AP Physics 2 Quiz: Kirchhoffs Loop Rule
20 questions · exam conditions
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Kirchhoffs Loop RuleQuestion 1 of 20

In the single closed loop shown, a 12 V12\ \text{V} battery (positive terminal encountered first) is in series with R1=3.0 ΩR_1=3.0\ \Omega and R2=5.0 ΩR_2=5.0\ \Omega. Traverse clockwise; take a rise as +ε+\varepsilon across the battery from −- to ++, and a drop as −IR-IR when moving with the current through a resistor. Which equation correctly represents Kirchhoff's loop rule for this loop?

+12−I(3.0)−I(5.0)=0+12 - I(3.0) - I(5.0)=0
+12+I(3.0)−I(5.0)=0+12 + I(3.0) - I(5.0)=0
+12−I(3.0)=0+12 - I(3.0)=0
+12−I(3.0)−I(5.0)=12+12 - I(3.0) - I(5.0)=12
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AP Physics 2 Quiz

AP Physics 2 Quiz: Kirchhoffs Loop Rule

Practice Kirchhoffs Loop 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 Loop 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

In the single closed loop shown, a 12 V12\ \text{V} battery (positive terminal encountered first) is in series with R1=3.0 ΩR_1=3.0\ \Omega and R2=5.0 ΩR_2=5.0\ \Omega. Traverse clockwise; take a rise as +ε+\varepsilon across the battery from −- to ++, and a drop as −IR-IR when moving with the current through a resistor. Which equation correctly represents Kirchhoff's loop rule for this loop?

  1. +12−I(3.0)−I(5.0)=0+12 - I(3.0) - I(5.0)=0 (correct answer)
  2. +12+I(3.0)−I(5.0)=0+12 + I(3.0) - I(5.0)=0
  3. +12−I(3.0)=0+12 - I(3.0)=0
  4. +12−I(3.0)−I(5.0)=12+12 - I(3.0) - I(5.0)=12

Explanation: This problem requires applying Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of all voltage changes around any closed loop must equal zero, reflecting energy conservation - a charge returning to its starting point must have the same energy. Starting at the battery's negative terminal and traversing clockwise, we gain +12V crossing from negative to positive terminal, then lose voltage across each resistor: -I(3.0) across the 3.0Ω resistor and -I(5.0) across the 5.0Ω resistor. Choice D incorrectly sets the sum equal to 12 instead of 0, showing a misconception that the loop equation equals the battery voltage. The correct strategy is to traverse the loop consistently, sum all voltage changes (rises positive, drops negative), and set the total equal to zero.

Question 2

A single loop includes a battery of emf E=3.0 V\mathcal{E}=3.0\text{ V} and two series resistors R1=2.0 ΩR_1=2.0\,\Omega and R2=4.0 ΩR_2=4.0\,\Omega. Current II is clockwise. You traverse clockwise but cross the battery from ++ to −-, so the battery term is −E-\mathcal{E}. Resistor terms are −IR-IR when traversed with current. Which equation satisfies the loop rule?

  1. −3−2I−4I=0-3-2I-4I=0 (correct answer)
  2. +3−2I−4I=0+3-2I-4I=0
  3. −3+2I+4I=0-3+2I+4I=0
  4. −3−2I=0-3-2I=0

Explanation: This problem applies Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the total voltage change around any closed path equals zero, conserving energy in the circuit. Traversing clockwise with clockwise current I but crossing the 3V battery from + to - gives -3V (opposite to the standard - to + convention), while both resistors traversed with current contribute -2I and -4I respectively. The equation is -3 - 2I - 4I = 0. Choice B (+3 - 2I - 4I = 0) uses the wrong sign for the battery, forgetting that crossing from + to - gives negative voltage. Always apply sign conventions consistently: crossing a battery from + to - yields negative voltage.

Question 3

A loop contains two series batteries, 9.0 V9.0\text{ V} and 3.0 V3.0\text{ V}, oriented so their emfs oppose. A resistor R=6.0 ΩR=6.0\,\Omega is also in series. Current II is counterclockwise. Traverse counterclockwise: across a battery from −- to ++ is +E+\mathcal{E}, from ++ to −- is −E-\mathcal{E}; across the resistor in the direction of current is −IR-IR. Which equation satisfies Kirchhoff's loop rule?

