College Physics Quiz: Kirchhoffs Junction Rule
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Kirchhoffs Junction RuleQuestion 1 of 20

In a three-junction network, junction C has I1=1 AI_1=1\,\text{A} in, I2=2 AI_2=2\,\text{A} in, and I3=5 AI_3=5\,\text{A} out. Find I4I_4 in.

I4=2 AI_4=2\,\text{A}
I4=8 AI_4=8\,\text{A}
I4=0 AI_4=0\,\text{A}
I4=4 AI_4=4\,\text{A}
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College Physics Quiz

College Physics Quiz: Kirchhoffs Junction Rule

Practice Kirchhoffs Junction Rule in College Physics 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 College Physics.

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

In a three-junction network, junction C has I1=1 AI_1=1\,\text{A} in, I2=2 AI_2=2\,\text{A} in, and I3=5 AI_3=5\,\text{A} out. Find I4I_4 in.

  1. I4=2 AI_4=2\,\text{A} (correct answer)
  2. I4=8 AI_4=8\,\text{A}
  3. I4=0 AI_4=0\,\text{A}
  4. I4=4 AI_4=4\,\text{A}
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction C, 1 A, 2 A, and I₄ enter while 5 A leaves, so 1 + 2 + I₄ = 5, giving I₄ = 2 A. The correct answer reflects this rule by accurately balancing the currents, indicating comprehension of current distribution. A common distractor assumes zero, reflecting a misunderstanding of current conservation. To help students, emphasize multiple ins. Reinforce practice in network junctions.

Question 2

Calculate missing current at junction C: I1=4 AI_1=4\,\text{A} out, I2=6 AI_2=6\,\text{A} out, I3=3 AI_3=3\,\text{A} in, I4I_4 in.

  1. I4=13 AI_4=13\,\text{A}
  2. I4=7 AI_4=7\,\text{A} (correct answer)
  3. I4=1 AI_4=1\,\text{A}
  4. I4=5 AI_4=5\,\text{A}
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction C, 3 A and I₄ enter while 4 A and 6 A leave, so 3 + I₄ = 4 + 6, giving I₄ = 7 A. The correct answer reflects this rule by accurately balancing the currents, indicating comprehension of current distribution. A common distractor subtracts wrongly, reflecting a misunderstanding of current conservation. To help students, emphasize isolating variables. Reinforce practice with multiple outputs.

Question 3

At junction D, Ia=3 AI_a=3\,\text{A} out, Ib=2 AI_b=2\,\text{A} out, and Ic=1 AI_c=1\,\text{A} in. Find total current into D.

  1. 4 A4\,\text{A}
  2. 5 A5\,\text{A} (correct answer)
  3. 6 A6\,\text{A}
  4. 2 A2\,\text{A}
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction D, total in equals 3 A + 2 A out, but with 1 A in given, total in = 5 A to balance. The correct answer reflects this rule by finding total entering current, indicating comprehension of current distribution. A common distractor miscounts outputs, reflecting a misunderstanding of current conservation. To help students, emphasize equating totals. Reinforce practice in partial information problems.

Question 4

A faulty branch is open: junction B has 5 A5\,\text{A} in and only 2 A2\,\text{A} out measured. Missing current equals?

  1. 3 A3\,\text{A} (correct answer)
  2. 7 A7\,\text{A}
  3. 2.5 A2.5\,\text{A}
  4. 10 A10\,\text{A}
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction B, 5 A enters while 2 A leaves, so missing current leaving is 5 - 2 = 3 A. The correct answer reflects this rule by finding the imbalance, indicating comprehension of current distribution. A common distractor ignores the fault, reflecting a misunderstanding of current conservation. To help students, emphasize applying the rule to diagnostics. Reinforce practice in troubleshooting scenarios.

Question 5

In a three-junction circuit, junction B has 6 A6\,\text{A} in; IxI_x out; 4 A4\,\text{A} out. Find IxI_x.

