Let's solve **part (c)** step by step: --- ### Step 1: Write the Equation for Charging a Capacitor The voltage across a charging capacitor in an RC circuit is given by: \[ V_C(t) = \mathcal{E} \left(1 - e^{-t/RC}\right) \] Given: - \(\mathcal{E} = 12\, \text{V}\) - \(V_C = 4\, \text{V}\) - \(C = 60\, \text{mF} = .060\, \text{F}\) - \(R = R_1 = 470\, \Omega\) Plug these into the equation: \[ 4 = 12 \left(1 - e^{-t/(470 \times .060)}\right) \] **Explanation:** We use the standard RC charging equation, substituting the given values for voltage, resistance, and capacitance. This gives us an equation in terms of \(t\). --- ### Step 2: Solve for Time \(t\) First, isolate the exponential: \[ \frac{4}{12} = 1 - e^{-t/(282)} \] \[ \frac{1}{3} = 1 - e^{-t/282} \] \[ e^{-t/282} = 1 - \frac{1}{3} = \frac{2}{3} \] Take the natural logarithm on both sides: \[ -t/282 = \ln\left(\frac{2}{3}\right) \] \[ t = -282 \times \ln\left(\frac{2}{3}\right) \] Calculate: \[ \ln\left(\frac{2}{3}\right) \approx -.405 \] \[ t = -282 \times (-.405) \approx 114 \text{ seconds} \] **Explanation:** By isolating the exponential and solving for \(t\), we find how long it takes for the capacitor to reach 4 V. --- ## **Final Answer** \[ \boxed{t \approx 114 \ \text{seconds}} \] This is the time it takes for the voltage across the capacitor to reach 4 V.

Let's solve **parts (c)** and **(d)** step by step: --- ## Part (c): Sketching Current vs. Time ### Step 1: Understand the Current Flow in the Circuit When the switch \(S\) is closed at \(t = 0.5\) seconds, the circuit consists of the resistor \(R_1\) and the capacitor \(C\). The current through \(R_1\) is given by: \[ I(t) = \frac{\mathcal{E} - V_C(t)}{R} \] Where: - \(I\) is the current through the resistor, - \(\mathcal{E}\) is the battery voltage, - \(V_C(t)\) is the voltage across the capacitor, - \(R\) is the resistor value. ### Step 2: Sketch the Currents \(I_R\) and \(I_C\) 1. **Before \(t = 0.5\):** - The capacitor is fully charged, so \(V_C = 12\, \text{V}\). - Current \(I_R = 0\, \text{A}\) (no current flows). 2. **At \(t = 0.5\):** - The switch is closed; current starts flowing. - \(I_R\) increases from 0 to its maximum as the capacitor discharges. 3. **After \(t = 0.5\):** - Current \(I_R\) will decrease exponentially as the capacitor discharges. ### Sketch: - Draw a time axis with a vertical axis for current. - At \(t < 0.5\), \(I_R = 0\). - At \(t = 0.5\), \(I_R\) starts at a maximum value and decreases exponentially towards zero. **Explanation:** This sketch represents how the current through resistor \(R_1\) behaves when the switch is closed, starting from zero when the capacitor is charged and decaying as the capacitor discharges. --- ## Part (d): Analyzing the Circuit Behavior ### Step 1: Current Behavior After Closing the Switch After closing the switch, the capacitor starts discharging through the resistor \(R_1\). The current can be described as: \[ I(t) = I_0 e^{-t/RC} \] Where: - \(I_0\) is the initial current at \(t = 0.5\), - \(R = 470\, \Omega\), - \(C = 60\, \text{mF} = 0.060\, \text{F}\). ### Step 2: Determine Initial Current \(I_0\) At \(t = 0.5\), the voltage across the capacitor \(V_C\) can be found from the previous calculations. Using \(V_C(t)\) at this time, we can find the initial current: \[ I_0 = \frac{\mathcal{E} - V_C(0.5)}{R} \] Assuming \(V_C(0.5) = 8\, \text{V}\) (for example): \[ I_0 = \frac{12 - 8}{470} \approx 0.00851\, \text{A} \text{ or } 8.51\, \text{mA} \] **Explanation:** This step shows how to calculate the initial current that flows immediately after the switch is closed based on the capacitor's voltage at that moment. --- ## **Final Summary** - The sketch illustrates the behavior of the current through the resistor when the switch is closed at \(t = 0.5\). - The discharge current follows an exponential decay pattern, starting from an initial value based on the voltage across the capacitor.