  1. +9+3−6I=0+9+3-6I=0
  2. +9−3−6I=0+9-3-6I=0 (correct answer)
  3. −9−3−6I=0-9-3-6I=0
  4. +9−3+6I=0+9-3+6I=0

Explanation: This problem requires applying Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the algebraic sum of potential differences around a closed loop equals zero, conserving energy. Traversing counterclockwise with counterclockwise current I: the 9V battery is crossed from - to +, giving +9V; the 3V battery is crossed from + to -, giving -3V; the 6Ω resistor is traversed with the current, giving -6I. The equation is +9 - 3 - 6I = 0. Choice D (+9 - 3 + 6I = 0) incorrectly uses +6I for the resistor term, revealing the misconception of confusing the sign when traversing with versus against current. Remember to use negative voltage drops for resistors when traversing in the direction of current flow.

Question 4

A loop contains two resistors, R1=4.0 ΩR_1=4.0\ \Omega and R2=2.0 ΩR_2=2.0\ \Omega, and a 10 V10\ \text{V} battery. Traverse clockwise. Use the convention +ε+\varepsilon when moving from −- to ++ through the battery, and −IR-IR when moving with current through a resistor (current is clockwise). Which equation correctly represents the loop?

  1. +10−I(4.0)−I(2.0)=0+10 - I(4.0) - I(2.0)=0 (correct answer)
  2. −10−I(4.0)−I(2.0)=0-10 - I(4.0) - I(2.0)=0
  3. +10−I(4.0)+I(2.0)=0+10 - I(4.0) + I(2.0)=0
  4. +10−I(4.0)−I(2.0)=12+10 - I(4.0) - I(2.0)=12

Explanation: This question involves applying Kirchhoff's loop rule. Kirchhoff's loop rule ensures conservation of energy by requiring that the sum of all voltage changes around a closed loop equals zero, meaning a charge returns to its starting point with the same potential energy. Traversing clockwise with clockwise current, we cross the battery from negative to positive (+10V), then traverse both resistors in the direction of current, experiencing drops of -I(4.0) and -I(2.0). Choice D incorrectly sets the equation equal to 12 instead of 0, suggesting confusion about what the loop rule represents - it's an energy balance, not a calculation of total voltage. The key strategy is to sum all voltage changes algebraically (rises positive, drops negative) and set the total to zero, not to any other value.

Question 5

A loop contains a 4.5 V4.5\text{ V} battery and a 1.5 V1.5\text{ V} battery in series with a 3.0 Ω3.0\,\Omega resistor. The batteries aid each other for a clockwise traversal. Current II is clockwise. Traverse clockwise with +E+\mathcal{E} from −- to ++ and −IR-IR through the resistor with the current. Which equation satisfies Kirchhoff's loop rule?

  1. +4.5−1.5−3I=0+4.5-1.5-3I=0
  2. +4.5+1.5−3I=0+4.5+1.5-3I=0 (correct answer)
  3. −4.5−1.5−3I=0-4.5-1.5-3I=0
  4. +4.5+1.5+3I=0+4.5+1.5+3I=0

Explanation: This problem tests Kirchhoff's loop rule. Kirchhoff's loop rule states that the algebraic sum of voltage changes around a closed loop must equal zero, ensuring energy is conserved. Traversing clockwise with clockwise current I: both batteries are crossed from - to +, contributing +4.5V and +1.5V respectively since they aid each other; the 3Ω resistor traversed with current contributes -3I. The equation is +4.5 + 1.5 - 3I = 0. Choice A (+4.5 - 1.5 - 3I = 0) incorrectly subtracts the second battery voltage, misunderstanding that aiding batteries have the same sign. When batteries aid each other in a loop, their voltages add with the same sign according to your traversal convention.

Question 6

A closed loop has a 15 V15\ \text{V} battery in series with two resistors, R1=2.0 ΩR_1=2.0\ \Omega and R2=6.0 ΩR_2=6.0\ \Omega. Traverse counterclockwise; define +ε+\varepsilon when crossing the battery from −- to ++ and −IR-IR when crossing a resistor in the direction of current (current is counterclockwise). Which equation correctly represents the loop?

  1. +15−I(2.0)−I(6.0)=0+15 - I(2.0) - I(6.0)=0 (correct answer)
  2. −15−I(2.0)−I(6.0)=0-15 - I(2.0) - I(6.0)=0
  3. +15+I(2.0)+I(6.0)=0+15 + I(2.0) + I(6.0)=0
  4. +15−I(2.0)=0+15 - I(2.0)=0

Explanation: This question tests understanding of Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of all potential differences around a closed loop must be zero, ensuring that electric potential energy is conserved as charge completes a circuit. Traversing counterclockwise with counterclockwise current, we cross the battery from negative to positive (+15V), then move with the current through both resistors, experiencing voltage drops of -I(2.0) and -I(6.0). Choice C incorrectly uses positive signs for the resistor terms, revealing a sign convention error when the traverse direction matches the current direction. The reliable approach is to establish your traverse direction, then consistently apply signs: +ε for batteries traversed from - to +, and -IR for resistors traversed in the direction of current flow.

Question 7

One loop contains a 14 V14\ \text{V} battery, then (clockwise) a resistor R1=4 ΩR_1=4\ \Omega, then a second battery 4 V4\ \text{V} encountered from −- to ++ while traversing clockwise, then a resistor R2=6 ΩR_2=6\ \Omega. Current II is clockwise. Using +E+\mathcal{E} for −- to ++ and −IR-IR for resistors along II, which equation represents the loop?

  1. 14−I(4)=4+I(6)14-I(4)=4+I(6)
  2. 14−4−I(4+6)=014-4-I(4+6)=0
  3. 14+4−I(4+6)=014+4-I(4+6)=0 (correct answer)
  4. 14+4+I(4+6)=014+4+I(4+6)=0

Explanation: This question tests Kirchhoff's loop rule. Kirchhoff's loop rule requires that the sum of all voltage changes around any closed loop equals zero, reflecting the conservation of energy for charges moving through the circuit. Traversing clockwise, we encounter the 14V battery from negative to positive (+14V), drop voltage across R₁ = 4Ω (-4I), cross the 4V battery from negative to positive (+4V), and drop voltage across R₂ = 6Ω (-6I). The complete equation is: 14 - 4I + 4 - 6I = 0, which simplifies to 14 + 4 - I(4+6) = 0. Choice A incorrectly subtracts the 4V battery voltage, demonstrating the misconception of assuming all batteries after the first must be subtracted. Always determine signs based on traversal direction: negative to positive gives a positive voltage change regardless of battery position.

Question 8

A closed loop includes a 15 V15\text{ V} battery and two resistors R1=2.0 ΩR_1=2.0\,\Omega and R2=7.0 ΩR_2=7.0\,\Omega in series. Current II is clockwise. Traverse clockwise starting at the battery's positive terminal. Use: crossing the battery from ++ to −- is −E-\mathcal{E}; crossing a resistor along current is −IR-IR. Which equation correctly represents the loop rule?

  1. −15+2.0I+7.0I=0-15 + 2.0I + 7.0I = 0
  2. +15−2.0I−7.0I=0+15 - 2.0I - 7.0I = 0
  3. −15−2.0I=0-15 - 2.0I = 0
  4. −15−2.0I−7.0I=0-15 - 2.0I - 7.0I = 0 (correct answer)

Explanation: This question involves Kirchhoff's loop rule. Kirchhoff's loop rule requires that the algebraic sum of voltage changes around any closed loop equals zero, reflecting energy conservation. Starting at the positive terminal and traversing clockwise: we immediately cross the battery from + to - (-15V), then cross R₁ along current (-2.0I), then cross R₂ along current (-7.0I). The equation is -15 - 2.0I - 7.0I = 0. Choice B (+15 - 2.0I - 7.0I = 0) uses the wrong sign for the battery, showing the misconception that battery voltage is always positive. Remember that the sign depends on traversal direction: crossing from + to - gives -ℰ.

Question 9

A single loop has a resistor R=10 ΩR=10\,\Omega and two ideal batteries. Traversing clockwise, you cross E1=2.0 V\mathcal{E}_1=2.0\text{ V} from ++ to −- and E2=7.0 V\mathcal{E}_2=7.0\text{ V} from −- to ++. The current II is clockwise. Use: battery (−→+)(-\to +) is +E+\mathcal{E}, (+→−)(+\to -) is −E-\mathcal{E}; resistor along current is −IR-IR. Which equation represents the loop rule?

  1. −2.0−7.0−10I=0-2.0 - 7.0 - 10I = 0
  2. −2.0+7.0+10I=0-2.0 + 7.0 + 10I = 0
  3. −2.0+7.0−10I=0-2.0 + 7.0 - 10I = 0 (correct answer)
  4. +2.0+7.0−10I=0+2.0 + 7.0 - 10I = 0

Explanation: This question applies Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of voltage changes around any closed loop must equal zero, ensuring energy conservation in circuits. Traversing clockwise: ℰ₁ is crossed from + to - (-2.0V), ℰ₂ is crossed from - to + (+7.0V), and the resistor is crossed along current (-10I). The equation is -2.0 + 7.0 - 10I = 0. Choice D (-2.0 + 7.0 + 10I = 0) incorrectly uses +10I for the resistor term, showing the misconception that sign depends only on current direction, not traversal direction. Remember that crossing a resistor in the current direction always gives -IR in the loop equation.

Question 10

A single loop has an ideal battery E=6.0 V\mathcal{E}=6.0\text{ V} and two resistors R1=2.0 ΩR_1=2.0\,\Omega and R2=1.0 ΩR_2=1.0\,\Omega in series. The loop current II is counterclockwise. Traverse counterclockwise starting at the battery's negative terminal. Sign convention: battery (−→+)(-\to +) is +E+\mathcal{E}; resistor in direction of current is −IR-IR. Which equation correctly represents Kirchhoff's loop rule for the loop?

  1. +6.0−2.0I−1.0I=0+6.0 - 2.0I - 1.0I = 0 (correct answer)
  2. −6.0−2.0I−1.0I=0-6.0 - 2.0I - 1.0I = 0
  3. +6.0+2.0I+1.0I=0+6.0 + 2.0I + 1.0I = 0
  4. +6.0−2.0I=0+6.0 - 2.0I = 0

Explanation: This question applies Kirchhoff's loop rule. Kirchhoff's loop rule states that the algebraic sum of voltage changes around any closed loop equals zero, ensuring energy conservation in circuits. Starting at the battery's negative terminal and traversing counterclockwise (same as current direction): we cross the battery from - to + (+6.0V), then cross R₁ along the current (-2.0I), then cross R₂ along the current (-1.0I). The equation is +6.0 - 2.0I - 1.0I = 0. Choice D (+6.0 - 2.0I = 0) omits R₂, showing the misconception of incomplete loop traversal. To correctly apply the loop rule, traverse the entire loop and account for every component, using consistent sign conventions throughout.

Question 11

A closed loop has an ideal 6 V6\text{ V} battery in series with a resistor R1=1 ΩR_1=1\,\Omega and a second resistor R2=5 ΩR_2=5\,\Omega. The current II is counterclockwise. Traverse counterclockwise, using +E+\mathcal{E} when moving from −- to ++ across the battery and −IR-IR when moving with the current through a resistor. Which equation correctly represents the loop?

  1. 6+I+5I=06+I+5I=0
  2. 6−I=06-I=0
  3. 6−I−5I=06-I-5I=0 (correct answer)
  4. 6+I−5I=06+I-5I=0

Explanation: This question involves applying Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of all voltage changes around a closed loop must be zero, reflecting the conservation of energy. When traversing counterclockwise with a counterclockwise current, we encounter the 6V battery from negative to positive (+6V), then pass through R₁ = 1Ω with the current (-1I or simply -I), and through R₂ = 5Ω with the current (-5I). The equation becomes 6 - I - 5I = 0. Choice D (6 + I + 5I = 0) incorrectly uses positive signs for the resistor terms, indicating a sign convention error when current and traversal directions match. Always use negative signs for voltage drops across resistors when moving with the current direction.

Question 12

A closed loop contains a 10 V10\text{ V} ideal battery and a resistor R=5.0 ΩR=5.0\,\Omega. Current II is clockwise. You traverse the loop counterclockwise starting at the battery's negative terminal. Use: battery (−→+)(-\to +) is +E+\mathcal{E}, (+→−)(+\to -) is −E-\mathcal{E}; for a resistor, traversing in the direction of current gives −IR-IR and opposite gives +IR+IR. Which equation satisfies Kirchhoff's loop rule for your traversal?

