Question

An a.c. voltage is applied to a pure inductor. The current in the inductor would be

• A. leading the voltage by \(\frac{\pi}{2}\)
• B. lagging the voltage by \(\frac{\pi}{2}\)
• C. leading the voltage by \(\frac{\pi}{4}\)
• D. in phase with the voltage.

Hint:

Think of \(V = L\,dI/dt\). Voltage is largest when the current changes fastest, which is when the current is zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a pure inductor there is no resistance. The only opposition to the current comes from the self-induced emf, which opposes any change in current.

Step 2: Key Formula or Approach:
Let the applied voltage be \(V = V_0 \sin \omega t\). The induced emf balances the applied voltage, so \(V = L \frac{dI}{dt}\).

Step 3: Detailed Explanation:
Integrate the relation to get the current.
\[ I = \frac{1}{L}\int V_0 \sin \omega t \, dt = -\frac{V_0}{\omega L}\cos \omega t \]
\[ I = \frac{V_0}{\omega L}\sin\left(\omega t - \frac{\pi}{2}\right) \]
The current has a phase of \(-\pi/2\) compared with the voltage. So the current reaches its peak a quarter cycle after the voltage does.

Step 4: Check the options:
Option 1 says the current leads. That is true for a capacitor, not an inductor. Option 3 gives a lead of \(\pi/4\), which happens for no pure element. Option 4 is true only for a pure resistor. Option 2 matches our result.

Final Answer:
In a pure inductor the current lags the voltage by \(\pi/2\). \[ \boxed{\text{Option 2: lagging the voltage by } \frac{\pi}{2}} \]