Let's solve the problem **step-by-step**: ### **Given:** - Voltage across capacitor, \( V(t) \) (graph shown) - Capacitance, \( C = 4\,\mu F = 4 \times 10^{-6} F \) - We are to find the current \( i(t) \) through the capacitor. ### **Formula:** For a capacitor, \[ i(t) = C \frac{dV(t)}{dt} \] --- ### **Step 1: Analyze the Voltage Graph** The graph is piecewise linear. Let's break it into intervals: #### **Interval 1: \(0 < t < 2\) ms** - Voltage rises linearly from 0 to 10V. \[ \text{Slope} = \frac{10\,V - 0\,V}{2\,ms - 0\,ms} = \frac{10}{2} = 5\, \text{V/ms} \] #### **Interval 2: \(2 < t < 4\) ms** - Voltage drops linearly from 10V to 0V. \[ \text{Slope} = \frac{0\,V - 10\,V}{4\,ms - 2\,ms} = \frac{-10}{2} = -5\, \text{V/ms} \] #### **Interval 3: \(4 < t < 6\) ms** - Voltage drops linearly from 0V to -10V. \[ \text{Slope} = \frac{-10\,V - 0\,V}{6\,ms - 4\,ms} = \frac{-10}{2} = -5\, \text{V/ms} \] #### **Interval 4: \(6 < t < 8\) ms** - Voltage rises linearly from -10V to 0V. \[ \text{Slope} = \frac{0\,V - (-10\,V)}{8\,ms - 6\,ms} = \frac{10}{2} = 5\, \text{V/ms} \] #### **Interval 5: \(t > 8\) ms** - Voltage stays constant at 0V. \[ \text{Slope} = 0 \] --- ### **Step 2: Find \(i(t)\) for Each Interval** Recall: \[ i(t) = C \frac{dV}{dt} \] #### **Convert Slope to V/s:** 1 ms = \(10^{-3}\) s So, \(5\,\text{V/ms} = 5000\,\text{V/s}\). --- #### **Interval 1: \(0 < t < 2\,ms\)** \[ i(t) = 4 \times 10^{-6} \times 5000 = 0.02\,\text{A} = 20\,\text{mA} \] --- #### **Interval 2: \(2 < t < 4\,ms\)** \[ i(t) = 4 \times 10^{-6} \times (-5000) = -0.02\,\text{A} = -20\,\text{mA} \] --- #### **Interval 3: \(4 < t < 6\,ms\)** \[ i(t) = 4 \times 10^{-6} \times (-5000) = -0.02\,\text{A} = -20\,\text{mA} \] --- #### **Interval 4: \(6 < t < 8\,ms\)** \[ i(t) = 4 \times 10^{-6} \times 5000 = 0.02\,\text{A} = 20\,\text{mA} \] --- #### **Interval 5: \(t > 8\,ms\)** \[ i(t) = 0 \] --- ### **Step 3: Final Answer (Piecewise Function)** \[ i(t) = \begin{cases} 20\,\text{mA} & 0 < t < 2\,ms \\ -20\,\text{mA} & 2 < t < 6\,ms \\ 20\,\text{mA} & 6 < t < 8\,ms \\ 0 & t > 8\,ms \\ \end{cases} \] --- ### **Explanation** - The current through the capacitor is proportional to the slope (rate of change) of the voltage. - Positive slope → positive current, negative slope → negative current, flat (zero slope) → zero current. --- ## **Final Answer:** \[ \boxed{ i(t) = \begin{cases} 20\,\text{mA} & 0 < t < 2\,ms \\ -20\,\text{mA} & 2 < t < 6\,ms \\ 20\,\text{mA} & 6 < t < 8\,ms \\ 0 & t > 8\,ms \\ \end{cases} } \]