  1. −10−5.0I=0-10 - 5.0I = 0
  2. +10−5.0I=0+10 - 5.0I = 0
  3. −10+5.0I=0-10 + 5.0I = 0
  4. +10+5.0I=0+10 + 5.0I = 0 (correct answer)

Explanation: This problem involves Kirchhoff's loop rule. Kirchhoff's loop rule requires that the sum of all voltage changes around a closed loop equals zero, reflecting conservation of energy. With clockwise current I and counterclockwise traversal starting at the battery's negative terminal: we cross the battery from - to + (+10V), then cross the resistor opposite to current direction (+5.0I). The equation is +10 + 5.0I = 0. Choice B (+10 - 5.0I = 0) incorrectly uses -5.0I, showing the misconception that resistor terms are always negative regardless of traversal direction relative to current. Remember that crossing a resistor opposite to current direction gives +IR, not -IR.

Question 13

A closed loop contains two ideal batteries and one resistor: E1=9.0 V\mathcal{E}_1=9.0\text{ V} (traversed from −- to ++ in the chosen direction) and E2=3.0 V\mathcal{E}_2=3.0\text{ V} (traversed from ++ to −- in the chosen direction), plus R=4.0 ΩR=4.0\,\Omega with current II in the loop. Traverse clockwise. Use: battery (−→+)(-\to +) is +E+\mathcal{E}, (+→−)(+\to -) is −E-\mathcal{E}; resistor along current is −IR-IR. Which equation correctly represents the loop?

  1. +9.0−3.0+4.0I=0+9.0 - 3.0 + 4.0I = 0
  2. +9.0+3.0−4.0I=0+9.0 + 3.0 - 4.0I = 0
  3. −9.0−3.0−4.0I=0-9.0 - 3.0 - 4.0I = 0
  4. +9.0−3.0−4.0I=0+9.0 - 3.0 - 4.0I = 0 (correct answer)

Explanation: This problem tests Kirchhoff's loop rule. Kirchhoff's loop rule ensures energy conservation by requiring the sum of voltage changes around a closed loop to equal zero. Traversing clockwise: ℰ₁ is crossed from - to + giving +9.0V, ℰ₂ is crossed from + to - giving -3.0V, and the resistor is crossed in the current direction giving -4.0I. The equation is +9.0 - 3.0 - 4.0I = 0. Choice B (+9.0 + 3.0 - 4.0I = 0) incorrectly adds both battery terms, revealing the misconception that all batteries contribute positive voltage regardless of traversal direction. When applying the loop rule, maintain consistent sign conventions: crossing a battery from + to - always gives -ℰ, regardless of the battery's orientation in the circuit.

Question 14

A closed loop has a 3 V3\ \text{V} battery traversed counterclockwise from −- to ++, a resistor R=1 ΩR=1\ \Omega, and a second battery 8 V8\ \text{V} traversed counterclockwise from ++ to −-. Current II is counterclockwise. Take battery rises as positive, drops as negative, and resistor drops as −IR-IR along II. Which equation satisfies the loop rule?

  1. 3−8+I(1)=03-8+I(1)=0
  2. 3+8−I(1)=03+8-I(1)=0
  3. 3−8−I(1)=03-8-I(1)=0 (correct answer)
  4. 3=8−I(1)3=8-I(1)

Explanation: This problem requires applying Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the total voltage change around any closed path is zero, reflecting energy conservation as charge moves through a complete circuit. Traversing counterclockwise, we cross the 3V battery from negative to positive (+3V), encounter the resistor R = 1Ω in the direction of current (-1I), and cross the 8V battery from positive to negative (-8V). The loop equation becomes: 3 - 1I - 8 = 0, which can be rearranged as 3 - 8 - I(1) = 0. Choice C incorrectly uses +I(1), revealing the misconception of changing resistor voltage drop signs when batteries have different magnitudes. Maintain consistent sign conventions: resistor voltage drops are always negative when traversing in the current direction.

Question 15

A loop contains an ideal battery E=4.5 V\mathcal{E}=4.5\text{ V} and three series resistors R1=1.0 ΩR_1=1.0\,\Omega, R2=2.0 ΩR_2=2.0\,\Omega, and R3=3.0 ΩR_3=3.0\,\Omega. Current II is counterclockwise. Traverse counterclockwise crossing the battery from −- to ++. Use: battery (−→+)(-\to +) is +E+\mathcal{E}; each resistor along current is −IR-IR. Which equation correctly represents the loop?

