In the figure R1 = 9.81 kΩ, R2 = 15.4 kΩ, C = 0.433 μF, and the ideal battery has emf ε = 18.0 V. First, the switch is closed a long time so that the steady state is reached. Then the switch is opened at time t = 0. What is the current in resistor 2 at t = 3.60 ms? The image depicts an electrical circuit diagram. Here are the key components and layout:1. **Battery (\(\mathcal{E}\))**: Represented by the standard symbol for a voltage source, it's positioned on the left side. The positive terminal is marked with a plus sign.2. **Resistor \(R_1\)**: Located immediately after the battery on the lower path of the circuit. It's annotated as \(R_1\).3. **Current Direction**: Illustrated by a red arrow pointing upwards between the battery’s terminals, indicating the conventional flow of current from positive to negative.4. **Switch**: Situated above \(R_1\) and before the junction that leads to other components of the circuit. It is shown in the open state.5. **Resistor \(R_2\)**: Connected in series with a capacitor in the upper path of the circuit. Labeled as \(R_2\).6. **Capacitor (C)**: Placed parallel to \(R_1\) and connected in series with \(R_2\). It's denoted by the symbol for a capacitor and marked with a letter "C".Overall, the circuit is a basic RC (resistor-capacitor) series circuit which can be used to study charging and discharging behavior of capacitors in electrical circuits. The switch allows for control over when the circuit is closed for these processes to occur.

Let R1 and R2 denote the two resistors, C denotes the capacitor, e denotes the emf of the battery, V denotes the voltage across the resistor R2 at the steady-state. At the steady-state, the capacitor does not flow any current and the voltage across the capacitor is equal to the voltage across the resistor R2. Therefore, apply the voltage division rule. Let I denotes the current at time t. Therefore, apply the formula for the capacitor discharging through the resistor.