Kirchhoff's Junction Rule

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AP Physics C: Electricity and Magnetism › Kirchhoff's Junction Rule

Questions 1 - 8
1

Solve for the unknown current at junction B using Kirchhoff’s Junction Rule: find $I_1$ entering B if $I_2=0.4,\text{A}$ and $I_3=1.3,\text{A}$ leave.

$I_1=1.7,\text{A}$ (entering B)

$I_1=0.9,\text{A}$ (entering B)

$I_1=-1.7,\text{A}$ (entering B)

$I_1=1.3,\text{A}$ (entering B)

Explanation

This question tests AP Physics C: Electricity and Magnetism, specifically Kirchhoff's Junction Rule and the conservation of charge at circuit junctions. Kirchhoff's Junction Rule states that the total current entering a junction equals the total current leaving it, reflecting the conservation of electric charge. At junction B, I₁ enters while both I₂ = 0.4 A and I₃ = 1.3 A leave the junction. Choice B is correct because applying the junction rule: I₁ = I₂ + I₃, so I₁ = 0.4 + 1.3 = 1.7 A entering B. Choice A appears to subtract one current from another, while Choice C incorrectly assigns a negative value to an entering current. To help students: Remember that Kirchhoff's Junction Rule is simply conservation of charge - charge cannot accumulate at a junction. Practice identifying all currents at a junction before writing the equation.

2

Calculate the currents at junction A and verify charge conservation: find $I_3$ leaving A if $I_1=1.8,\text{A}$ enters and $I_2=0.5,\text{A}$ leaves.

$I_3=2.3,\text{A}$ (leaving A)

$I_3=-1.3,\text{A}$ (leaving A)

$I_3=0.5,\text{A}$ (leaving A)

$I_3=1.3,\text{A}$ (leaving A)

Explanation

This question tests AP Physics C: Electricity and Magnetism, specifically Kirchhoff's Junction Rule and the conservation of charge at circuit junctions. Kirchhoff's Junction Rule states that the total current entering a junction equals the total current leaving it, reflecting the conservation of electric charge. At junction A, I₁ = 1.8 A enters while both I₂ = 0.5 A and I₃ leave the junction. Choice A is correct because applying the junction rule: I₁ = I₂ + I₃, so 1.8 = 0.5 + I₃, giving I₃ = 1.3 A leaving A. Choice B incorrectly adds the entering and one leaving current, while Choice D appears to be the difference between the two given values. To help students: The phrase 'verify charge conservation' reminds us that the junction rule is fundamentally about charge conservation. Practice stating the physical principle before applying the mathematical rule.

3

Calculate the currents at junction B and verify conservation: $I_1=1.10,\text{A}$ enters, $I_2=0.40,\text{A}$ leaves, and $I_3$ leaves.

$I_3=0.70,\text{A}$ because $I_1=I_2+I_3$

$I_3=0.30,\text{A}$ because $I_1-I_3=I_2$

$I_3=0.40,\text{A}$ because $I_3=I_2$

$I_3=1.50,\text{A}$ because $I_1+I_2=I_3$

Explanation

This question tests AP Physics C: Electricity and Magnetism, specifically Kirchhoff's Junction Rule and the conservation of charge at circuit junctions. Kirchhoff's Junction Rule states that the total current entering a junction equals the total current leaving it, reflecting the conservation of electric charge. At junction B, I₁ = 1.10 A enters while I₂ = 0.40 A and I₃ both leave the junction. Applying the junction rule: I₁ = I₂ + I₃, which gives us 1.10 = 0.40 + I₃, solving for I₃ = 0.70 A. Choice A is correct because it properly applies the equation I₁ = I₂ + I₃ to find the unknown current. Choice B incorrectly adds the entering and leaving currents, which would mean 1.50 A leaves through I₃ alone, violating conservation. To help students: Always verify your answer by checking that total current in equals total current out. Watch for common mistakes like confusing junction rule with loop rule equations.

4

Using Kirchhoff's Junction Rule, determine $I_2$ leaving junction D if $I_1=0.80,\text{A}$ enters and $I_3=0.25,\text{A}$ and $I_4=0.15,\text{A}$ leave.

$I_2=0.40,\text{A}$ because $I_1-I_3-I_4=I_2$

$I_2=0.70,\text{A}$ because $I_1-I_3+I_4=I_2$

$I_2=1.20,\text{A}$ because $I_1+I_3+I_4=I_2$

$I_2=0.10,\text{A}$ because $I_3-I_4=I_2$

Explanation

This question tests AP Physics C: Electricity and Magnetism, specifically Kirchhoff's Junction Rule and the conservation of charge at circuit junctions. Kirchhoff's Junction Rule states that the total current entering a junction equals the total current leaving it, reflecting the conservation of electric charge. At junction D, I₁ = 0.80 A enters while I₂, I₃ = 0.25 A, and I₄ = 0.15 A all leave the junction. Applying the junction rule: I₁ = I₂ + I₃ + I₄, which gives us 0.80 = I₂ + 0.25 + 0.15, solving for I₂ = 0.40 A. Choice A is correct because it properly applies the conservation principle with the correct equation I₁ - I₃ - I₄ = I₂. Choice B incorrectly adds all currents without considering their directions relative to the junction. To help students: Draw a clear diagram with arrows showing current directions before writing equations. Remember that Kirchhoff's rule is simply stating that charge cannot accumulate at a junction.

5

Calculate the current at junction A: $I_T$ enters, then splits into $I_1=0.90,\text{A}$ and $I_2=0.65,\text{A}$ leaving along two branches.

