Current is the time derivative of charge, $$I(t) = dQ/dt$$. Differentiating the given function: $$I(t) = \frac{d}{dt} [(10 , \mu\text{C})(1 - e^{-t/(2.0\text{ s})})] = (10 , \mu\text{C}) (-(-\frac{1}{2.0\text{ s}})e^{-t/(2.0\text{ s})}) = (5.0 , \mu\text{A}) e^{-t/(2.0\text{ s})}$$. To find the current at $$t=0$$, we evaluate this expression at $$t=0$$: $$I(0) = (5.0 , \mu\text{A}) e^{0} = 5.0 , \mu\text{A}$$.

At time $$t=0$$, the uncharged capacitor acts as a short circuit (a wire with zero resistance). Resistor 2 is in parallel with this short circuit, so the equivalent resistance of the parallel branch is 0. The total resistance of the circuit is just $$R_1 = 2 , \Omega$$. The initial current through resistor 1 is $$I_1(0) = E/R_1 = 12 \text{ V} / 2 , \Omega = 6.0 \text{ A}$$. After a very long time ($$t \to \infty$$), the capacitor is fully charged and acts as an open circuit. No current flows through the capacitor's branch. The circuit then behaves as a simple series circuit with resistor 1 and resistor 2. The total resistance is $$R_{total} = R_1 + R_2 = 2 , \Omega + 4 , \Omega = 6 , \Omega$$. The final current through resistor 1 is $$I_1(\infty) = E/R_{total} = 12 \text{ V} / 6 , \Omega = 2.0 \text{ A}$$.

The total energy supplied by the battery is the total charge that flows, $$Q_{final} = CE$$, multiplied by the battery emf $$E$$, which is $$W_{batt} = Q_{final}E = (CE)E = CE^2$$. The energy stored in the fully charged capacitor is $$U_C = \frac{1}{2}CV^2 = \frac{1}{2}CE^2$$. By conservation of energy, the energy dissipated in the resistor is the difference between the energy supplied by the battery and the energy stored in the capacitor: $$E_{dissipated} = W_{batt} - U_C = CE^2 - \frac{1}{2}CE^2 = \frac{1}{2}CE^2$$.

For capacitors in parallel, the equivalent capacitance is the sum of the individual capacitances: $$C_{eq} = C_1 + C_2 = 3.0 , \mu\text{F} + 6.0 , \mu\text{F} = 9.0 , \mu\text{F}$$. The time constant of an RC circuit is given by the product of the resistance and the equivalent capacitance: $$τ = RC_{eq} = (2.0 , \Omega)(9.0 , \mu\text{F}) = 18 , \mu\text{s}$$.

During the discharge of a capacitor through a resistor in a simple loop, the potential difference across the capacitor decreases exponentially according to $$V_C(t) = V_0 e^{-t/RC}$$. By Kirchhoff's loop rule for this circuit, the magnitude of the potential difference across the resistor must be equal to the potential difference across the capacitor at all times. Thus, $$V_R(t) = V_C(t) = V_0 e^{-t/RC}$$.

This question tests AP Physics C: Electricity and Magnetism skills, specifically analyzing capacitor discharge in RC circuits. During discharge, voltage follows V(t) = V₀e^(-t/RC), where V₀ is the initial voltage across the capacitor. With R = 3.0 kΩ and C = 100 μF, the time constant is τ = RC = (3.0×10³)(100×10⁻⁶) = 0.30 s, and the capacitor starts at V₀ = 6.0 V. Choice A is correct because after t = 0.60 s (two time constants), V_C = 6.0×e^(-0.60/0.30) = 6.0×e^(-2) = 6.0×0.135 = 0.81 V. Choice B might result from using only one time constant (6.0×e^(-1) ≈ 2.2 V) or calculation error, while choice D might incorrectly use the charging formula. To help students: remember that after two time constants, voltage drops to about 13.5% of initial value (e^(-2) ≈ 0.135), and practice recognizing multiples of the time constant. Watch for confusion between charging and discharging scenarios and ensure proper exponential calculations.

