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

Question:

Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, Land are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.

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 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: