Question

A parallel plate capacitor of capacitance \(5\,\mu\text{F}\) is connected to an AC source. If the relation between the potential difference \(V\) (in volt) across the plates of the capacitor and time \(t\) (in second) is \[ V=7\cos(100\pi t), \] then the displacement current between the plates of the capacitor at a time of \(5\,\text{ms}\) is:

• A. \(5\,\text{mA}\)
• B. \(9\,\text{mA}\)
• C. \(7\,\text{mA}\)
• D. \(11\,\text{mA}\)

Hint:

For a capacitor, displacement current is obtained directly from \(I_d=C\dfrac{dV}{dt}\). Differentiate first and then substitute the required time.

Answer 1

The Correct Option is D

Concept: The displacement current through a capacitor is given by \[ I_d=C\frac{dV}{dt} \] where \(C\) is the capacitance and \(V\) is the instantaneous voltage across the capacitor.

Step 1: Differentiate the given voltage equation. Given, \[ V=7\cos(100\pi t) \] Differentiating with respect to time, \[ \frac{dV}{dt} = -7(100\pi)\sin(100\pi t) \] \[ \frac{dV}{dt} = -700\pi \sin(100\pi t) \]

Step 2: Evaluate at \(t=5\text{ ms}\). \[ t=5\times10^{-3}\text{ s} \] Therefore, \[ 100\pi t = 100\pi(5\times10^{-3}) = \frac{\pi}{2} \] Hence, \[ \sin\left(\frac{\pi}{2}\right)=1 \] Thus, \[ \left|\frac{dV}{dt}\right| = 700\pi \]

Step 3: Calculate displacement current. \[ I_d = 5\times10^{-6}\times700\pi \] \[ = 3500\pi\times10^{-6} \] \[ \approx 11\times10^{-3}\text{ A} \] \[ I_d\approx11\text{ mA} \] \[ \boxed{11\text{ mA}} \]