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