**Question:**

Keeping the source frequency equal to the resonating frequency of the series *LCR *circuit, if the three elements, *L*, *C *and *R *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 *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: