Kirchhoff's Loop Rule - AP Physics C Electricity and Magnetism

1. Which equation is a valid application of Kirchhoff's loop rule for the loop containing the battery, resistor $R_1$, and resistor $R_3$?
   Answer: $\mathcal{E} - I_1 R_1 - I_3 R_3 = 0$ For the specified loop, we sum the potential changes. Starting from the negative terminal of the battery, we gain $\mathcal{E}$. Then, we traverse $R_1$ in the direction of current $I_1$, so the potential drops by $-I_1 R_1$. Finally, we traverse $R_3$ in the direction of current $I_3$, so the potential drops by $-I_3 R_3$. Setting the sum to zero gives $\mathcal{E} - I_1 R_1 - I_3 R_3 = 0$.
2. According to Kirchhoff's loop rule, what must be true about the potential drops across the resistors in this same clockwise traversal?
   Answer: The sum of the potential drops across the resistors must also be $+15\,\text{V}$. Kirchhoff's loop rule states that the sum of all potential changes around a closed loop is zero: $\sum \Delta V = 0$. This can be written as $\sum \mathcal{E} + \sum (\text{potential changes across resistors}) = 0$. If the sum of EMFs is $+15\,\text{V}$, then the sum of potential changes across resistors must be $-15\,\text{V}$. However, when we speak of 'potential drops,' we refer to the magnitude of the voltage decrease, which is a positive quantity. Therefore, the sum of potential drops across the resistors is $+15\,\text{V}$.
3. Which equation correctly represents Kirchhoff's loop rule for the loop consisting of the top and middle branches, traversed clockwise starting from the leftmost junction?
   Answer: $+\mathcal{E}_1 - I_1 R_1 + I_2 R_2 = 0$ Traversing the top branch clockwise (to the right) from the left junction, we go through the battery ($+\mathcal{E}_1$) and the resistor ($-I_1 R_1$). Then, traversing the middle branch clockwise (to the left), we go against the direction of current $I_2$, resulting in a potential gain of $+I_2 R_2$. Summing these gives $+\mathcal{E}_1 - I_1 R_1 + I_2 R_2 = 0$.
4. Which differential equation correctly describes the circuit for $t>0$ based on Kirchhoff's loop rule?
   Answer: $\mathcal{E} - IR - L\frac{dI}{dt} = 0$ Applying the loop rule in the direction of the current: the battery provides a potential gain of $\mathcal{E}$. The resistor causes a potential drop of $IR$. As the current is increasing, the inductor generates a back EMF that opposes this change, resulting in a potential drop of $L(dI/dt)$. The sum of potential changes is zero: $\mathcal{E} - IR - L(dI/dt) = 0$.
5. After the switch has been closed for a very long time (in the steady state), which equation correctly describes the circuit based on Kirchhoff's loop rule?
   Answer: $\mathcal{E} - \frac{Q_{max}}{C} = 0$, where $Q_{max}$ is the final charge on the capacitor. After a very long time, the circuit reaches a steady state. The inductor acts as a short circuit (a wire with zero resistance), so the potential drop across it is zero. The capacitor becomes fully charged and acts as an open circuit, allowing no DC current to flow through its branch. Therefore, the current through the inductor branch becomes zero. The potential difference across the parallel branches must be equal. The potential across the inductor branch is $V_L + V_{R1} = 0 + I R_1 = 0$, so I=0. The loop containing the battery and the capacitor gives $+\mathcal{E} - V_C = 0$, or $\mathcal{E} - Q_{max}/C = 0$.
6. Which aspect of this graph is a direct illustration of Kirchhoff's loop rule?
   Answer: The sum of the magnitudes of the potential drops across the resistors is equal to the battery's EMF. Kirchhoff's loop rule states the sum of potential changes around a loop is zero ($\sum \Delta V = 0$). In this circuit, this means $+\mathcal{E} - \sum IR_i = 0$, or $\mathcal{E} = \sum IR_i$. The graph shows the potential gain from the battery ($\mathcal{E}$) is exactly cancelled by the sum of the potential drops across the resistors ($\sum IR_i$) as the potential returns to its starting value. This directly illustrates the rule.
7. What is the error in the student's equation?
   Answer: The term for the inductor should be negative because its induced EMF opposes the current increase. When traversing a circuit in the direction of an increasing current, the inductor creates a back-EMF that opposes this increase. This back-EMF corresponds to a potential drop. Therefore, the term for the inductor should be $-L(dI/dt)$. The student has the wrong sign for this term.