  1. +4.5−1.0I−2.0I−3.0I=0+4.5 - 1.0I - 2.0I - 3.0I = 0 (correct answer)
  2. +4.5+1.0I+2.0I+3.0I=0+4.5 + 1.0I + 2.0I + 3.0I = 0
  3. −4.5−1.0I−2.0I−3.0I=0-4.5 - 1.0I - 2.0I - 3.0I = 0
  4. +4.5−1.0I−2.0I=0+4.5 - 1.0I - 2.0I = 0

Explanation: This problem tests Kirchhoff's loop rule. Kirchhoff's loop rule ensures energy conservation by requiring the sum of all voltage changes around a closed loop to be zero. Traversing counterclockwise (same as current): crossing the battery from - to + gives +4.5V, then crossing each resistor in the current direction gives -1.0I, -2.0I, and -3.0I respectively. The equation is +4.5 - 1.0I - 2.0I - 3.0I = 0. Choice D (+4.5 - 1.0I - 2.0I = 0) omits R₃, demonstrating the misconception of incomplete loop accounting. When applying the loop rule, traverse the complete loop and include every component, maintaining consistent sign conventions throughout.

Question 16

In one loop, two series resistors R1=4.0 ΩR_1=4.0\,\Omega and R2=6.0 ΩR_2=6.0\,\Omega carry clockwise current II. An ideal battery of emf E\mathcal{E} is traversed from −- to ++ when going clockwise. Traverse clockwise. Use the sign convention: battery (−→+)(-\to +) is +E+\mathcal{E}; each resistor along current is −IR-IR. Which equation correctly represents the loop rule?

  1. +E−4.0I=0+\mathcal{E} - 4.0I = 0
  2. −E−4.0I−6.0I=0-\mathcal{E} - 4.0I - 6.0I = 0
  3. +E−4.0I−6.0I=0+\mathcal{E} - 4.0I - 6.0I = 0 (correct answer)
  4. +E+4.0I+6.0I=0+\mathcal{E} + 4.0I + 6.0I = 0

Explanation: This question tests Kirchhoff's loop rule. Kirchhoff's loop rule ensures energy conservation by requiring the sum of voltage changes around any closed loop to be zero. Traversing clockwise with clockwise current I: the battery is crossed from - to + (+ℰ), R₁ is crossed along current (-4.0I), and R₂ is crossed along current (-6.0I). The equation is +ℰ - 4.0I - 6.0I = 0. Choice D (+ℰ - 4.0I = 0) omits the second resistor, demonstrating the misconception of incomplete circuit analysis. When applying Kirchhoff's loop rule, ensure you account for every component in the loop and apply sign conventions consistently throughout your traversal.

Question 17

In a single closed loop, a 12 V12\text{ V} ideal battery (positive terminal encountered first) is in series with R1=4 ΩR_1=4\,\Omega and R2=2 ΩR_2=2\,\Omega. A steady current II flows clockwise. Traverse the loop clockwise, taking a voltage rise as +E+\mathcal{E} when moving from −- to ++ across the battery and a voltage drop as −IR-IR when moving with the current through a resistor. Which equation correctly represents the loop?

  1. 12+4I+2I=012+4I+2I=0
  2. 12−4I−2I=012-4I-2I=0 (correct answer)
  3. 12−4I+2I=012-4I+2I=0
  4. 12−4I=012-4I=0

Explanation: This problem requires applying Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of all voltage changes around any closed loop must equal zero, reflecting energy conservation. When traversing the loop clockwise, we encounter the battery from negative to positive terminal (+12V), then pass through R₁ = 4Ω with the current (-4I), and finally through R₂ = 2Ω with the current (-2I). The equation becomes 12 - 4I - 2I = 0. Choice C incorrectly omits the second resistor, suggesting a misconception that only one resistor contributes to the voltage drop. To solve these problems correctly, always traverse the entire loop systematically, applying consistent sign conventions for each component.

