Question

An a.c. e.m.f. of peak value $230\text{ V}$ and frequency $50\text{ Hz}$ is connected to a circuit with $R = 11.5\Omega, L = 2.5\text{H}$ and a capacitor all in series. The value of capacitance is ' $C$ ' for the current in the circuit to be maximum. The value of ' $C$ ' and maximum current are respectively ($\pi^2 = 10$ )}

• A. $4\mu\text{F}, 20\text{ A}$
• B. $5\mu\text{F}, 10\text{ A}$
• C. $2\mu\text{F}, 20\text{ A}$
• D. $8\mu\text{F}, 12\text{ A}$

Hint:

At resonance in a series \(RLC\) circuit: \[ C=\frac{1}{\omega^2L} \quad \text{and} \quad I_{\max}=\frac{V_0}{R} \]

Answer 1

The Correct Option is B

Concept:
In a series \(RLC\) circuit, current is maximum at resonance: \[ X_L=X_C \] So, \[ \omega L=\frac{1}{\omega C} \] which gives \[ C=\frac{1}{\omega^2 L} \] At resonance, impedance becomes: \[ Z=R \] ip

Step 1: Find angular frequency.
\[ f=50\text{ Hz} \] \[ \omega=2\pi f=100\pi \] ip

Step 2: Find capacitance at resonance.
\[ C=\frac{1}{\omega^2L} = \frac{1}{(100\pi)^2(2.5)} \] Using \(\pi^2=10\): \[ (100\pi)^2=10000\pi^2=100000 \] So, \[ C=\frac{1}{100000\times 2.5} =\frac{1}{250000} =4\times10^{-6}\text{ F} \] \[ C=4\mu\text{F} \] ip

Step 3: Find maximum current.
At resonance: \[ Z=R=11.5\Omega \] Peak current: \[ I_{\max}=\frac{V_0}{R}=\frac{230}{11.5}=20\text{ A} \] ip Hence, the correct answer is:
\[ \boxed{(A)\ 4\mu\text{F},\ 20\text{ A}} \]