AP Physics 2 Quiz: Kirchhoffs Junction Rule

20 questions · exam conditions

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Kirchhoffs Junction RuleQuestion 1 of 20

At junction $J$ , currents $I_A=0.40\,\text{A}$ and $I_B=0.70\,\text{A}$ enter from the left and from below. Currents $I_C$ leaves upward and $I_D=0.50\,\text{A}$ leaves to the right (directions as labeled). Which current relationship must be true?

I_C=I_A+I_B-I_D

I_C=I_D-I_A-I_B

I_C=I_A+I_B+I_D

I_C=I_A=I_B=I_D

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AP Physics 2 Quiz

AP Physics 2 Quiz: Kirchhoffs Junction Rule

Practice Kirchhoffs Junction Rule in AP Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Kirchhoffs Junction Rule, 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.

All questions

Question 1

$I_C=I_A+I_B-I_D$ (correct answer)
$I_C=I_D-I_A-I_B$
$I_C=I_A+I_B+I_D$
$I_C=I_A=I_B=I_D$

Explanation: This question applies Kirchhoff's junction rule. The junction rule embodies charge conservation: at any junction, the total current flowing in must equal the total current flowing out, as charge cannot accumulate at a point. Here, currents I_A = 0.40 A and I_B = 0.70 A enter junction J, providing total entering current of 1.10 A. Currents I_C and I_D = 0.50 A leave the junction, so their sum must equal 1.10 A. Choice C incorrectly adds all currents regardless of direction, demonstrating a fundamental misunderstanding of how current conservation works at junctions. Applying the rule: I_A + I_B = I_C + I_D, which gives I_C = I_A + I_B - I_D. Remember: always separate entering from leaving currents before applying the junction rule.

Question 2

At junction $J$ in a DC circuit, current $I_1=2.0\,\text{A}$ enters $J$ from the left and $I_2=1.5\,\text{A}$ enters from above. Current $I_3$ leaves $J$ to the right, and current $I_4=0.5\,\text{A}$ leaves downward. Directions are as labeled. Which current relationship must be true?

$I_3=I_1+I_2-I_4$ (correct answer)
$I_3=I_1-I_2-I_4$
$I_3=I_1=I_2=I_4$
$I_3=I_1+I_2+I_4$

Explanation: This problem tests understanding of Kirchhoff's junction rule. At any junction in a circuit, the total current entering must equal the total current leaving, which reflects conservation of charge—charge cannot accumulate at a junction. Here, currents I₁ = 2.0 A and I₂ = 1.5 A enter junction J, giving a total entering current of 3.5 A. Currents I₃ and I₄ = 0.5 A leave the junction, so the total leaving current must also be 3.5 A. Choice C incorrectly assumes all currents are equal, which represents a misconception that current is the same everywhere in a circuit. Setting entering currents equal to leaving currents: I₁ + I₂ = I₃ + I₄, which gives I₃ = I₁ + I₂ - I₄. Remember: at any junction, sum of currents in equals sum of currents out.

Question 3

At junction $J$ , current $I_1=3.0\,\text{A}$ enters from the left. Currents $I_2=1.0\,\text{A}$ and $I_3$ leave upward and to the right, respectively, and current $I_4=0.5\,\text{A}$ enters from below (directions as labeled). Which statement correctly applies Kirchhoff's junction rule?

$I_3=I_1+I_4-I_2$ (correct answer)
$I_3=I_1-I_4-I_2$
$I_3=I_1+I_2+I_4$
$I_3=I_1=I_2=I_4$

Explanation: This problem involves Kirchhoff's junction rule. At a junction, charge conservation requires that the rate of charge entering (total current in) equals the rate of charge leaving (total current out), preventing charge accumulation. In this scenario, currents I₁ = 3.0 A and I₄ = 0.5 A enter junction J, giving total entering current of 3.5 A. Currents I₂ = 1.0 A and I₃ leave the junction, so their sum must also equal 3.5 A. Choice B incorrectly subtracts entering currents from leaving currents, revealing confusion about the direction-dependent nature of the junction rule. Setting entering equal to leaving: I₁ + I₄ = I₂ + I₃, which rearranges to I₃ = I₁ + I₄ - I₂. Key strategy: carefully track current directions and balance entering versus leaving currents.