8. A battery with emf $\varepsilon=18\,\text{V}$ and internal resistance $r=2.0\,\Omega$ is connected in series with an external resistor $R=7.0\,\Omega$ to form one closed loop. Using the loop rule, what is the potential difference across the battery terminals while the circuit is operating?
   Answer: 14.0 V This question tests AP Physics C concepts of applying Kirchhoff's Loop Rule in circuit analysis. Kirchhoff's Loop Rule states that the algebraic sum of voltage changes around a closed loop must equal zero. In this circuit, a battery with emf ε=18V and internal resistance r=2.0Ω connects to an external resistor R=7.0Ω. Choice A (14.0 V) is correct because the current is I = ε/(R+r) = 18/(7+2) = 2A, making the terminal voltage = ε - Ir = 18 - 2(2) = 14V. Choice B (18.0 V) is incorrect as it ignores the internal resistance voltage drop. To help students: Stress that terminal voltage equals emf minus the internal voltage drop. Practice problems with varying internal resistance values to build intuition.
9. Which fundamental physical principle is Kirchhoff's loop rule a direct consequence of?
   Answer: The conservation of energy for a charge moving in a closed path. Kirchhoff's loop rule states that the sum of the electric potential differences around any closed circuit loop is zero. This is a statement of conservation of energy. If a charge moves around a closed loop and returns to its starting point, its potential energy must be the same, so the net work done on the charge by the electric field (and thus the net change in potential) must be zero.
10. Given that $I(t) = dQ/dt$, which of the following differential equations correctly describes this LC circuit based on Kirchhoff's loop rule?
    Answer: $\frac{Q}{C} + L\frac{d^2Q}{dt^2} = 0$ Applying Kirchhoff's loop rule, the sum of potential differences around the loop must be zero. The potential difference across the capacitor is $Q/C$ and across the inductor is $L(dI/dt)$. Following standard sign conventions for a loop traversal in the direction of current flow: $\frac{Q}{C} + L\frac{dI}{dt} = 0$. Since $I = dQ/dt$, we have $dI/dt = d^2Q/dt^2$, giving us $\frac{Q}{C} + L\frac{d^2Q}{dt^2} = 0$.
11. A battery with emf $\varepsilon=15\,\text{V}$ and internal resistance $r=0.50\,\Omega$ is connected in series with an external resistor $R=4.5\,\Omega$. Using the loop rule, what is the potential difference across the battery terminals during operation?
    Answer: 13.5 V This question tests AP Physics C concepts of applying Kirchhoff's Loop Rule in circuit analysis. Kirchhoff's Loop Rule requires that the sum of all voltage changes around a closed loop equals zero. In this circuit, a battery with emf ε=15V and internal resistance r=0.50Ω connects to external resistor R=4.5Ω. Choice A (13.5 V) is correct because the current is I = ε/(R+r) = 15/(4.5+0.5) = 3A, making terminal voltage = ε - Ir = 15 - 3(0.5) = 13.5V. Choice B (15.0 V) is incorrect as it neglects the internal resistance effect. To help students: Reinforce that terminal voltage is always less than emf when current flows. Practice calculating both current and terminal voltage in circuits with internal resistance.
12. Which of the following is the correct expression for Kirchhoff's loop rule applied to this circuit, accounting for the internal resistance?
    Answer: $\mathcal{E} - I(R+r) = 0$ The internal resistance $r$ acts as a resistor in series with the ideal EMF source. Applying the loop rule, we have a potential gain of $\mathcal{E}$ from the ideal source, a potential drop of $-Ir$ across the internal resistance, and a potential drop of $-IR$ across the external resistor. The equation is $\mathcal{E} - Ir - IR = 0$, which can be rewritten as $\mathcal{E} - I(R+r) = 0$.
13. In the context of the discharging process where current $I$ flows away from the positive plate, the current is related to the charge $Q$ by $I = -\frac{dQ}{dt}$. Which differential equation accurately describes the charge $Q$ on the capacitor?
    Answer: $-R\frac{dQ}{dt} - \frac{Q}{C} = 0$ The question presents the loop rule equation as $IR - Q/C = 0$. This implies a traversal direction opposite to the standard one, but it is internally consistent. If we substitute $I = -dQ/dt$ into this given equation, we get $(-dQ/dt)R - Q/C = 0$, which is $-R(dQ/dt) - Q/C = 0$. Note that the standard formulation $+Q/C - IR = 0$ leads to $+Q/C - (-dQ/dt)R = 0$, or $Q/C + R(dQ/dt) = 0$, which is mathematically equivalent.
14. A battery with emf $\varepsilon=18\,\text{V}$ and internal resistance $r=2.0\,\Omega$ is connected to an external load resistor $R=7.0\,\Omega$ in series (single loop). Using Kirchhoff's Loop Rule, what is the potential difference across the battery terminals while delivering current to the load?
