Kirchhoff's Loop Rule - AP Physics 2

A single loop has a $6\ \text{V}$ battery and three series resistors $R_1=1\ \Omega$ , $R_2=2\ \Omega$ , and $R_3=3\ \Omega$ . Current $I$ is clockwise. Traverse clockwise; count the battery as $+6\ \text{V}$ when going from $-$ to $+$ and each resistor as $-IR$ along $I$ . Which equation correctly represents the loop?
Answer: $6-I(1+2+3)=0$ This question involves Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the total energy change around any closed path is zero, reflecting conservation of energy in electrical circuits. Traversing clockwise, we cross the 6V battery from negative to positive (+6V), then encounter three resistors in series: R₁ = 1Ω (-1I), R₂ = 2Ω (-2I), and R₃ = 3Ω (-3I), all producing voltage drops since we traverse in the current direction. The complete loop equation is: 6 - 1I - 2I - 3I = 0, which simplifies to 6 - I(1+2+3) = 0. Choice D incorrectly treats the resistors as if they have different currents, revealing the misconception that series resistors can have different currents flowing through them. Remember that in a series circuit, the same current flows through all components.
A loop contains a $1.2\text{ V}$ battery and one resistor $R=0.60\,\Omega$ in series. Conventional current $I$ is counterclockwise. Traverse counterclockwise: across the battery from $-$ to $+$ is $+\mathcal{E}$ and across the resistor with current is $-IR$ . Which equation correctly represents the closed loop?
Answer: $+1.2-0.60I=0$ This question tests Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of voltage changes around a complete loop must be zero, ensuring energy conservation in electrical circuits. Traversing counterclockwise with counterclockwise current I: crossing the 1.2V battery from - to + contributes +1.2V, and traversing the 0.60Ω resistor with the current contributes -0.60I. The equation is +1.2 - 0.60I = 0. Choice B (+1.2 + 0.60I = 0) incorrectly uses a positive sign for the resistor term, revealing confusion about voltage drops across resistors. Remember that resistors always cause voltage drops (negative terms) when traversed in the direction of current flow.
In a closed loop, a $12\ \text{V}$ battery is oriented so that traversing counterclockwise crosses it from $+$ to $-$ . The loop also contains $R_1=4\ \Omega$ and $R_2=8\ \Omega$ in series, and current $I$ is counterclockwise. Use $-\mathcal{E}$ for $+$ to $-$ across the battery and $-IR$ along $I$ across resistors. Which loop equation is correct?
Answer: $-12-I(4+8)=0$ This problem tests Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of all voltage changes around a closed loop must equal zero, ensuring energy is conserved as charge moves around the circuit. Traversing counterclockwise, we cross the 12V battery from positive to negative (-12V), then encounter R₁ = 4Ω in the direction of current (-4I), followed by R₂ = 8Ω also in the current direction (-8I). The loop equation becomes: -12 - 4I - 8I = 0, which simplifies to -12 - I(4+8) = 0. Choice B incorrectly uses +12V, demonstrating the misconception of always treating battery voltage as positive regardless of traversal direction. Consistently apply the rule: crossing a battery from positive to negative gives a negative voltage change.
In a single closed loop, a $12\text{ V}$ battery (positive terminal on the left) is in series with $R_1=3.0\,\Omega$ and $R_2=5.0\,\Omega$ . Conventional current $I$ is clockwise. Traverse the loop clockwise starting at the battery's negative terminal. Use the sign convention: crossing a battery from $-$ to $+$ is $+\mathcal{E}$ , from $+$ to $-$ is $-\mathcal{E}$ ; crossing a resistor in the direction of current is $-IR$ , opposite is $+IR$ . Which equation correctly represents Kirchhoff's loop rule for this loop (energy conservation)?
Answer: $+12 - 3.0I - 5.0I = 0$ 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. Starting at the battery's negative terminal and traversing clockwise: we cross the battery from negative to positive (+12V), then cross R₁ in the direction of current (-3.0I), then cross R₂ in the direction of current (-5.0I). The equation becomes +12 - 3.0I - 5.0I = 0. Choice B (+12 + 3.0I + 5.0I = 0) incorrectly adds the resistor terms, suggesting the misconception that voltage increases across resistors. To solve loop problems systematically, choose a consistent traversal direction and apply sign conventions carefully: batteries give +ℰ when crossed from - to +, and resistors give -IR when crossed in the current direction.