Question 4

At junction $J$ , currents $I_1=1.2\,\text{A}$ and $I_2=0.6\,\text{A}$ leave $J$ to the left and upward (directions as labeled). Current $I_3=0.9\,\text{A}$ enters from the right, and current $I_4$ enters from below. Which current relationship must be true?

$I_4=I_1+I_2-I_3$ (correct answer)
$I_4=I_3-I_1-I_2$
$I_4=I_1+I_2+I_3$
$I_4=I_1=I_2=I_3$

Explanation: This question tests Kirchhoff's junction rule. The junction rule ensures charge conservation by requiring that the total current entering any junction equals the total current leaving it—charge cannot build up or disappear at a junction. Here, currents I₃ = 0.9 A and I₄ enter junction J, while currents I₁ = 1.2 A and I₂ = 0.6 A leave, giving a total leaving current of 1.8 A. Therefore, the total entering current must also be 1.8 A. Choice C incorrectly adds all currents together, showing a misconception that fails to distinguish between entering and leaving currents. Applying conservation: I₃ + I₄ = I₁ + I₂, which gives I₄ = I₁ + I₂ - I₃. Strategy: list entering currents on one side and leaving currents on the other, then solve.

Question 5

At junction $J$ , currents $I_A=1.8\,\text{A}$ and $I_B=0.4\,\text{A}$ enter from the left and from above (directions as labeled). Currents $I_C=0.9\,\text{A}$ and $I_D$ leave downward and to the right. Which current relationship must be true?

$I_D=I_A+I_B-I_C$ (correct answer)
$I_D=I_A-I_B-I_C$
$I_D=I_A+I_B+I_C$
$I_D=I_A=I_B=I_C$

Explanation: This question tests Kirchhoff's junction rule. The junction rule ensures charge conservation at circuit junctions: the total current entering must equal the total current leaving, preventing charge accumulation. Here, currents I_A = 1.8 A and I_B = 0.4 A enter junction J, giving total entering current of 2.2 A. Currents I_C = 0.9 A and I_D leave the junction, so their sum must also equal 2.2 A. Choice C incorrectly adds all currents without considering direction, a common error when students don't recognize that entering and leaving currents play opposite roles. Applying the rule: I_A + I_B = I_C + I_D, which gives I_D = I_A + I_B - I_C. Strategy: list all entering currents, all leaving currents, then set their sums equal.

Question 6

A junction $J$ has three labeled currents. $I_1=0.30\,\text{A}$ enters from below, $I_2=0.10\,\text{A}$ enters from the left, and $I_3$ leaves upward, as labeled. Which current relationship must be true at $J$ ?

$I_1=I_2=I_3$
$I_3=I_1+I_2$ (correct answer)
$I_3=I_2-I_1$
$I_3=I_1-I_2$

Explanation: This question applies Kirchhoff's junction rule to three currents. Kirchhoff's junction rule states that charge is conserved at junctions—current flowing in must equal current flowing out. At junction J, currents I₁ = 0.30 A and I₂ = 0.10 A enter, while I₃ leaves upward. Therefore: I₁ + I₂ = I₃, giving 0.30 + 0.10 = I₃, so I₃ = 0.40 A. Choice B (I₃ = I₁ - I₂) incorrectly subtracts the two entering currents, misunderstanding that both contribute to the outgoing current. Remember: at any junction, add all entering currents and set equal to the sum of leaving currents.

