The current wave form in figure below is applied across an ideal capacitor of 100 μF which is initially uncharged. What is the voltage across it at t = 5 ms

Concept:

The voltage across a capacitor is given by: \( V(t) = \frac{1}{C} \int_0^t i(t) \, dt \)

Where \( C \) is capacitance, and \( i(t) \) is the current through the capacitor.

Given:

Capacitance, \( C = 100~\mu F = 100 \times 10^{-6}~F \)

Current waveform is a triangle from 0 to 10 mA over 4 ms

We are to find voltage at \( t = 5~ms \)

Calculation:

From 0 to 4 ms, current increases linearly ⇒ it's a triangle

Area under the triangle (charge): \( Q = \frac{1}{2} \cdot base \cdot height = \frac{1}{2} \cdot 4 \times 10^{-3} \cdot 10 \times 10^{-3} = 20 \times 10^{-6}~C \)

From 4 ms to 5 ms, current = 0 ⇒ no additional charge

Voltage: \( V = \frac{Q}{C} = \frac{20 \times 10^{-6}}{100 \times 10^{-6}} = 0.2~V = 200~mV \)

Answer:

Option 3) 200 mV