Question:medium

A resistor of 100 \(\Omega\), inductor of self inductance \((\frac{4}{π^2})\) H and a capacitor of unknown capacity are connected in series to an a.c. source of 200 V and 50 Hz. When the current and voltage are in phase, the value of capacity is

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Current and voltage are in phase at resonance, so X_L = X_C.
Updated On: Oct 1, 2026
  • \(40 μF\)
  • \(50 μF\)
  • \(20 μF\)
  • \(25 μF\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Find the reactance of the inductor.
$X_L = \omega L = 100\pi\times\dfrac{4}{\pi^2} = \dfrac{400}{\pi}\ \Omega$.

Step 2: Equate to the capacitor.
$X_C = \dfrac{1}{\omega C} = X_L$, so $C = \dfrac{1}{\omega X_L} = \dfrac{\pi}{100\pi\times 400} = \dfrac{1}{40000}$ F.

Step 3: Convert.
$\dfrac{1}{40000}$ F $= 25\times10^{-6}$ F $= 25\ \mu$F.

Final Answer:
Option (D). \[ \boxed{25\ \mu\text{F}} \]
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