This question tests understanding of resistor-capacitor (RC) circuits. During charging, current flows from the battery through the resistor to build up charge on the capacitor plates, creating a voltage across the capacitor that opposes the battery. As time progresses, this opposing voltage increases, reducing the current until equilibrium is reached when the capacitor voltage equals the battery voltage and current stops flowing. At this steady state, the capacitor acts like an open circuit with no current flow, maintaining the battery voltage indefinitely. Choice C incorrectly suggests unlimited voltage increase, representing the misconception that capacitors can somehow generate voltage beyond their source—in reality, a capacitor can never exceed its charging source voltage in a DC circuit. When analyzing RC circuits at long times, remember that capacitors block DC current once fully charged, creating an open circuit condition.

This question tests understanding of resistor-capacitor (RC) circuits. When a charged capacitor discharges through a resistor, the capacitor initially maintains its full voltage (12V in this case), which appears entirely across the resistor since they are in a simple loop. As the capacitor discharges, its stored charge decreases, causing both the capacitor voltage and the resistor voltage to decrease exponentially toward zero. The current flows in one direction only, depleting the capacitor's charge until no voltage remains across either component. Choice A incorrectly assumes the resistor somehow maintains a fixed voltage, which represents the misconception that resistors generate or maintain voltages independently rather than having voltage drops proportional to current (V = IR). To solve RC problems, remember that in a discharging circuit, both the capacitor and resistor voltages decay together following the same exponential time constant.

This question tests understanding of resistor-capacitor (RC) circuits. When charging a capacitor from a battery, the capacitor voltage increases exponentially according to V_C = V_battery(1 - e^(-t/RC)), starting from 0V and asymptotically approaching the battery voltage. As V_C approaches 9.0V, the voltage across the resistor (V_R = V_battery - V_C) approaches zero, which means the current (I = V_R/R) also approaches zero. Once fully charged, the capacitor acts like an open circuit with no current flow. Choice C incorrectly suggests V_C can exceed the battery voltage, violating energy conservation—a capacitor cannot store more energy than the source provides. Remember that in steady state (t → ∞), capacitors act like open circuits with V_C = V_battery and I = 0.

This question tests understanding of resistor-capacitor (RC) circuits. When a charged capacitor discharges through a resistor, current flows from the positive plate through the resistor to the negative plate, maintaining one consistent direction throughout the discharge. The current magnitude starts at I₀ = V_C/R and decreases exponentially as I = I₀e^(-t/RC) because the capacitor voltage driving the current also decreases exponentially. The current never reverses direction because the capacitor polarity remains the same—it just loses charge magnitude. Choice A incorrectly suggests oscillating current, confusing RC circuits with LC circuits which can oscillate. In RC circuits, all quantities (V_C, I, and Q) decay exponentially without oscillation—think of it as energy dissipating in the resistor, not bouncing back and forth.

This question tests understanding of resistor-capacitor (RC) circuits. During charging, the current follows I = (V_battery/R)e^(-t/RC), starting at its maximum value I₀ = V_battery/R when the capacitor has zero voltage and acts like a short circuit. As time progresses and the capacitor voltage builds up to oppose the battery, the current decreases exponentially, approaching zero at long times when the capacitor is fully charged and acts like an open circuit. The current is always positive (same direction) but decreases monotonically from maximum to zero. Choice B incorrectly reverses the time behavior, representing the misconception that capacitors initially block current like they do in steady-state DC circuits—in reality, uncharged capacitors offer no initial opposition to current flow. To remember RC behavior, think of capacitors as initially transparent to current when uncharged but eventually blocking DC current when fully charged.

This question tests understanding of resistor-capacitor (RC) circuits. During discharge, the current at any time is I = (V_C/R)e^(-t/RC), where the time constant τ = RC determines the decay rate. With doubled resistance (2R), the initial current is halved (V_C/2R versus V_C/R) and the time constant doubles (2RC versus RC), making the exponential decay twice as slow. This means at any given time t > 0, the current with 2R is less than half the current with R because it started smaller and decays more slowly. Choice B incorrectly suggests faster decay with larger current, representing the misconception that higher resistance somehow increases current flow—Ohm's law clearly shows resistance opposes current. To analyze modified RC circuits, consider how changes affect both the initial value and the time constant of the exponential behavior.

This question tests understanding of resistor-capacitor (RC) circuits. When a charged capacitor discharges through a resistor, the initial current is determined by the capacitor's voltage and the resistance (I₀ = V_C/R), flowing in the direction that depletes the capacitor's charge. As charge leaves the capacitor, its voltage decreases, which proportionally decreases the current through the resistor, creating an exponential decay toward zero current. The current maintains the same direction throughout the discharge process because charge always flows from higher to lower potential. Choice A incorrectly suggests alternating current, representing the misconception that capacitors inherently create AC behavior—this only occurs in AC circuits, not in DC discharge scenarios. To analyze discharge circuits, remember that current direction is set by initial charge distribution and magnitude always decreases monotonically.

This question tests understanding of resistor-capacitor (RC) circuits. When a charged capacitor discharges through a resistor, it initially has some voltage V₀ and drives current through the resistor. As charge flows off the capacitor plates, the voltage across the capacitor decreases exponentially according to V_C = V₀e^(-t/RC). Since there's no battery to maintain voltage, the capacitor continues losing charge until it's completely discharged and V_C approaches 0V. Choice A incorrectly suggests the capacitor voltage approaches battery voltage, but there's no battery in a discharging circuit—only the resistor provides a path for current. When analyzing RC circuits, remember that discharging means the capacitor is the only energy source, and it will eventually deplete completely.

This question tests understanding of resistor-capacitor (RC) circuits. During discharge, the charged capacitor acts as a voltage source with the positive plate at higher potential than the negative plate. Conventional current always flows from higher to lower potential through the external circuit (the resistor), so current flows from the positive plate, through the resistor, to the negative plate, gradually neutralizing the charge separation. This current direction is the same as when the capacitor was charging, just with the battery removed—positive charges effectively move from the positive plate to neutralize negative charges on the other plate. The current magnitude decreases exponentially but maintains the same direction throughout the discharge. Choice A incorrectly reverses the current direction, representing the misconception that current flows from negative to positive or that discharge current opposes charging current. To determine current direction in RC circuits, identify which terminal is at higher potential and remember that conventional current flows from high to low potential through resistors.

This question tests understanding of resistor-capacitor (RC) circuits. At t = 0+, an uncharged capacitor acts like a short circuit (zero voltage), so the full battery voltage appears across the resistor, giving initial current I₀ = V/R. Circuit 1 with resistance R has initial current V/R, while Circuit 2 with resistance 2R has initial current V/(2R), which is half as large. Choice D incorrectly assumes zero initial current, confusing the final steady state (when capacitors are fully charged) with the initial transient state. To find initial conditions in RC circuits, treat uncharged capacitors as short circuits and fully charged capacitors as open circuits.