Question 7

In the circuit shown, three currents meet at junction $J$ . Current $I_1=2.0\,\text{A}$ enters $J$ from the left, and current $I_2=1.5\,\text{A}$ enters $J$ from above. Current $I_3$ leaves $J$ to the right, and current $I_4=0.5\,\text{A}$ leaves $J$ downward. Directions are as labeled. Which current relationship must be true at junction $J$ ?

$I_3=I_1=I_2=I_4$
$I_1+I_2=I_4$
$I_3=I_1+I_2+I_4$
$I_3=I_1+I_2-I_4$ (correct answer)

Explanation: This problem requires applying Kirchhoff's junction rule. Kirchhoff's junction rule states that the total current entering a junction must equal the total current leaving the junction, which reflects conservation of charge. At junction J, currents I₁ = 2.0 A and I₂ = 1.5 A enter, while currents I₃ and I₄ = 0.5 A leave. Setting current in equal to current out: I₁ + I₂ = I₃ + I₄, which gives 2.0 + 1.5 = I₃ + 0.5, so I₃ = 3.0 A. Choice C incorrectly assumes all currents are equal, which violates the physical constraint that charge cannot accumulate at a junction. Remember: at any junction, sum of currents in equals sum of currents out.

Question 8

At junction $J$ , current $I_1=0.60\,\text{A}$ enters from the left and current $I_2=0.10\,\text{A}$ leaves upward (directions as labeled). Currents $I_3$ and $I_4=0.20\,\text{A}$ leave to the right and downward. Which statement correctly applies Kirchhoff's junction rule?

$I_3=I_1-I_2-I_4$ (correct answer)
$I_3=I_1+I_2+I_4$
$I_3=I_2+I_4-I_1$
$I_3=I_1=I_2=I_4$

Explanation: This problem applies Kirchhoff's junction rule. The junction rule reflects charge conservation: at any junction, the sum of entering currents must equal the sum of leaving currents, as charge cannot accumulate at a point. In this scenario, only I₁ = 0.60 A enters junction J, while currents I₂ = 0.10 A, I₃, and I₄ = 0.20 A all leave the junction. The total leaving current must equal 0.60 A. Choice B incorrectly adds all currents together, showing confusion about the directional nature of current flow at junctions. Setting entering equal to leaving: I₁ = I₂ + I₃ + I₄, which rearranges to I₃ = I₁ - I₂ - I₄. Remember: always separate currents by direction (in vs. out) before applying the junction rule.

Question 9

At junction $J$ , current $I_1=2.0\,\text{A}$ enters from the left and $I_2=1.5\,\text{A}$ enters from above. Current $I_3$ leaves to the right and $I_4=0.5\,\text{A}$ leaves downward, as labeled by the arrows. Which current relationship must be true?

$I_3=I_1+I_2-I_4$ (correct answer)
$I_3=I_1-I_2-I_4$
$I_3=I_1+I_2+I_4$
$I_3=I_1=I_2=I_4$

Explanation: This problem tests Kirchhoff's junction rule. At any junction, the total current entering must equal the total current leaving, which expresses conservation of charge—charge cannot accumulate at a junction. Here, currents $I_1 = 2.0$ A and $I_2 = 1.5$ A enter the junction, while $I_3$ and $I_4 = 0.5$ A leave. Setting current in equal to current out: $I_1 + I_2 = I_3 + I_4$ , which gives $2.0 + 1.5 = I_3 + 0.5$ , so $I_3 = 3.0$ A. Choice D incorrectly assumes all currents are equal, ignoring that different branches can carry different currents. Remember: at any junction, sum of currents in equals sum of currents out.

Question 10

At junction $J$ , $I_1=3.0\,\text{A}$ enters from the left, while $I_2$ leaves upward and $I_3=1.2\,\text{A}$ leaves to the right. A fourth current $I_4=0.8\,\text{A}$ enters from below (directions shown by arrows). Which statement correctly applies Kirchhoff's junction rule?