    Answer: 14.0 V This question tests AP Physics C concepts of applying Kirchhoff's Loop Rule with internal resistance. Kirchhoff's Loop Rule states that the sum of all voltage changes around a closed loop must equal zero. In this circuit, an 18V battery with 2Ω internal resistance connects to a 7Ω load. Applying the loop rule: 18V - I(2Ω) - I(7Ω) = 0, giving I = 18V/9Ω = 2A. The terminal voltage equals the emf minus the internal voltage drop: V_terminal = 18V - (2A)(2Ω) = 14V, making choice A correct. Choice B (18.0 V) is incorrect as it ignores the internal resistance effect. To help students: Emphasize that real batteries have internal resistance affecting terminal voltage. Practice problems involving power delivery and efficiency with internal resistance.
15. Immediately after the switch is closed ($t=0^+$), what is the correct application of Kirchhoff's loop rule for the outer loop containing the battery, $R_1$, and $R_2$?
    Answer: $\mathcal{E} - I_1 R_1 - I_2 R_2 = 0$ Immediately after the switch is closed, the current through the inductor $I_L$ must be zero, because current in an inductor cannot change instantaneously. By Kirchhoff's junction rule, the total current $I_1$ must therefore pass entirely through resistor $R_2$, so $I_1 = I_2$. The loop rule for the outer loop is $+\mathcal{E} - I_1 R_1 - I_2 R_2 = 0$. This equation is valid at all times, including at $t=0^+$.
16. By applying Kirchhoff's loop rule, what is the magnitude of the current flowing through the resistor?
    Answer: $1.5\,\text{A}$ Let's assume the current $I$ flows in the direction driven by the larger battery, $\mathcal{E}_1$ (counter-clockwise). Traversing the loop in this direction, we get a potential gain $+\mathcal{E}_1$, a drop $-IR$, and a drop $- \mathcal{E}_2$ (since we traverse it from positive to negative). The loop rule is $+\mathcal{E}_1 - \mathcal{E}_2 - IR = 0$. Solving for $I$: $I = (\mathcal{E}_1 - \mathcal{E}_2) / R = (12\,\text{V} - 3\,\text{V}) / 6\,\Omega = 9\,\text{V} / 6\,\Omega = 1.5\,\text{A}$.
17. Which of the following differential equations is a correct application of Kirchhoff's loop rule to this charging circuit for $t>0$?
    Answer: $\mathcal{E} - R \frac{dQ}{dt} - \frac{Q}{C} = 0$ Applying the loop rule, we sum the potential changes around the circuit. The battery provides a potential gain of $\mathcal{E}$. The resistor causes a potential drop of $IR = R(dQ/dt)$. The capacitor causes a potential drop of $Q/C$. Setting the sum to zero gives $\mathcal{E} - R(dQ/dt) - Q/C = 0$.
18. Which of the following equations correctly represents Kirchhoff's loop rule for this discharging circuit?
    Answer: $+\frac{Q}{C} - IR = 0$ Starting at the negative plate of the capacitor and traversing through it to the positive plate, there is a potential gain of $+Q/C$. Continuing through the resistor in the direction of the current $I$, there is a potential drop of $-IR$. Summing these potential changes around the loop to zero gives $+Q/C - IR = 0$. Note that for discharging, $I = -dQ/dt$.
19. Which of the following equations correctly represents Kirchhoff's loop rule for this circuit when traversing the loop clockwise, starting from the negative terminal of the battery?
    Answer: $+\mathcal{E} - IR - \frac{Q}{C} = 0$ Traversing the loop clockwise: the battery provides a potential gain of $+\mathcal{E}$. The resistor causes a potential drop of $-IR$ when traversed in the direction of the current. The capacitor, being charged by the clockwise current, has its positive plate encountered first, resulting in a potential drop of $-\frac{Q}{C}$. Summing these potential changes to zero gives $+\mathcal{E} - IR - \frac{Q}{C} = 0$.
20. Which equation correctly applies Kirchhoff's loop rule to this AC circuit?
    Answer: $\mathcal{E}_0 \cos(\omega t) - I(t)R = 0$ Kirchhoff's loop rule applies at any instant in time. At any time $t$, the potential gain from the generator is $\mathcal{E}(t) = \mathcal{E}_0 \cos(\omega t)$. The potential drop across the resistor is $I(t)R$. Summing these potential changes around the loop to zero gives $\mathcal{E}_0 \cos(\omega t) - I(t)R = 0$.

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