches containing R₂ = 4Ω and R₃ = 8Ω. In parallel branches, current divides inversely with resistance: lower resistance carries more current. Since R₂ has half the resistance of R₃, it carries twice the current of R₃ according to I = V/R where both experience the same voltage. If the voltage across the parallel section is V_parallel, then I₂ = V_parallel/4Ω while I₃ = V_parallel/8Ω, making I₂ > I₃. Choice D incorrectly assumes R₁ "consumes" current, misunderstanding current conservation. Remember: in parallel branches, current inversely follows resistance ratios.

Question 13

1. The voltage across $R_3$ is greater than the total voltage across Branch 1.
2. The voltage across $R_3$ equals the total voltage across Branch 1. (correct answer)
3. The voltage across $R_3$ is zero because current divides at J.
4. The voltage across $R_3$ is less than the total voltage across Branch 1.

Explanation: This problem tests understanding of compound DC circuits. Between junctions J and K, Branch 1 contains R₁ + R₂ = 3Ω + 6Ω = 9Ω total, while Branch 2 contains R₃ = 9Ω. Since these parallel branches have equal total resistance and connect to the same junction points, they experience identical voltage drops across their entire lengths. The 15V battery voltage is divided between any series elements before J and the parallel section J-K, but whatever voltage appears across J-K applies equally to both branches. Choice D incorrectly assumes voltage becomes zero when current divides, misunderstanding voltage behavior in parallel circuits. Remember: parallel branches between the same junctions always have equal voltage drops.

Question 14

1. The voltage across $R_2$ is greater than the voltage across $R_3$.
2. The voltage across $R_2$ is zero because current splits at J.
3. The voltage across $R_2$ is less than the voltage across $R_3$.
4. The voltage across $R_2$ equals the voltage across $R_3$. (correct answer)

Explanation: This problem tests understanding of compound DC circuits. Between junctions J and K, R₂ and R₃ are in parallel, meaning they connect to the same two junction points. The fundamental principle of parallel circuits is that all elements between the same junctions experience identical voltage drops, regardless of their individual resistances or the currents they carry. This voltage equality holds whether R₂ and R₃ have equal or different resistance values. Choice D incorrectly assumes voltage becomes zero when current splits at a junction, confusing current division with voltage behavior. Key strategy: recognize that parallel elements always share the same voltage across their terminals.

Question 15

1. The current through $R_1$ is less than the current through $R_2$.
2. The current through $R_1$ equals the current through $R_2$.
3. The current through $R_1$ is greater than the current through $R_2$. (correct answer)
4. The current through $R_1$ is zero because current is consumed in the parallel branches.

Explanation: This problem tests understanding of compound DC circuits. R₁ is in series with a parallel combination of R₂ and R₃ (both 8.0 Ω). Since R₁ is in series with the entire parallel section, it carries the total current that then splits between the parallel branches. With R₂ = R₃, the current splits equally, so each branch carries half the total current. Therefore, the current through R₁ equals I₂ + I₃ = 2I₂, making the current through R₁ twice that through R₂. Choice B incorrectly assumes series and parallel components carry equal currents, missing the fundamental principle of current conservation at junctions. To analyze compound circuits, trace the current path and apply Kirchhoff's current law: current entering a junction equals current leaving it.

Question 16

1. The current through $R_2$ is greater than the current through $R_1$.
2. The current through $R_2$ equals the current through $R_1$.
3. The current through $R_2$ is less than the current through $R_1$. (correct answer)
4. The current through $R_2$ is zero because current splits and is consumed.

Explanation: This problem examines current relationships in compound DC circuits. In this configuration, R₁ carries the total circuit current before it splits at junction J into two parallel branches containing R₂ and R₃. By Kirchhoff's current law, the current through R₁ equals the sum of currents through R₂ and R₃, making the current through R₁ greater than the current through either parallel branch alone. Since current divides between parallel branches, the current through R₂ must be less than the current through R₁. Choice D incorrectly suggests current is "consumed" when it splits, misunderstanding that current is conserved at junctions. When analyzing compound circuits, trace the current path and remember that series elements carry the same current while parallel branches divide the total current.

Question 17

1. The current in Branch 1 is zero because the two resistors use up the current.
2. The current in Branch 1 equals the current in Branch 2. (correct answer)
3. The current in Branch 1 is greater than the current in Branch 2.
4. The current in Branch 1 is less than the current in Branch 2.

Explanation: This problem tests understanding of compound DC circuits. The parallel section between J and K has Branch 1 with total resistance R₁ + R₂ = 2Ω + 2Ω = 4Ω, and Branch 2 with R₃ = 4Ω. Since both branches have equal total resistance (4Ω) and experience the same voltage (10V across J-K), they carry equal currents of 10V/4Ω = 2.5A each. The fact that Branch 1 contains two resistors while Branch 2 has one is irrelevant; only the total branch resistance matters for current division. Choice D reflects the misconception that resistors "use up" current, violating conservation principles. Strategy: always compare total branch resistances when analyzing parallel current division.

Question 18

1. The current through $R_1$ equals the current through $R_2$.
2. The current through $R_1$ is zero because the current splits at J.
3. The current through $R_1$ is less than the current through $R_2$.
4. The current through $R_1$ is greater than the current through $R_2$. (correct answer)

Explanation: This problem tests understanding of compound DC circuits. In this configuration, R₁ is in series with the entire parallel combination of R₂ and R₃, meaning all current through R₁ must split between the two parallel branches at junction J. By Kirchhoff's current law, the current through R₁ equals the sum of currents through R₂ and R₃, making I₁ greater than I₂ alone. Since R₂ = 2Ω is less than R₃ = 4Ω, more current flows through R₂ than R₃, but both branch currents are less than the total current through R₁. Choice D incorrectly assumes current becomes zero when splitting, violating current conservation. Remember: series elements before a junction carry the total current that later divides among parallel branches.

Question 19

1. The current through $R_2$ equals the current through $R_3$.
2. The current through $R_2$ is zero because current is used up in $R_1$.
3. The current through $R_2$ is greater than the current through $R_3$. (correct answer)
4. The current through $R_2$ is less than the current through $R_3$.

Explanation: This problem tests understanding of compound DC circuits. In this circuit, R₁ is in series with a parallel combination of R₂ and R₃, where R₂ = 6Ω and R₃ = 12Ω are connected between junctions J and K. In a parallel connection, both branches experience the same voltage drop, but current divides inversely with resistance according to Ohm's law (I = V/R). Since R₂ has lower resistance than R₃, more current flows through R₂ than through R₃. Choice D incorrectly assumes current is "used up" in R₁, which reflects the misconception that current is consumed as it passes through circuit elements. Remember: identify series and parallel regions before reasoning about current or voltage distribution.

Question 20

1. The charge per second entering J equals the total charge per second leaving J. (correct answer)
2. The charge per second entering J is greater because resistors consume charge.
3. The charge per second entering J is less because current splits at J.
4. The charge per second leaving J is zero because the branches are parallel.

Explanation: This problem tests understanding of compound DC circuits. At any junction in a circuit, charge conservation requires that charge per second entering equals charge per second leaving—this is Kirchhoff's current law. Current (charge per second) entering junction J through R₁ must equal the sum of currents leaving through R₂ and R₃, as charge cannot accumulate or disappear at a junction. Choice B incorrectly claims resistors "consume" charge, confusing energy dissipation with charge flow—resistors transform electrical energy to heat but don't destroy charge carriers. When analyzing junctions, always apply conservation of charge: the sum of currents entering any junction equals the sum of currents leaving.