$I_2=I_1-I_4-I_3$
$I_2=I_1+I_4-I_3$ (correct answer)
$I_2=I_1+I_4+I_3$
$I_2=I_1=I_3=I_4$

Explanation: This problem applies Kirchhoff's junction rule. The junction rule states that the sum of currents entering a junction equals the sum of currents leaving, ensuring charge conservation. At junction J, currents $I_1 = 3.0$ A and $I_4 = 0.8$ A enter, while $I_2$ and $I_3 = 1.2$ A leave. Applying the rule: $I_1 + I_4 = I_2 + I_3$ , which gives $3.0 + 0.8 = I_2 + 1.2$ , so $I_2 = 2.6$ A. Choice C incorrectly adds all currents together, failing to distinguish between entering and leaving currents. Always identify current directions first, then apply: currents in = currents out.

Question 11

At junction $J$ , $I_1=5.0\,\text{A}$ enters from the left, and currents $I_2=2.0\,\text{A}$ , $I_3=1.0\,\text{A}$ , and $I_4$ all leave along the top, right, and bottom branches (directions shown). Which current relationship must be true?

$I_4=I_1=I_2=I_3$
$I_4=I_1+I_2-I_3$
$I_4=I_1-I_2-I_3$ (correct answer)
$I_4=I_2+I_3-I_1$

Explanation: This problem involves Kirchhoff's junction rule. The junction rule expresses conservation of charge: the sum of currents entering a junction must equal the sum leaving. Here, only $I_1 = 5.0$ A enters junction J, while $I_2 = 2.0$ A, $I_3 = 1.0$ A, and $I_4$ all leave. Setting current in equal to currents out: $I_1 = I_2 + I_3 + I_4$ , which gives $5.0 = 2.0 + 1.0 + I_4$ , so $I_4 = 2.0$ A. Choice C incorrectly rearranges the equation as if $I_1$ were leaving rather than entering. Strategy: always verify current directions from arrows before applying the junction rule.

Question 12

At junction $J$ , $I_1$ enters from the left and splits into two currents leaving: $I_2=0.90\,\text{A}$ upward and $I_3=0.30\,\text{A}$ to the right. A fourth current $I_4=0.20\,\text{A}$ enters from below (arrow directions shown). Which current relationship must be true?

$I_1=I_2-I_3-I_4$
$I_1=I_2+I_3+I_4$
$I_1=I_2=I_3=I_4$
$I_1=I_2+I_3-I_4$ (correct answer)

Explanation: This question applies Kirchhoff's junction rule. The junction rule expresses charge conservation: the total current flowing into a junction must equal the total current flowing out. Here, $I_1$ and $I_4 = 0.20$ A enter the junction, while $I_2 = 0.90$ A and $I_3 = 0.30$ A leave. Setting currents in equal to currents out: $I_1 + I_4 = I_2 + I_3$ , giving $I_1 + 0.20 = 0.90 + 0.30$ , so $I_1 = 1.0$ A. Choice B incorrectly subtracts all leaving currents from $I_1$ , misunderstanding that $I_4$ enters the junction. Key strategy: identify arrow directions first, then apply current conservation.

Question 13

At junction $J$ , three currents enter: $I_1=1.0\,\text{A}$ from the left, $I_2=0.40\,\text{A}$ from above, and $I_3=0.60\,\text{A}$ from below. A single current $I_4$ leaves to the right (directions shown). Which current relationship must be true?

$I_4=I_1-I_2-I_3$
$I_4=I_1+I_2+I_3$ (correct answer)
$I_4=I_1=I_2=I_3$
$I_4=I_1+I_2-I_3$

Explanation: This problem tests Kirchhoff's junction rule. The junction rule ensures that charge is conserved at every junction—no charge can accumulate or disappear. Three currents enter junction J: $I_1 = 1.0$ A, $I_2 = 0.40$ A, and $I_3 = 0.60$ A, while only $I_4$ leaves. Applying the conservation principle: sum of entering currents equals sum of leaving currents, so $I_1 + I_2 + I_3 = I_4$ , giving $1.0 + 0.40 + 0.60 = I_4 = 2.0$ A. Choice B incorrectly subtracts entering currents, confusing the sign convention. Remember: at any junction, total current in must equal total current out.

