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

Step 1: Understanding the Concept:
The current in a series LCR circuit is maximum at the resonant frequency, where inductive reactance equals capacitive reactance ($X_L = X_C$). In this state, the impedance $Z$ is purely resistive ($Z = R$).
Step 2: Key Formula or Approach:
1. Resonance Condition: $\omega^2 = \frac{1}{LC}$
2. Peak Current: $I_0 = \frac{V_0}{R}$
Step 3: Detailed Explanation:
Given $V_0 = 230\text{ V}, f = 50\text{ Hz}, R = 11.5\Omega, L = 2.5\text{ H}$.
1. Calculate Peak Current:
\[ I_0 = \frac{230}{11.5} = 20\text{ A} \]
2. Calculate Capacitance $C$:
$\omega = 2\pi f = 100\pi$.
$\omega^2 = (100\pi)^2 = 10000 \times \pi^2 = 10000 \times 10 = 10^5$.
From $C = \frac{1}{\omega^2 L}$:
\[ C = \frac{1}{10^5 \times 2.5} = \frac{0.4}{10^5} = 4 \times 10^{-6} \text{ F} = 4\mu\text{F} \]
Step 4: Final Answer:
Capacitance is $4\mu\text{F}$ and current is $20\text{ A}$.