Let's solve the problem step by step. --- ### **Step 1: Find the current \( I(t) \) as a function of time \( t \)** **Explanation:** - Initially, **one capacitor is charged** to \( V_0 \), the other is uncharged. - At \( t=0 \), the switch is closed, connecting both capacitors and the resistor in series. #### **Step 1.1: Initial conditions and equivalent circuit** - Capacitance of each capacitor: \( C \) - Resistance: \( R \) - Initial voltage across the charged capacitor: \( V_0 \) - The two capacitors are now **in parallel** (after closing the switch). - The total charge initially: \( Q_0 = CV_0 \) #### **Step 1.2: Final voltage after equilibrium** When switch is closed, the charge redistributes equally: - Total charge remains the same: \( Q_0 \) - Total capacitance (in parallel): \( C_{eq} = 2C \) - Final voltage: \( V_f = \frac{Q_0}{2C} = \frac{CV_0}{2C} = \frac{V_0}{2} \) #### **Step 1.3: Write the differential equation** Let \( q(t) \) be the charge on the first capacitor at time \( t \). The voltage across the resistor is the difference in voltages across the capacitors: But since they are in parallel, let's use the **loop rule**: \[ \text{Current } I(t) = \frac{dq}{dt} \] \[ \text{Voltage across resistor: } V_R = I(t)R \] \[ \text{Voltage difference across capacitors: } V_1 - V_2 \] Let \( q_1 \) = charge on first capacitor, \( q_2 \) = charge on second. - At \( t=0 \): \( q_1 = CV_0 \), \( q_2 = 0 \) - At \( t \): \( q_1 \), \( q_2 \) - Current flows from \( C_1 \) to \( C_2 \), so \( I = -\frac{dq_1}{dt} = \frac{dq_2}{dt} \) The loop equation: \[ V_{C1} - V_{C2} = IR \] \[ \frac{q_1}{C} - \frac{q_2}{C} = IR \] But \( q_1 + q_2 = CV_0 \implies q_2 = CV_0 - q_1 \) \[ \frac{q_1}{C} - \frac{CV_0 - q_1}{C} = IR \] \[ \frac{q_1 - (CV_0 - q_1)}{C} = IR \] \[ \frac{2q_1 - CV_0}{C} = IR \] \[ 2q_1 - CV_0 = I R C \] But \( I = -\frac{dq_1}{dt} \): \[ 2q_1 - CV_0 = -RC \frac{dq_1}{dt} \] \[ \frac{dq_1}{dt} + \frac{2}{RC}q_1 = \frac{V_0}{R} \] This is a first-order linear ODE. #### **Step 1.4: Solve the ODE** The general solution is: \[ q_1(t) = A e^{-2t/(RC)} + \text{Particular solution} \] Particular solution: set \( dq_1/dt = 0 \) (steady state), \[ \frac{2}{RC}q_1 = \frac{V_0}{R} \implies q_1 = \frac{CV_0}{2} \] So, \[ q_1(t) = \frac{CV_0}{2} + A e^{-2t/(RC)} \] At \( t = 0 \), \( q_1(0) = CV_0 \) \[ CV_0 = \frac{CV_0}{2} + A \implies A = \frac{CV_0}{2} \] So, \[ q_1(t) = \frac{CV_0}{2} + \frac{CV_0}{2} e^{-2t/(RC)} \] The current is: \[ I(t) = -\frac{dq_1}{dt} = -\frac{CV_0}{2} \cdot \left(-\frac{2}{RC}\right) e^{-2t/(RC)} = \frac{CV_0}{RC} e^{-2t/(RC)} \] --- ### **Step 2: Find the generated heat as a function of \( I(t) \)** **Explanation:** The heat \( Q \) generated in the resistor is the energy dissipated: \[ Q = \int_0^\infty I^2(t) R dt \] Substitute the result from part (a): \[ I(t) = \frac{CV_0}{RC} e^{-2t/(RC)} \] \[ I^2(t) = \left(\frac{CV_0}{RC}\right)^2 e^{-4t/(RC)} \] \[ Q = \int_0^\infty \left(\frac{CV_0}{RC}\right)^2 e^{-4t/(RC)} R dt \] \[ = \frac{C^2 V_0^2}{R^2 C^2} R \int_0^\infty e^{-4t/(RC)} dt \] \[ = \frac{C V_0^2}{R} \int_0^\infty e^{-4t/(RC)} dt \] Let \( u = \frac{4t}{RC} \implies dt = \frac{RC}{4} du \): \[ Q = \frac{C V_0^2}{R} \cdot \frac{RC}{4} \int_0^\infty e^{-u} du \] \[ = \frac{C^2 V_0^2}{4} [1] \] So, \[ \boxed{Q = \frac{C V_0^2}{4}} \] --- ## **Final Answers** ### **(a) Current as a function of time:** \[ \boxed{I(t) = \frac{CV_0}{RC} e^{-2t/(RC)}} \] ### **(b) Total heat generated:** \[ \boxed{Q = \frac{C V_0^2}{4}} \] **Each step includes the necessary explanations and derivations.**

Certainly! Based on the problem description, after the switch is closed at \( t=0 \), the circuit involves two capacitors \( C \) connected in parallel with a resistor \( R \). **Initial conditions:** - Capacitor 1 (with voltage \( V_1 \)) initially charged to \( V_0 \). - Capacitor 2 (with voltage \( V_2 \)) initially uncharged (0 V). **At \( t=0 \), when the switch is closed:** - The charged capacitor \( C_1 \) (initially at \( V_0 \)) shares charge with \( C_2 \), which was initially at 0 V. - The voltages \( V_1(t) \) and \( V_2(t) \) evolve over time as the charges redistribute. --- ### **Sketch of the voltages \( V_1(t) \) and \( V_2(t) \):** - **At \( t=0 \):** - \( V_1(0) = V_0 \) - \( V_2(0) = 0 \) - **As \( t \to \infty \):** - Both voltages equalize at half the initial voltage: \[ V_1(\infty) = V_2(\infty) = \frac{V_0}{2} \] - This is because the total charge is conserved and the total capacitance is effectively doubled when in parallel. --- ### **Qualitative sketch:** ``` V_1(t) |-------------------\___________ | / | / V_0 |---------------/ | / | / V_2(t) |__________/ | / | / | / | / |/ +--------------------------- t=0 t → ∞ ``` - **\( V_1(t) \)** starts at \( V_0 \) and decreases exponentially toward \( V_0/2 \). - **\( V_2(t) \)** starts at 0 V and increases exponentially toward \( V_0/2 \). ### **Mathematical expressions:** From the derivation: \[ V_1(t) = \frac{V_0}{2} + \frac{V_0}{2} e^{-2t/(RC)} \] \[ V_2(t) = \frac{V_0}{2} - \frac{V_0}{2} e^{-2t/(RC)} \] - \( V_1(t) \) decreases from \( V_0 \) to \( V_0/2 \). - \( V_2(t) \) increases from 0 to \( V_0/2 \). --- **Summary:** - **Initial voltages:** \[ V_1(0) = V_0,\quad V_2(0) = 0 \] - **Voltages at time \( t \):** \[ V_1(t) = \frac{V_0}{2} + \frac{V_0}{2} e^{-2t/(RC)} \] \[ V_2(t) = \frac{V_0}{2} - \frac{V_0}{2} e^{-2t/(RC)} \] This describes how the voltages evolve as the charge redistributes after switching.