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.

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 x 10^–6F, R = 40Ω

Potential of the voltage source, V = 230 V                                                                                                    Impedance (Z) of the given parallel LCR circuit is given as:

1/Z = √(1/R² +(1/ωL -ωC)²)

Where, ω = Angular frequency                                  ^Z

At resonance, (1/ωL – ωC) = 0

Therefore, ω = 1/√(LC)   =  1 / √(5 x 80 x 10^–6)  = 50 rad/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= V/ωL = 230 /(50 x 5) = 0.92 A

rms current flowing through capacitor C is given as :

I_C = V/(1/ωC) =ωCV = 50 x 80 x 10^–6 x 230 =0.92A

rms current flowing through resistor R is given as :

I_R= V/R = 230/40 = 5.75A