Question 14

At junction $J$ , $I_1=4.0\,\text{A}$ enters from the left. Currents $I_2=1.0\,\text{A}$ and $I_3$ leave upward and to the right, and $I_4=0.5\,\text{A}$ enters from below (directions indicated). Which current relationship must be true?

$I_3=I_1-I_4-I_2$
$I_3=I_2=I_1$
$I_3=I_1-I_4+I_2$
$I_3=I_1+I_4-I_2$ (correct answer)

Explanation: This problem involves Kirchhoff's junction rule. The junction rule states that charge cannot accumulate at a junction, so the sum of entering currents must equal the sum of leaving currents. At junction J, currents $I_1 = 4.0$ A and $I_4 = 0.5$ A enter, while $I_2 = 1.0$ A and $I_3$ leave. Applying conservation: $I_1 + I_4 = I_2 + I_3$ , which gives $4.0 + 0.5 = 1.0 + I_3$ , so $I_3 = 3.5$ A. Choice B incorrectly treats $I_4$ as leaving rather than entering, a common sign error. Always verify current directions from the problem statement before applying: sum in = sum out.

Question 15

At junction $J$ , $I_1=0.70\,\text{A}$ leaves to the right and $I_2=0.20\,\text{A}$ leaves upward. Currents $I_3$ enters from the left and $I_4=0.10\,\text{A}$ enters from below (arrows shown). Which current relationship must be true?

$I_3=I_1+I_2-I_4$ (correct answer)
$I_3=I_1-I_2-I_4$
$I_3=I_4-I_1-I_2$
$I_3=I_1=I_2=I_4$

Explanation: This question tests Kirchhoff's junction rule. The junction rule states that charge cannot accumulate at a junction, so total current in must equal total current out. At junction J, $I_3$ and $I_4 = 0.10$ A enter, while $I_1 = 0.70$ A and $I_2 = 0.20$ A leave. Applying conservation: $I_3 + I_4 = I_1 + I_2$ , which gives $I_3 + 0.10 = 0.70 + 0.20$ , so $I_3 = 0.80$ A. Choice C incorrectly subtracts entering currents from leaving current $I_4$ , reversing the proper relationship. Remember: identify current directions carefully, then apply the rule that currents in equal currents out.

Question 16

At junction $J$ , $I_1=2.5\,\text{A}$ enters from the left and $I_2=0.5\,\text{A}$ leaves upward. Currents $I_3$ and $I_4=1.0\,\text{A}$ both leave to the right and downward (arrows shown). Which current relationship must be true?

$I_3=I_1-I_2+I_4$
$I_3=I_1+I_2-I_4$
$I_3=I_1-I_2-I_4$ (correct answer)
$I_3=I_2+I_4-I_1$

Explanation: This question involves Kirchhoff's junction rule. The junction rule states that the algebraic sum of currents at any junction must be zero, reflecting charge conservation. At junction J, $I_1 = 2.5$ A enters, while $I_2 = 0.5$ A, $I_3$ , and $I_4 = 1.0$ A all leave. Setting entering current equal to leaving currents: $I_1 = I_2 + I_3 + I_4$ , which gives $2.5 = 0.5 + I_3 + 1.0$ , so $I_3 = 1.0$ A. Choice C incorrectly adds $I_4$ to $I_1$ instead of recognizing both $I_3$ and $I_4$ leave the junction. Strategy: group currents by direction, then apply conservation.

Question 17

At junction $J$ , currents $I_1=0.80\ \text{A}$ and $I_2=1.20\ \text{A}$ enter. Current $I_3$ leaves as labeled. Which statement correctly applies Kirchhoff's junction rule?

