Question

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

• A. \(40 μF\)
• B. \(50 μF\)
• C. \(20 μF\)
• D. \(25 μF\)

Hint:

Current and voltage are in phase at resonance, so X_L = X_C.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In a series LCR circuit, current and voltage are in phase when the inductive reactance equals the capacitive reactance (resonance).

Step 2: Key Formula or Approach:
\[ \omega^2 LC = 1 \Rightarrow C = \frac{1}{\omega^2L} \]

Step 3: Calculate.
\(\omega = 2\pi f = 2\pi\times 50 = 100\pi\) rad/s, and \(L = \dfrac{4}{\pi^2}\) H.
\[ \omega^2L = 10^4\pi^2\times\frac{4}{\pi^2} = 4\times 10^4 \]
\[ C = \frac{1}{4\times 10^4} = 2.5\times 10^{-5}\text{ F} = 25\ \mu\text{F} \]

Step 4: Check the options.
Only 25 \(\mu\)F matches. The resistance and the supply voltage do not matter at resonance.

Final Answer:
The capacitance is 25 \(\mu\)F, option (D). \[ \boxed{25\ \mu\text{F}} \]