Question

A resistor of 5 $\Omega$, inductor of self inductance $\left(\frac{2}{\pi}\right)$ H and a capacitor of unknown capacity are connected in series to an a.c. source of 100 V, 50 Hz supply. When the voltage and current are in phase, the value of capacitance is ______.

• A. $\frac{10}{\pi}$ $\mu$F
• B. $\frac{20}{\pi}$ $\mu$F
• C. $\frac{40}{\pi}$ $\mu$F
• D. $\frac{50}{\pi}$ $\mu$F

Hint:

"Voltage and current are in phase" is the examiner's code phrase for Resonance ($X_L = X_C$). The resistor value (5 $\Omega$) and the supply voltage (100 V) are distractor values and completely irrelevant to finding the capacitance!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given an LCR series circuit where the voltage and current are exactly in phase. This strictly implies the circuit is in electrical resonance. We must use the resonance condition to find the unknown capacitance $C$.

Step 2: Detailed Explanation:
When the voltage and current in an LCR circuit are completely in phase, the phase angle $\phi = 0^\circ$.
This only occurs at resonance, where the inductive reactance ($X_L$) perfectly equals and cancels out the capacitive reactance ($X_C$).
$X_L = X_C$
We know the formulas for reactance:
$X_L = \omega L = 2\pi f L$
$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$
First, let's calculate $X_L$:
Given frequency $f = 50$ Hz and Inductance $L = \frac{2}{\pi}$ H.
$X_L = 2\pi \times 50 \times \left(\frac{2}{\pi}\right)$
The $\pi$ cancels out:
$X_L = 100 \times 2 = 200 \ \Omega$
At resonance, $X_C$ must also be exactly $200 \ \Omega$.
$X_C = 200 \implies \frac{1}{2\pi f C} = 200$
Substitute $f = 50$:
$\frac{1}{2\pi(50) C} = 200$
$\frac{1}{100\pi C} = 200$
Rearrange to solve for $C$:
$C = \frac{1}{200 \times 100\pi}$
$C = \frac{1}{20000\pi} \text{ Farads}$
To convert Farads to microFarads ($\mu$F), multiply by $10^6$:
$C = \frac{10^6}{20000\pi} \mu\text{F}$
$C = \frac{1000000}{20000\pi} \mu\text{F}$
Cancel four zeros from top and bottom:
$C = \frac{100}{2\pi} \mu\text{F} = \frac{50}{\pi} \mu\text{F}$

Step 3: Final Answer:
The capacitance is $\frac{50}{\pi} \mu\text{F}$, matching option (d).