Question

Keeping the source frequency equal to the resonating frequency ofthe series LCR circuit, if the three elements, L, C and R are arrangedin parallel, show that the total current in the parallel LCR circuit isminimum at this frequency. Obtain the current rms value in eachbranch of the circuit for the elements and source specified inExercise 7.11 for this frequency.

Answer 1

Given: The inductance of inductor is 5 H, the capacitance of the capacitor is 80 μF, the resistance of resistor is 40 Ω and the applied potential difference is 230 V and the frequency of signal is 50 Hz.

The impedance of the given parallel LCR circuit is given as,

1 Z = 1 R 2 + ( 1 ωL −ωC ) 2

Where, the resistance of the resistor is R, the inductance is L, the capacitance is C and the angular frequency is ω.

The angular frequency at resonance is given as,

ω= 1 LC

By substituting the given values in the above equation, we get

ω= 1 5×80× 10 −6 =50  rad/s

The rms current flowing through inductor is given as,

I L = V ωL

Where, the applied potential difference is V.

By substituting the given values in the above equation, we get

I L = 230 50×5 =0.92 A

The rms current flowing through inductor is given as,

I C = V 1 ωC

By substituting the given values in the above equation, we get

I C = 230 1 50×80× 10 −6 =230×50×80× 10 −6 =0.92 A

The rms current flowing through inductor is given as,

I R = V R

By substituting the given values in the above equation, we get

I R = 230 40 =5.75 A

Thus, the rms currents through inductor, capacitor and resistor are 0.92 A, 0.92 A and 5.75 A respectively.