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 $$\mathcal{E} - Q_{max}/C = 0$$.

First, find the total resistance $$R_{tot} = 2,\Omega + 4,\Omega = 6,\Omega$$. The current is $$I = \mathcal{E}/R_{tot} = 12,\text{V} / 6,\Omega = 2,\text{A}$$. Starting from the negative terminal (0 V), the potential at the positive terminal is $$+12,\text{V}$$. Traversing through the $$2,\Omega$$ resistor in the direction of the current, the potential drops by $$IR = (2,\text{A})(2,\Omega) = 4,\text{V}$$. The potential at the point between the resistors is $$12,\text{V} - 4,\text{V} = 8,\text{V}$$.

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 $$I$$: $$I = (\mathcal{E}_1 - \mathcal{E}_2) / R = (12,\text{V} - 3,\text{V}) / 6,\Omega = 9,\text{V} / 6,\Omega = 1.5,\text{A}$$.

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.

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.

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.

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.

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.

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 $$+\mathcal{E} - IR - \frac{Q}{C} = 0$$.

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 $$+\mathcal{E}_1 - I_1 R_1 + I_2 R_2 = 0$$.