$I_T=1.55,\text{A}$ because $I_T=I_1+I_2$

$I_T=1.75,\text{A}$ because $I_T=I_1+I_2+0.20$

$I_T=0.25,\text{A}$ because $I_T=I_1-I_2$

$I_T=0.90,\text{A}$ because $I_T=I_1$

Explanation

This question tests AP Physics C: Electricity and Magnetism, specifically Kirchhoff's Junction Rule and the conservation of charge at circuit junctions. Kirchhoff's Junction Rule states that the total current entering a junction equals the total current leaving it, reflecting the conservation of electric charge. At junction A, the total current I_T enters and then splits into two branches with I₁ = 0.90 A and I₂ = 0.65 A leaving. Applying the junction rule: I_T = I₁ + I₂ = 0.90 + 0.65 = 1.55 A. Choice A is correct because it recognizes that the entering current must equal the sum of the two leaving currents. Choice B incorrectly subtracts the currents, violating the principle that current is conserved at a junction. To help students: Visualize the junction as a pipe splitting into two smaller pipes - the flow rate in must equal the total flow rate out. Always check that your answer makes physical sense by verifying conservation of charge.

6

Using Kirchhoff's Junction Rule, determine current $I_1$ entering junction A if $I_2=0.35,\text{A}$ enters and $I_3=0.90,\text{A}$ leaves plus $I_4=0.15,\text{A}$ leaves.

$I_1=0.70,\text{A}$ because $I_1+I_2=I_3+I_4$

$I_1=0.40,\text{A}$ because $I_1=I_3+I_4-I_2$

$I_1=-0.40,\text{A}$ because $I_1=I_2-I_3-I_4$

$I_1=1.40,\text{A}$ because $I_1=I_2+I_3+I_4$

Explanation

This question tests AP Physics C: Electricity and Magnetism, specifically Kirchhoff's Junction Rule and the conservation of charge at circuit junctions. Kirchhoff's Junction Rule states that the total current entering a junction equals the total current leaving it, reflecting the conservation of electric charge. At junction A, I₁ and I₂ = 0.35 A enter the junction, while I₃ = 0.90 A and I₄ = 0.15 A leave (total 1.05 A leaving). Applying the junction rule: I₁ + I₂ = I₃ + I₄, which gives us I₁ + 0.35 = 0.90 + 0.15, solving for I₁ = 0.70 A. Choice A is correct because it properly balances entering and leaving currents with the equation I₁ + I₂ = I₃ + I₄. Choice C incorrectly suggests a negative current entering, which would mean current actually leaves through that wire. To help students: Remember that the problem states I₁ 'enters' - this fixes its direction. Always check that your mathematical result matches the physical description given.

7

Solve for the unknown current $I_3$ entering junction C if $I_1=1.20,\text{A}$ and $I_2=0.45,\text{A}$ leave junction C, and $I_4=0.30,\text{A}$ enters.

$I_3=0.75,\text{A}$ because $I_3=I_1-I_2+I_4$

$I_3=0.45,\text{A}$ because $I_3=I_2$

$I_3=1.35,\text{A}$ because $I_3+I_4=I_1+I_2$

$I_3=0.15,\text{A}$ because $I_3=I_1+I_2-I_4$

Explanation

This question tests AP Physics C: Electricity and Magnetism, specifically Kirchhoff's Junction Rule and the conservation of charge at circuit junctions. Kirchhoff's Junction Rule states that the total current entering a junction equals the total current leaving it, reflecting the conservation of electric charge. At junction C, I₃ and I₄ = 0.30 A enter the junction, while I₁ = 1.20 A and I₂ = 0.45 A leave the junction. Applying the junction rule: I₃ + I₄ = I₁ + I₂, which gives us I₃ + 0.30 = 1.20 + 0.45, solving for I₃ = 1.35 A. Choice A is correct because it properly identifies that the sum of entering currents equals the sum of leaving currents. Choice D incorrectly adds entering and leaving currents together without proper sign convention. To help students: Create a consistent sign convention (entering positive, leaving negative) and stick to it throughout the problem. Practice setting up the equation before substituting numbers to avoid sign errors.

8

Solve for unknown current $I_6$ at junction C if $I_4=0.25,\text{A}$ enters, $I_5=0.10,\text{A}$ enters, and $I_6$ leaves with $I_7=0.05,\text{A}$ leaving.

$I_6=-0.10,\text{A}$ because $I_6=I_7-I_4-I_5$

$I_6=0.40,\text{A}$ because $I_6=I_4+I_5+I_7$

$I_6=0.10,\text{A}$ because $I_6=I_5$

$I_6=0.30,\text{A}$ because $I_4+I_5=I_6+I_7$

Explanation

This question tests AP Physics C: Electricity and Magnetism, specifically Kirchhoff's Junction Rule and the conservation of charge at circuit junctions. Kirchhoff's Junction Rule states that the total current entering a junction equals the total current leaving it, reflecting the conservation of electric charge. At junction C, I₄ = 0.25 A and I₅ = 0.10 A enter (total 0.35 A entering), while I₆ and I₇ = 0.05 A leave. Applying the junction rule: I₄ + I₅ = I₆ + I₇, which gives us 0.25 + 0.10 = I₆ + 0.05, solving for I₆ = 0.30 A. Choice A is correct because it properly applies the conservation equation I₄ + I₅ = I₆ + I₇. Choice D incorrectly suggests a negative current, which would mean I₆ enters rather than leaves as stated. To help students: Create a table listing entering currents (+) and leaving currents (-) to organize your work. This systematic approach prevents sign errors and makes complex junctions manageable.