$I_3=I_1-I_2$
$I_3=I_1+I_2$ (correct answer)
$I_3=\dfrac{I_1+I_2}{2}$
$I_3=I_1=I_2$

Explanation: This problem tests understanding of Kirchhoff's junction rule. The junction rule states that the algebraic sum of currents at any junction must be zero, reflecting conservation of charge. In this case, currents I₁ = 0.80 A and I₂ = 1.20 A both enter the junction, while current I₃ leaves. Applying conservation: total current in = total current out, so I₁ + I₂ = I₃, which gives I₃ = 0.80 + 1.20 = 2.00 A. Choice C incorrectly averages the entering currents, suggesting a misconception that currents split equally rather than conserving total charge. Always identify which currents enter and which leave, then apply: sum in = sum out.

Question 18

At junction $J$ , $I_1=1.2\,\text{A}$ enters, $I_2=0.4\,\text{A}$ leaves, $I_3=0.5\,\text{A}$ leaves, and $I_4$ leaves. Which current relationship must be true?

$I_4=2.1\,\text{A}$
$I_4=0.3\,\text{A}$ (correct answer)
$I_4=0.7\,\text{A}$
$I_4=1.2\,\text{A}$

Explanation: This question applies Kirchhoff's junction rule. Kirchhoff's junction rule states that the algebraic sum of currents at any junction must be zero, reflecting the conservation of electric charge. Here, $I_1 = 1.2$ A enters, while $I_2 = 0.4$ A, $I_3 = 0.5$ A, and $I_4$ all leave the junction. Setting up the conservation equation: current in = current out gives us $1.2 = 0.4 + 0.5 + I_4$ , which simplifies to $1.2 = 0.9 + I_4$ , yielding $I_4 = 0.3$ A. Choice B ( $I_4 = 2.1$ A) likely results from adding all currents instead of balancing inflow and outflow. Remember: at any junction, the sum of entering currents must equal the sum of leaving currents.

Question 19

At junction $J$ , currents $I_1=0.30\ \text{A}$ and $I_2=0.50\ \text{A}$ enter, and currents $I_3=0.40\ \text{A}$ and $I_4$ leave as labeled. Which statement correctly applies Kirchhoff's junction rule?

$I_4=\dfrac{I_1+I_2+I_3}{3}$
$I_4=I_1+I_2+I_3$
$I_4=I_1+I_2-I_3$ (correct answer)
$I_4=I_3-(I_1+I_2)$

Explanation: This problem requires careful application of Kirchhoff's junction rule. The junction rule states that charge is conserved at every junction: total current entering equals total current leaving. Two currents enter (I₁ = 0.30 A and I₂ = 0.50 A) while two leave (I₃ = 0.40 A and I₄). By conservation: I₁ + I₂ = I₃ + I₄, which gives 0.30 + 0.50 = 0.40 + I₄, so I₄ = 0.40 A or I₄ = I₁ + I₂ - I₃. Choice B incorrectly adds all currents together, ignoring their directions relative to the junction. Remember to separate entering and leaving currents, then apply: total in = total out.

Question 20

At junction $J$ , current $I_1=4.0\ \text{A}$ leaves, while currents $I_2=1.5\ \text{A}$ and $I_3$ enter as labeled. Which current relationship must be true?

$I_3=I_1+I_2$
$I_3=I_1=I_2$
$I_3=I_1-I_2$ (correct answer)
$I_3=I_2-I_1$

Explanation: This question applies Kirchhoff's junction rule to find an unknown entering current. The junction rule ensures that electric charge is conserved: no charge builds up at any junction in steady state. Current I₁ = 4.0 A leaves the junction, while currents I₂ = 1.5 A and I₃ enter. Applying conservation: total in = total out, so I₂ + I₃ = I₁, which gives 1.5 + I₃ = 4.0, therefore I₃ = 2.5 A or I₃ = I₁ - I₂. Choice B would give I₃ = 5.5 A, incorrectly treating I₁ as entering rather than leaving the junction. Always verify current directions before applying: sum entering = sum leaving.