A single loop contains an ideal $8\text{ V}$ battery and two resistors in series, $R_1=2\,\Omega$ and $R_2=2\,\Omega$ . The current $I$ is clockwise. Traverse clockwise: across the battery from $-$ to $+$ is $+8$ , and across each resistor with the current is $-IR$ . Which equation correctly represents energy conservation around the loop?
Answer: $8-2I-2I=0$ This problem involves applying Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the sum of all voltage changes around a closed loop equals zero, reflecting energy conservation in electrical circuits. Traversing clockwise with clockwise current, we cross the 8V battery from negative to positive (+8V), then R₁ = 2Ω with the current (-2I), and R₂ = 2Ω with the current (-2I). The equation is 8 - 2I - 2I = 0. Choice B (8 - 2I + 2I = 0) incorrectly uses opposite signs for identical resistors, suggesting a misconception that resistors can have different sign conventions in the same loop. Apply the same sign convention consistently: voltage drops are negative when moving with the current.
A loop includes a $5.0\ \text{V}$ battery and a $2.0\ \text{V}$ battery aiding each other, in series with $R=7.0\ \Omega$ . Traverse counterclockwise; treat a battery crossing from $-$ to $+$ as $+\varepsilon$ and a resistor crossing with current as $-IR$ (current is counterclockwise). Which equation correctly represents the loop?
Answer: $+5.0 +2.0 - I(7.0)=0$ This problem tests understanding of Kirchhoff's loop rule. Kirchhoff's loop rule states that the algebraic sum of voltage changes around any closed loop must equal zero, reflecting the conservation of electric potential energy. Traversing counterclockwise with counterclockwise current, we cross both batteries from negative to positive (since they aid each other): +5.0V and +2.0V, then drop -I(7.0) across the resistor traversed with the current. Choice A incorrectly subtracts the second battery voltage, misunderstanding that aiding batteries have the same polarity orientation and both contribute positive voltage when traversed from - to +. The correct approach is to identify whether batteries aid (same orientation) or oppose (opposite orientation) before applying sign conventions consistently.
A closed loop has a $15\ \text{V}$ battery in series with two resistors, $R_1=2.0\ \Omega$ and $R_2=6.0\ \Omega$ . Traverse counterclockwise; define $+\varepsilon$ when crossing the battery from $-$ to $+$ and $-IR$ when crossing a resistor in the direction of current (current is counterclockwise). Which equation correctly represents the loop?
Answer: $+15 - I(2.0) - I(6.0)=0$ 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.
A closed loop has a $20\text{ V}$ battery, a $5.0\,\Omega$ resistor, and a second battery of emf $8.0\text{ V}$ that opposes the first for a clockwise traversal. Current $I$ is clockwise. Traverse clockwise using $+\mathcal{E}$ from $-$ to $+$ and $-\mathcal{E}$ from $+$ to $-$ , and use $-IR$ across the resistor with current. Which equation correctly represents the loop?
Answer: $+20-8-5I=0$ This question involves Kirchhoff's loop rule. Kirchhoff's loop rule requires the sum of all potential differences around a closed loop to equal zero, representing energy conservation. Traversing clockwise with clockwise current I: the 20V battery crossed from - to + gives +20V, the 8V battery crossed from + to - (opposing) gives -8V, and the 5Ω resistor traversed with current gives -5I. The equation is +20 - 8 - 5I = 0. Choice A (+20 + 8 - 5I = 0) incorrectly adds both battery voltages, missing that opposing batteries subtract. When batteries oppose in a circuit, pay careful attention to their polarities relative to your traversal direction.
A closed loop contains a $10\text{ V}$ ideal battery and a resistor $R=5.0\,\Omega$ . Current $I$ is clockwise. You traverse the loop counterclockwise starting at the battery's negative terminal. Use: battery $(-\to +)$ is $+\mathcal{E}$ , $(+\to -)$ is $-\mathcal{E}$ ; for a resistor, traversing in the direction of current gives $-IR$ and opposite gives $+IR$ . Which equation satisfies Kirchhoff's loop rule for your traversal?
Answer: $+10 + 5.0I = 0$ 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.
