Keeping the source frequency equal to the resonating frequency of the series LCR circuit,

Solution:

An inductor (L), a capacitor (C), and a resistor (R) is connected in parallel with each other in a circuit where,

L = 5.0 H

C = 80 μF = 80 × 10−6 F

R = 40 Ω

$\frac{1}{Z}=\sqrt{\frac{1}{R^{2}}+\left(\frac{1}{\omega L}-\omega C\right)^{2}}$

Where,

ω = Angular frequency

At resonance, $\frac{1}{\omega L}-\omega C=0$

$\therefore \omega=\frac{1}{\sqrt{L C}}$

$=\frac{1}{\sqrt{5 \times 80 \times 10^{-6}}}=50 \mathrm{rad} / \mathrm{s}$

Hence, the magnitude of Z is the maximum at 50 rad/s. As a result, the total current is minimum.

Rms current flowing through inductor L is given as:

$I_{L}=\frac{V}{\omega L}$

$=\frac{230}{50 \times 5}=0.92 \mathrm{~A}$

Rms current flowing through capacitor C is given as:

$I_{C}=\frac{V}{\frac{1}{\omega C}}=\omega C V$

$=50 \times 80 \times 10^{-6} \times 230=0.92 \mathrm{~A}$

Rms current flowing through resistor R is given as:

$I_{R}=\frac{V}{R}$

$=\frac{230}{40}=5.75 \mathrm{~A}$

Potential of the voltage source, V = 230 V

Impedance (Z) of the given parallel LCR circuit is given as: