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: Recall the phasor picture:
Use a phasor diagram. The voltage across a pure inductor is $V_L = I X_L$, and it is drawn $90^\circ$ ahead of the current phasor. This is the standard result for an inductive circuit.

Step 2: Turn the statement around:
If the voltage is 90 degrees ahead of the current, the current must be 90 degrees behind the voltage. A quarter of a cycle is $\pi/2$ radian.

Step 3: Energy argument:
During the first quarter cycle the voltage is at its maximum when the current is zero. The inductor then stores energy in its magnetic field as the current builds up. This delay in the current is the reason it lags.

Step 4: Compare with the other choices:
Leading by $\pi/2$ belongs to a capacitor. In phase belongs to a resistor. A phase of $\pi/4$ needs a mix of resistance and reactance, so it cannot describe a pure inductor.

Final Answer:
The phase difference is exactly $\pi/2$ with current behind voltage. \[ \boxed{I \text{ lags } V \text{ by } \frac{\pi}{2}} \]