  1. Ix=10 AI_x=10\,\text{A}
  2. Ix=2 AI_x=2\,\text{A} (correct answer)
  3. Ix=1.5 AI_x=1.5\,\text{A}
  4. Ix=6 AI_x=6\,\text{A}
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction B, 6 A enters while I_x and 4 A leave, so 6 = I_x + 4, giving I_x = 2 A. The correct answer reflects this rule by accurately balancing the currents, indicating comprehension of current distribution. A common distractor is adding extras, reflecting a misunderstanding of current conservation. To help students, emphasize basic subtraction. Reinforce practice in three-junction setups.

Question 6

At junction A, 9 A9\,\text{A} enters and splits into IxI_x and 3 A3\,\text{A} leaving. Using Kirchhoff, find IxI_x.

  1. Ix=12 AI_x=12\,\text{A}
  2. Ix=6 AI_x=6\,\text{A} (correct answer)
  3. Ix=3 AI_x=3\,\text{A}
  4. Ix=9 AI_x=9\,\text{A}
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction A, 9 A enters and splits into Iₓ and 3 A leaving, so 9 = Iₓ + 3, resulting in Iₓ = 6 A. The correct answer reflects this rule by accurately balancing the currents, indicating comprehension of current distribution. A common distractor is halving the current, reflecting a misunderstanding of current conservation. To help students, emphasize simple arithmetic and direction arrows. Reinforce practice in splitting currents at junctions.

Question 7

Calculate missing current IqI_q at junction D: 2 A2\,\text{A} in, IqI_q out, 1 A1\,\text{A} out, 5 A5\,\text{A} in.

  1. Iq=8 AI_q=8\,\text{A}
  2. Iq=6 AI_q=6\,\text{A} (correct answer)
  3. Iq=2 AI_q=2\,\text{A}
  4. Iq=4 AI_q=4\,\text{A}
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction D, 2 A and 5 A enter while I_q and 1 A leave, so 2 + 5 = I_q + 1, giving I_q = 6 A. The correct answer reflects this rule by accurately balancing the currents, indicating comprehension of current distribution. A common distractor miscounts ins, reflecting a misunderstanding of current conservation. To help students, emphasize total sums. Reinforce practice with mixed values.

Question 8

At junction D, Iin=3 AI_{in}=3\,\text{A} and Iout=1 A+IxI_{out}=1\,\text{A}+I_x. Using junction rule, find IxI_x.

  1. Ix=1 AI_x=1\,\text{A}
  2. Ix=2 AI_x=2\,\text{A} (correct answer)
  3. Ix=3 AI_x=3\,\text{A}
  4. Ix=4 AI_x=4\,\text{A}
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction D, 3 A enters while 1 A and I_x leave, so 3 = 1 + I_x, giving I_x = 2 A. The correct answer reflects this rule by accurately balancing the currents, indicating comprehension of current distribution. A common distractor adds extras, reflecting a misunderstanding of current conservation. To help students, emphasize simple equations. Reinforce practice in basic junctions.

Question 9

At junction C, 8 A8\,\text{A} leaves; 3 A3\,\text{A} enters; IkI_k enters. Using junction rule, find IkI_k.

  1. Ik=5 AI_k=5\,\text{A} (correct answer)
  2. Ik=11 AI_k=11\,\text{A}
  3. Ik=2.7 AI_k=2.7\,\text{A}
  4. Ik=8 AI_k=8\,\text{A}
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction C, 3 A and I_k enter while 8 A leaves, so 3 + I_k = 8, giving I_k = 5 A. The correct answer reflects this rule by accurately balancing the currents, indicating comprehension of current distribution. A common distractor is subtracting outputs wrongly, reflecting a misunderstanding of current conservation. To help students, emphasize isolating the unknown. Reinforce practice with single output junctions.

Question 10

Calculate the missing current at junction C: I4=7 AI_4=7\,\text{A} in, I5=1 AI_5=1\,\text{A} in, I6=5 AI_6=5\,\text{A} out, I7I_7 out.

