Let's solve the circuit step by step. --- ## **Given:** - **Voltage across capacitor:** \( v(t) = 5(1 - 2e^{-2t}) \) V - **Capacitance:** \( C = 10\,\mu\text{F} = 10 \times 10^{-6} \) F - **Resistor:** \( R = 200\,k\Omega = 200 \times 10^3 \) Ω --- ## **Step 1: Express \(i(t)\) in terms of \(v(t)\)** The current \(i(t)\) is the total current entering the parallel RC network. This current splits into two branches: - Through the resistor (\(i_R\)) - Through the capacitor (\(i_C\)) So, \[ i(t) = i_R + i_C \] Where: - \(i_R = \frac{v(t)}{R}\) - \(i_C = C \frac{dv}{dt}\) --- ## **Step 2: Calculate each component** ### **A. Current through the resistor** \[ i_R = \frac{v(t)}{R} = \frac{5(1 - 2e^{-2t})}{200 \times 10^3} \] ### **B. Current through the capacitor** \[ i_C = C \frac{dv}{dt} \] First, find \(\frac{dv}{dt}\): \[ v(t) = 5(1 - 2e^{-2t}) \] \[ \frac{dv}{dt} = 5 \cdot - 5 \cdot 2 \cdot \frac{d}{dt}(e^{-2t}) = -10 \cdot (-2) e^{-2t} = 20 e^{-2t} \] So, \[ i_C = 10 \times 10^{-6} \cdot 20 e^{-2t} = 200 \times 10^{-6} e^{-2t} = 2 \times 10^{-4} e^{-2t} \ \text{A} \] --- ## **Step 3: Add them up** \[ i(t) = i_R + i_C \] \[ i(t) = \frac{5(1 - 2e^{-2t})}{200 \times 10^3} + 2 \times 10^{-4} e^{-2t} \] Let's simplify \(\frac{5}{200 \times 10^3}\): \[ \frac{5}{200 \times 10^3} = \frac{1}{40 \times 10^3} = \frac{1}{40000} = 2.5 \times 10^{-5} \] So, \[ i_R = 2.5 \times 10^{-5} (1 - 2e^{-2t}) \ \text{A} \] Now, combine: \[ i(t) = 2.5 \times 10^{-5} (1 - 2e^{-2t}) + 2 \times 10^{-4} e^{-2t} \] \[ = 2.5 \times 10^{-5} - 5 \times 10^{-5} e^{-2t} + 2 \times 10^{-4} e^{-2t} \] \[ = 2.5 \times 10^{-5} + (2 \times 10^{-4} - 5 \times 10^{-5}) e^{-2t} \] \[ = 2.5 \times 10^{-5} + 1.5 \times 10^{-4} e^{-2t} \] --- ## **Final Answer** \[ \boxed{ i(t) = 2.5 \times 10^{-5} + 1.5 \times 10^{-4} e^{-2t} \quad \text{A} } \] --- ### **Summary of Steps** 1. Write \(i(t) = \frac{v(t)}{R} + C \frac{dv}{dt}\) 2. Plug in \(v(t)\) and compute \(\frac{dv}{dt}\) 3. Substitute values for \(R\) and \(C\) 4. Simplify to get the final answer!