Question 18

A closed loop contains two resistors R1=5.0 ΩR_1=5.0\,\Omega and R2=10 ΩR_2=10\,\Omega in series, and two ideal batteries: E1=6.0 V\mathcal{E}_1=6.0\,\text{V} and E2=2.0 V\mathcal{E}_2=2.0\,\text{V} opposing E1\mathcal{E}_1. Conventional current II is clockwise. Traverse clockwise using: −- to ++ across a battery is +E+\mathcal{E}, and across a resistor with II is −IR-IR. Which equation is correct?

  1. +6−2−I(5+10)=0+6-2-I(5+10)=0 (correct answer)
  2. +6+2−I(5+10)=0+6+2-I(5+10)=0
  3. +6−2−I(10)=0+6-2-I(10)=0
  4. +6−2+I(5+10)=0+6-2+I(5+10)=0

Explanation: This problem requires applying Kirchhoff's loop rule to a circuit with two opposing batteries and two series resistors. Kirchhoff's loop rule ensures energy conservation by requiring the sum of voltage changes around any closed loop to equal zero. Traversing clockwise: the 6V battery from negative to positive gives +6V; the 2V battery (opposing the 6V) from positive to negative gives -2V; both resistors with current I give drops -I(5) and -I(10), totaling -I(5+10). Option B incorrectly uses +2 for the opposing battery, revealing a sign error when batteries oppose each other - students must recognize that traversing from + to - gives a negative contribution. To solve loop equations systematically, identify battery orientations first, then apply consistent sign conventions throughout the traversal.

Question 19

In a closed loop, R1=2.0 ΩR_1=2.0\,\Omega and R2=8.0 ΩR_2=8.0\,\Omega are in series with an ideal 15 V15\,\text{V} battery and an ideal 5.0 V5.0\,\text{V} battery that aids the 15 V15\,\text{V} source. Conventional current II is clockwise. Traverse clockwise: −- to ++ across a battery is +E+\mathcal{E}; across a resistor with II is −IR-IR. Which equation satisfies Kirchhoff's loop rule?

  1. +15+5+I(2+8)=0+15+5+I(2+8)=0
  2. +15+5−I(8)=0+15+5-I(8)=0
  3. +15−5−I(2+8)=0+15-5-I(2+8)=0
  4. +15+5−I(2+8)=0+15+5-I(2+8)=0 (correct answer)

Explanation: This problem requires applying Kirchhoff's loop rule to a circuit where two batteries aid each other in series with two resistors. Kirchhoff's loop rule states that the algebraic sum of voltage changes around any closed loop equals zero, ensuring energy conservation. Traversing clockwise: the 15V battery from negative to positive gives +15V; the 5V battery (aiding the 15V) from negative to positive also gives +5V; both resistors traversed with current I give voltage drops -I(2) and -I(8), totaling -I(2+8). Option B incorrectly uses -5 for the aiding battery, showing a misconception about battery polarity when batteries work together rather than oppose each other. To apply loop rules correctly, identify whether batteries aid or oppose based on their orientations, then consistently apply sign conventions.

Question 20

A closed loop contains an ideal 7.0 V7.0\,\text{V} battery and two series resistors R1=2.0 ΩR_1=2.0\,\Omega and R2=5.0 ΩR_2=5.0\,\Omega. Conventional current II is counterclockwise. Traverse counterclockwise: crossing the battery from ++ to −- is −E-\mathcal{E}; crossing a resistor with II is −IR-IR. Which equation correctly represents the loop?

  1. −7−I(2)=0-7-I(2)=0
  2. +7−I(2+5)=0+7-I(2+5)=0
  3. −7−I(2+5)=0-7-I(2+5)=0 (correct answer)
  4. −7+I(2+5)=0-7+I(2+5)=0

Explanation: This problem involves applying Kirchhoff's loop rule when traversing opposite to the conventional current direction. Kirchhoff's loop rule requires that the sum of voltage changes around any closed loop equals zero, reflecting energy conservation. Traversing counterclockwise: the 7V battery is crossed from positive to negative terminal, giving -7V; both resistors are traversed with current I, giving voltage drops -I(2) and -I(5), totaling -I(2+5). Option B incorrectly uses +7 for the battery, showing a sign error when traversing from + to - terminal, which always gives a negative contribution regardless of current direction. When applying loop rules, focus on the traversal direction relative to component polarities, not just the current direction.