This question tests AP Physics C: Electricity and Magnetism skills, specifically determining the time required for a capacitor to reach a specific voltage during charging. For a charging capacitor, V(t) = V₀(1-e^(-t/RC)), which can be rearranged to solve for time: t = -RC×ln(1-V/V₀). With R = 5.0 kΩ, C = 100 μF, the time constant is τ = RC = (5.0×10³)(100×10⁻⁶) = 0.50 s, and we need the time to reach 6.0 V from 0 V with V₀ = 10 V. Choice A is correct because t = -0.50×ln(1-6.0/10) = -0.50×ln(0.40) = -0.50×(-0.916) = 0.46 s. Choice B might result from incorrectly using t = τ×(V/V₀) = 0.50×(6/10) = 0.30 s, assuming linear rather than exponential growth. To help students: practice rearranging exponential equations, use natural logarithms correctly, and understand that reaching 60% of final voltage takes less than one time constant. Watch for algebraic errors when solving exponential equations and confusion about when to use ln versus other operations.

This question tests AP Physics C: Electricity and Magnetism skills, specifically calculating voltage during capacitor discharge. For a discharging capacitor, V(t) = V₀e^(-t/RC), where V₀ is the initial voltage. With R = 10 kΩ and C = 10 μF, the time constant is τ = RC = (10×10³)(10×10⁻⁶) = 0.10 s, and the capacitor starts at V₀ = 12 V. Choice A is correct because after t = 0.050 s (half a time constant), V_C = 12×e^(-0.050/0.10) = 12×e^(-0.5) = 12×0.606 = 7.3 V. Choice B might result from using the wrong exponential value or calculation error, while choice C might incorrectly apply the charging formula. To help students: memorize that after half a time constant of discharge, voltage drops to about 60.6% of initial value, and practice quick mental calculations with common exponential values. Watch for confusion between charging and discharging formulas and ensure accurate exponential calculations.

This question tests AP Physics C: Electricity and Magnetism skills, specifically analyzing RC circuits during capacitor discharge. For a discharging capacitor, the voltage follows V(t) = V₀e^(-t/RC), where V₀ is the initial voltage and τ = RC is the time constant. In this circuit with R = 1.5 kΩ and C = 2.0×10⁻⁴ F, the time constant is τ = (1.5×10³)(2.0×10⁻⁴) = 0.30 s, with initial voltage V₀ = 9.0 V. Choice A is correct because after t = 0.30 s (one time constant), V_C = 9.0×e^(-0.30/0.30) = 9.0×e^(-1) = 9.0×0.368 = 3.3 V. Choice B might result from incorrectly using the charging formula (9.0×(1-e^(-1)) ≈ 5.7 V) instead of the discharging formula. To help students: emphasize the difference between charging and discharging equations, and remember that after one time constant of discharge, voltage drops to about 36.8% of initial value. Watch for sign errors and confusion about which formula applies to which scenario.

This question tests AP Physics C: Electricity and Magnetism skills, specifically analyzing RC circuits during capacitor discharge. For a discharging capacitor, the voltage decreases exponentially according to V(t) = V₀e^(-t/RC), where V₀ is the initial voltage. In this circuit, with R = 4.7×10³ Ω and C = 4.7×10⁻⁶ F, the time constant is τ = RC = 0.0221 s, and we need the voltage after t = 0.022 s. Choice A is correct because V(0.022) = 10×e^(-0.022/0.0221) = 10×e^(-0.995) = 10×0.370 ≈ 3.7 V. Choice B is incorrect because it appears to use the charging formula V = V₀(1-e^(-t/RC)) instead of the discharging formula, giving approximately 6.3 V. To help students: clearly distinguish between charging and discharging formulas, emphasize that discharge starts from initial voltage and decays to zero, and practice identifying initial conditions. Watch for: mixing up charging and discharging equations, which is the most common error in RC circuit problems.