A closed loop contains two ideal batteries and one resistor: $\mathcal{E}_1=9.0\text{ V}$ (traversed from $-$ to $+$ in the chosen direction) and $\mathcal{E}_2=3.0\text{ V}$ (traversed from $+$ to $-$ in the chosen direction), plus $R=4.0\,\Omega$ with current $I$ in the loop. Traverse clockwise. Use: battery $(-\to +)$ is $+\mathcal{E}$ , $(+\to -)$ is $-\mathcal{E}$ ; resistor along current is $-IR$ . Which equation correctly represents the loop?
Answer: $+9.0 - 3.0 - 4.0I = 0$ 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.
A loop has a $12\ \text{V}$ battery and a resistor $R=6.0\ \Omega$ . Traverse clockwise, but the labeled current $I$ is counterclockwise. Use $+\varepsilon$ across the battery from $-$ to $+$ and $+IR$ when traversing a resistor opposite the current. Which equation correctly represents the loop?
Answer: $+12 + I(6.0)=0$ This problem involves applying Kirchhoff's loop rule. Kirchhoff's loop rule states that the sum of voltage changes around any closed loop must equal zero, ensuring conservation of energy as charge completes a circuit. Traversing clockwise while current flows counterclockwise, we cross the battery from negative to positive (+12V), then traverse the resistor opposite to the current direction, giving +I(6.0) rather than -I(6.0). Choice A incorrectly uses -I(6.0), failing to recognize that traversing against the current produces a voltage rise, not a drop. The key strategy is to distinguish between traverse direction and current direction: when they oppose, use +IR for resistors; when they align, use -IR.
A loop contains two series batteries, $9.0\text{ V}$ and $3.0\text{ V}$ , oriented so their emfs oppose. A resistor $R=6.0\,\Omega$ is also in series. Current $I$ is counterclockwise. Traverse counterclockwise: across a battery from $-$ to $+$ is $+\mathcal{E}$ , from $+$ to $-$ is $-\mathcal{E}$ ; across the resistor in the direction of current is $-IR$ . Which equation satisfies Kirchhoff's loop rule?
Answer: $+9-3-6I=0$ 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.
In one loop, two series resistors $R_1=4.0\,\Omega$ and $R_2=6.0\,\Omega$ carry clockwise current $I$ . An ideal battery of emf $\mathcal{E}$ is traversed from $-$ to $+$ when going clockwise. Traverse clockwise. Use the sign convention: battery $(-\to +)$ is $+\mathcal{E}$ ; each resistor along current is $-IR$ . Which equation correctly represents the loop rule?
Answer: $+\mathcal{E} - 4.0I - 6.0I = 0$ 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.
A closed loop contains a $9\ \text{V}$ battery and two resistors $R_1=3\ \Omega$ and $R_2=6\ \Omega$ in series. Current $I$ is counterclockwise. Traverse counterclockwise; treat crossing the battery from $-$ to $+$ as $+9\ \text{V}$ and each resistor in the direction of $I$ as $-IR$ . Which equation satisfies Kirchhoff's loop rule?
Answer: $9-I(3+6)=0$ This problem involves applying Kirchhoff's loop rule. Kirchhoff's loop rule ensures energy conservation by requiring that the sum of all voltage changes around a closed loop equals zero. Traversing counterclockwise with counterclockwise current I, we cross the 9V battery from negative to positive (+9V), then encounter R₁ = 3Ω in the direction of current (-3I), followed by R₂ = 6Ω also in the current direction (-6I). The loop equation becomes: 9 - 3I - 6I = 0, which simplifies to 9 - I(3+6) = 0. Choice B incorrectly uses +I(3+6), revealing the misconception of treating resistor voltage drops as positive when traversing in the current direction. Remember to consistently apply sign conventions: voltage drops across resistors are always negative when traversing in the direction of current flow.
A loop contains an ideal $15\text{ V}$ battery in series with three resistors: $R_1=1\,\Omega$ , $R_2=2\,\Omega$ , and $R_3=3\,\Omega$ . The current $I$ is clockwise. Traverse clockwise: battery from $-$ to $+$ is $+15$ , and each resistor with the current is a drop $-IR$ . Which equation correctly represents the loop?
Answer: $15-(1+2+3)I=0$ This problem tests understanding of Kirchhoff's loop rule. Kirchhoff's loop rule ensures that the algebraic sum of all voltage changes around a closed loop equals zero, conserving energy. Traversing clockwise with clockwise current, we cross the 15V battery from negative to positive (+15V), then encounter three resistors in series, each producing a voltage drop: R₁ (-1I), R₂ (-2I), and R₃ (-3I). The equation can be written as 15 - 1I - 2I - 3I = 0 or 15 - (1+2+3)I = 0. Choice C (15 - 1I - 2I = 0) omits the third resistor, revealing a misconception about accounting for all components. Always include every element in the loop, and remember that series resistors can be combined algebraically.
A single loop has an ideal $18\,\text{V}$ battery in series with $R_1=3.0\,\Omega$ and $R_2=6.0\,\Omega$ . Conventional current $I$ is clockwise. Traverse clockwise: crossing the battery from $-$ to $+$ is $+18\,\text{V}$ ; each resistor with current gives a drop $-IR$ . Which equation correctly represents the loop?
Answer: $+18-I(3+6)=0$ This problem involves applying Kirchhoff's loop rule to a simple series circuit with one battery and two resistors. Kirchhoff's loop rule requires that the sum of all voltage changes around a closed loop equals zero, reflecting the conservation of energy. Traversing clockwise: crossing the 18V battery from negative to positive terminal gives +18V; crossing each resistor in the direction of current I produces voltage drops -I(3) and -I(6), which combine to -I(3+6) for series resistors. Option D incorrectly includes only R₁ in the current term, demonstrating a misconception that series resistors contribute separately to the loop equation rather than having their resistances sum. When solving loop problems, remember that series resistors share the same current and their total resistance is the sum of individual resistances.
A loop contains two resistors, $R_1=4.0\ \Omega$ and $R_2=2.0\ \Omega$ , and a $10\ \text{V}$ battery. Traverse clockwise. Use the convention $+\varepsilon$ when moving from $-$ to $+$ through the battery, and $-IR$ when moving with current through a resistor (current is clockwise). Which equation correctly represents the loop?
Answer: $+10 - I(4.0) - I(2.0)=0$ 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.
A single loop has an ideal $18\ \text{V}$ battery and two series resistors, $R_1=9.0\ \Omega$ and $R_2=3.0\ \Omega$ . Traverse counterclockwise; take $-\varepsilon$ when crossing the battery from $+$ to $-$ and $-IR$ when crossing a resistor with the current (counterclockwise). Which equation correctly represents the loop?
Answer: $-18 - I(9.0) - I(3.0)=0$ 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 conservation of electric potential energy. Traversing counterclockwise with counterclockwise current, we cross the battery from positive to negative (-18V), then traverse both resistors with the current, experiencing drops of -I(9.0) and -I(3.0). Choice C incorrectly uses positive signs for the resistor terms, confusing the sign convention when traversing with the current versus against it. The reliable strategy is to establish clear sign conventions before starting: crossing a battery from + to - gives -ε, and traversing resistors with the current gives -IR.
A closed loop has two ideal batteries and one resistor: $\mathcal{E}_1=5\text{ V}$ , $\mathcal{E}_2=7\text{ V}$ , and $R=4\,\Omega$ . The current $I$ is counterclockwise. Traverse counterclockwise: moving from $-$ to $+$ across a battery is $+\mathcal{E}$ , from $+$ to $-$ is $-\mathcal{E}$ , and through the resistor with the current is $-IR$ . If both batteries are traversed from $-$ to $+$ , which equation satisfies the loop rule?
Answer: $5+7-4I=0$ This question requires applying Kirchhoff's loop rule. Kirchhoff's loop rule states that the total voltage change around any closed loop must be zero, ensuring energy conservation. Traversing counterclockwise with counterclockwise current, both batteries are crossed from negative to positive (ℰ₁: +5V, ℰ₂: +7V), and the 4Ω resistor is traversed with the current (-4I). The equation becomes 5 + 7 - 4I = 0. Choice B (5 - 7 - 4I = 0) incorrectly subtracts the second battery voltage, indicating confusion about applying sign conventions when both batteries aid the current. Remember that the sign depends on how you traverse each component, not on their relative values.
One loop contains a $14\ \text{V}$ battery, then (clockwise) a resistor $R_1=4\ \Omega$ , then a second battery $4\ \text{V}$ encountered from $-$ to $+$ while traversing clockwise, then a resistor $R_2=6\ \Omega$ . Current $I$ is clockwise. Using $+\mathcal{E}$ for $-$ to $+$ and $-IR$ for resistors along $I$ , which equation represents the loop?
Answer: $14+4-I(4+6)=0$ 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.

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