  1. I7=11 AI_7=11\,\text{A}
  2. I7=1 AI_7=1\,\text{A}
  3. I7=3 AI_7=3\,\text{A} (correct answer)
  4. I7=13 AI_7=13\,\text{A}
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction C, 7 A and 1 A enter while 5 A and I₇ leave, so 7 + 1 = 5 + I₇, giving I₇ = 3 A. The correct answer reflects this rule by accurately balancing the currents, indicating comprehension of current distribution. A common distractor involves ignoring one input, reflecting a misunderstanding of current conservation. To help students, emphasize labeling all currents and verifying totals. Reinforce practice in multi-input junctions to build confidence.

Question 11

How does current entering junction A compare to leaving if Iin=9 AI_{in}=9\,\text{A} and outgoing branches are 4 A4\,\text{A} and 5 A5\,\text{A}?

  1. Entering equals leaving: 9 A=4 A+5 A9\,\text{A}=4\,\text{A}+5\,\text{A} (correct answer)
  2. Leaving exceeds entering by 1 A1\,\text{A}
  3. Entering exceeds leaving by 1 A1\,\text{A}
  4. Equality holds only if a battery is present
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction A, 9 A enters while 4 A and 5 A leave, so 9 = 4 + 5, showing entering equals leaving. The correct answer reflects this rule by confirming balance, indicating comprehension of current distribution. A common distractor assumes excess, reflecting a misunderstanding of current conservation. To help students, emphasize verification without extras. Reinforce practice in equality checks.

Question 12

At junction P in a circuit, four wires meet. The currents in three of the wires are: 3.2 A entering the junction, 1.8 A leaving the junction, and 2.1 A entering the junction. If a fourth wire carries current I₄, which statement correctly describes I₄?

  1. I₄ = 3.5 A leaving the junction (correct answer)
  2. I₄ = 3.5 A entering the junction
  3. I₄ = 7.1 A leaving the junction
  4. I₄ = 0.7 A entering the junction
  5. I₄ = 0.7 A leaving the junction
Explanation: When you encounter a junction problem in circuit analysis, you're dealing with Kirchhoff's Current Law (KCL), which states that the total current entering a junction must equal the total current leaving it. This reflects the conservation of electric charge—current can't accumulate or disappear at a junction point. To solve this systematically, first identify what's entering versus leaving the junction. You have 3.2 A and 2.1 A entering, for a total of 5.3 A flowing into junction P. Meanwhile, 1.8 A is leaving the junction. Since current in must equal current out, you need: Current in=Current out\text{Current in} = \text{Current out} 5.3 A=1.8 A+I45.3 \text{ A} = 1.8 \text{ A} + I_4 Solving for I4I_4: I4=5.3−1.8=3.5 A leavingI_4 = 5.3 - 1.8 = 3.5 \text{ A leaving} This confirms answer A is correct—3.5 A must leave the junction to balance the equation. Answer B suggests 3.5 A entering, which would create a total of 8.8 A entering but only 1.8 A leaving, violating KCL. Answer C proposes 7.1 A leaving, which would mean 8.9 A total leaving versus 5.3 A entering—again impossible. Answer D suggests 0.7 A entering, giving 6.0 A total entering but only 1.8 A leaving, another KCL violation. Remember: at any junction, treat it like a balance scale. Whatever current flows in must flow out. Set up your equation accordingly, being careful about the direction of each current, and KCL problems become straightforward algebra.

Question 13

Two junctions A and B are connected by three parallel branches in a circuit. At junction A, the total current entering from external sources is 12.0 A, and the current leaving through one external wire is 5.0 A. If the three parallel branches carry currents of 3.0 A, 2.5 A, and I₃ respectively (all flowing from A to B), what is I₃?

  1. I₃ = 1.5 A from A to B (correct answer)
  2. I₃ = 1.5 A from B to A
  3. I₃ = 6.5 A from A to B
  4. I₃ = 17.5 A from A to B
  5. I₃ = 10.5 A from A to B
Explanation: When you encounter circuit problems involving junctions, you're dealing with Kirchhoff's Current Law (KCL), which states that the total current entering a junction must equal the total current leaving it. Let's apply KCL to junction A. Current entering: 12.0 A from external sources. Current leaving: 5.0 A through one external wire, plus the currents through the three parallel branches (3.0 A, 2.5 A, and I3I_3). Setting up the equation: 12.0=5.0+3.0+2.5+I312.0 = 5.0 + 3.0 + 2.5 + I_3 Solving: 12.0=10.5+I312.0 = 10.5 + I_3, so I3=1.5 AI_3 = 1.5 \text{ A} Since we defined all branch currents as flowing from A to B, and our calculation gives a positive value, I3=1.5I_3 = 1.5 A flows from A to B. Choice A is correct: I3=1.5I_3 = 1.5 A from A to B. Choice B gives the right magnitude but wrong direction—this would violate KCL since we'd have insufficient current leaving junction A. Choice C (6.5 A) likely comes from adding instead of subtracting: 5.0+3.0+2.5=10.55.0 + 3.0 + 2.5 = 10.5, but this ignores the constraint that total outflow must equal inflow. Choice D (17.5 A) appears to sum all given currents, completely misunderstanding KCL. Remember: In junction problems, always write KCL equations carefully, defining current directions consistently. The algebraic sign of your answer tells you the actual direction—positive means your assumed direction was correct, negative means the current flows opposite to your assumption.

Question 14

In a circuit analysis, junction P has four branches. When applying Kirchhoff's junction rule with the convention that currents toward P are positive, the equation becomes: I₁ + I₂ + I₃ + I₄ = 0. If I₁ = +3.0 A, I₂ = -1.5 A, and I₃ = -2.8 A, what is the value and physical meaning of I₄?

  1. I₄ = +1.3 A, meaning 1.3 A flows toward junction P (correct answer)
  2. I₄ = +1.3 A, meaning 1.3 A flows away from junction P
  3. I₄ = -1.3 A, meaning 1.3 A flows toward junction P
  4. I₄ = -1.3 A, meaning 1.3 A flows away from junction P
  5. I₄ = +7.3 A, meaning 7.3 A flows toward junction P
Explanation: When you encounter Kirchhoff's junction rule problems, remember that this fundamental principle states that the algebraic sum of all currents at any junction must equal zero—essentially, charge cannot accumulate at a point. Given the sign convention that currents toward junction P are positive, you can solve directly using the junction equation: I1+I2+I3+I4=0I_1 + I_2 + I_3 + I_4 = 0. Substituting the known values: 3.0+(−1.5)+(−2.8)+I4=03.0 + (-1.5) + (-2.8) + I_4 = 0. This simplifies to −1.3+I4=0-1.3 + I_4 = 0, so I4=+1.3 AI_4 = +1.3 \text{ A}. Since I4I_4 is positive and the convention defines positive currents as flowing toward P, this means 1.3 A flows toward the junction. Looking at the incorrect options: Choice B gives the right numerical value but misinterprets the sign convention—a positive current means flow toward P, not away from it. Choice C makes an algebraic error, likely from incorrectly adding the given currents, and compounds this with a sign convention error. Choice D contains the same algebraic mistake as C but correctly interprets what a negative current would mean (flow away from the junction). The key strategy for junction rule problems is to establish your sign convention clearly at the start, then stick to it consistently. Always double-check both your algebra and your physical interpretation of the final sign. Remember: positive and negative are arbitrary choices, but once you've chosen a convention, the physics must match that choice.

Question 15

A student measures currents at a junction and obtains: 4.7 A, 2.3 A, 1.8 A, and 3.2 A. When applying Kirchhoff's junction rule, the student finds that no combination of directions for these four currents satisfies current conservation. What can be concluded?

  1. The measurements contain experimental error and should be repeated
  2. Kirchhoff's junction rule doesn't apply to junctions with four branches
  3. Additional current paths exist that weren't measured
  4. The currents are AC values and require different analysis methods
  5. Both measurement error and additional current paths are possible explanations (correct answer)
Explanation: When analyzing current flow at electrical junctions, Kirchhoff's current law (KCL) states that the sum of currents entering a junction must equal the sum of currents leaving. This fundamental principle reflects charge conservation—electrons can't accumulate or disappear at a junction point. Let's check if these measurements can satisfy KCL. The four currents are 4.7 A, 2.3 A, 1.8 A, and 3.2 A. For any valid assignment of directions (in or out), we need: currents in = currents out. Testing different combinations:
  • If 4.7 A flows in: 4.7 = 2.3 + 1.8 + 3.2 = 7.3 (doesn't work)
  • If 4.7 A and 2.3 A flow in: 4.7 + 2.3 = 7.0, while 1.8 + 3.2 = 5.0 (doesn't work)
  • Other combinations also fail to balance
Since no directional assignment works, there must be additional current paths that weren't measured. Real circuits often have more complexity than initially apparent—perhaps there's a fifth wire, a connection through the mounting hardware, or a path through what appeared to be insulation. Answer A assumes the measurements are wrong, but if they're accurate, this doesn't solve the fundamental imbalance. Answer B incorrectly suggests KCL has limitations based on branch count—it applies universally. Answer D misunderstands the issue; AC vs DC analysis doesn't change the basic requirement that current be conserved at junctions. Study tip: When KCL appears violated with accurate measurements, always consider whether you've identified all current paths in the circuit. Hidden connections are more common than you might expect.

Question 16

Which statement correctly applies Kirchhoff's Junction Rule at junction A with 6 A6\,\text{A} out and 2 A,4 A2\,\text{A},4\,\text{A} in?

  1. Outgoing equals incoming: 6=2+46=2+4 (correct answer)
  2. Incoming equals outgoing times resistance
  3. Currents add only in series, not at junctions
  4. Voltage divides at junction, so current need not balance
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction A, 2 A and 4 A enter while 6 A leaves, so 2 + 4 = 6, matching the rule. The correct answer reflects this rule by stating outgoing equals incoming, indicating comprehension of current distribution. A common distractor confuses with voltage, reflecting a misunderstanding of current conservation. To help students, emphasize current focus. Reinforce practice in rule application statements.

Question 17

A complex junction has six branches with currents I₁ through I₆. Five of these currents are known: I₁ = 3.0 A (in), I₂ = 1.8 A (out), I₃ = 2.2 A (in), I₄ = 1.5 A (out), I₅ = 2.1 A (in). If the junction rule is satisfied, which statement about I₆ is correct?

  1. I₆ = 0.4 A flowing into the junction
  2. I₆ = 0.4 A flowing out of the junction
  3. I₆ = 4.0 A flowing out of the junction (correct answer)
  4. I₆ = 10.6 A flowing into the junction
  5. I₆ = 7.3 A flowing out of the junction
Explanation: When you encounter a junction problem in circuits, you're dealing with Kirchhoff's Current Law (KCL), which states that the total current flowing into a junction must equal the total current flowing out. This reflects the principle of charge conservation—electrons can't accumulate at a point. To solve this, first calculate the currents flowing in each direction. Currents flowing into the junction: I1+I3+I5=3.0+2.2+2.1=7.3 AI_1 + I_3 + I_5 = 3.0 + 2.2 + 2.1 = 7.3 \text{ A}. Currents flowing out: I2+I4=1.8+1.5=3.3 AI_2 + I_4 = 1.8 + 1.5 = 3.3 \text{ A}. Since more current flows in (7.3 A) than out (3.3 A), the sixth current I6I_6 must flow out of the junction to balance the equation. Using KCL: Current in=Current out\text{Current in} = \text{Current out}, so 7.3=3.3+I67.3 = 3.3 + I_6. Therefore, I6=4.0 AI_6 = 4.0 \text{ A} flowing out of the junction. Answer choice A gives the wrong magnitude (0.4 A instead of 4.0 A) and wrong direction. This likely results from calculation errors. Answer choice B has the correct magnitude but wrong direction—this would violate current conservation. Answer choice D has an impossibly large value (10.6 A) flowing in the wrong direction, suggesting the student added all currents without considering direction. The correct answer is C: I6=4.0 AI_6 = 4.0 \text{ A} flowing out of the junction. Study tip: Always set up KCL systematically by separating inward and outward currents first, then use the balance equation. Double-check that your answer makes physical sense—the missing current should compensate for any imbalance.

Question 18

Calculate missing current ImI_m at junction D: ImI_m in, 4 A4\,\text{A} in, 1 A1\,\text{A} out, 6 A6\,\text{A} out.

  1. Im=2 AI_m=2\,\text{A}
  2. Im=11 AI_m=11\,\text{A}
  3. Im=3 AI_m=3\,\text{A} (correct answer)
  4. Im=7 AI_m=7\,\text{A}
Explanation: This question tests understanding of Kirchhoff's Junction Rule in DC circuits, which states that the sum of currents entering a junction equals the sum of currents leaving. Kirchhoff's Junction Rule is a fundamental principle in circuit analysis, ensuring charge conservation at junctions. At junction D, Iₘ and 4 A enter while 1 A and 6 A leave, so Iₘ + 4 = 1 + 6, giving Iₘ = 3 A. The correct answer reflects this rule by accurately balancing the currents, indicating comprehension of current distribution. A common distractor is reversing directions, reflecting a misunderstanding of current conservation. To help students, emphasize consistent labeling of ins and outs. Reinforce practice with mixed direction problems.

Question 19

In the circuit shown, three resistors are connected to form two junctions. If the current through the 6-ohm resistor is 2.0 A flowing toward junction A, and the current through the 4-ohm resistor is 1.5 A flowing away from junction A, what is the current through the 8-ohm resistor?

  1. 0.5 A flowing toward junction A (correct answer)
  2. 0.5 A flowing away from junction A
  3. 3.5 A flowing toward junction A
  4. 3.5 A flowing away from junction A
  5. Cannot be determined without knowing the applied voltage
Explanation: Kirchhoff's junction rule states that the sum of currents entering a junction equals the sum of currents leaving. At junction A: currents entering = 2.0 A (from 6-ohm) + I₈ (from 8-ohm if positive), currents leaving = 1.5 A (through 4-ohm). So 2.0 + I₈ = 1.5, giving I₈ = -0.5 A. The negative sign means the current flows toward junction A at 0.5 A. Choice B has wrong direction, C and D ignore conservation of current, E is incorrect since junction rule doesn't depend on voltage values.

Question 20

In the junction shown, four currents meet at point S. The currents are: I₁ = 5.5 A (entering), I₂ = 2.3 A (entering), I₃ = 4.1 A (leaving), and I₄ (direction unknown). After applying Kirchhoff's junction rule, what additional information is needed to solve for both the magnitude and direction of I₄?

  1. No additional information is needed; I₄ = 3.7 A leaving the junction (correct answer)
  2. No additional information is needed; I₄ = 3.7 A entering the junction
  3. The resistance values of each branch are required
  4. The voltage across the junction is required
  5. The power dissipated in each branch is required
Explanation: Kirchhoff's junction rule provides complete information: current in = current out. Total entering = 5.5 + 2.3 = 7.8 A. Current leaving = 4.1 A + I₄. Therefore: 7.8 = 4.1 + I₄, giving I₄ = 3.7 A leaving. The direction is determined by the sign when we apply conservation. Choice B has incorrect direction, while C, D, and E incorrectly suggest that junction rule analysis requires additional circuit parameters beyond current conservation.