HC Verma Chapter 39 Problem 2

Problem Statement

A series RLC circuit has $R = 15\,\Omega$, $L = 0.15\,\text{H}$, $C = 30\,\mu$F connected to $V_{rms} = 220\,\text{V}$ at $f = 50\,\text{Hz}$. Find (a) impedance, (b) rms current, (c) resonant frequency.

Given Information

$R = 15\,\Omega$
$L = 0.15\,\text{H}$
$C = 30\,\mu\text{F}$
$f = 50\,\text{Hz}$
$V_{rms} = 220\,\text{V}$

Physical Concepts & Formulas

In a series RLC circuit, inductive reactance $X_L = \omega L$ and capacitive reactance $X_C = 1/(\omega C)$ partially cancel. The net reactance is $X = X_L – X_C$. Impedance $Z = \sqrt{R^2 + X^2}$. Resonance occurs at $f_0 = 1/(2\pi\sqrt{LC})$ where $Z = R_{min}$.

$X_L = \omega L$, $X_C = 1/(\omega C)$
$Z = \sqrt{R^2+(X_L-X_C)^2}$
$f_0 = 1/(2\pi\sqrt{LC})$

Step-by-Step Solution

Step 1 — Reactances: $$X_L = 2\pi\times50\times0.15 = 47.12\,\Omega;\quad X_C = \frac{1}{2\pi\times50\times30\times10^{-6}} = 106.10\,\Omega$$

Step 2 — Impedance: $$Z = \sqrt{15^2+(47.12-106.10)^2} = \sqrt{225+3478.57} = 60.86\,\Omega$$

Step 3 — Current and resonance: $$I_{rms} = 220/60.86 = 3.615\,\text{A};\quad f_0 = \frac{1}{2\pi\sqrt{0.15\times30\times10^{-6}}} = 75.03\,\text{Hz}$$

Worked Calculation

$$Z = 60.86\,\Omega;\quad I_{rms} = 3.615\,\text{A};\quad f_0 = 75.03\,\text{Hz}$$

Answer

$$\boxed{Z = 60.86\,\Omega,\quad I_{rms} = 3.615\,\text{A},\quad f_0 = 75.03\,\text{Hz}}$$

At 50 Hz the impedance is 60.86Ω. At resonance (75.03 Hz), Z drops to 15Ω and current rises to 14.67A — below resonance now. Resonance is exploited in radio tuning and power factor correction.

Physical Interpretation