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.
Answer:
Inductor L = 5 H
Capacitor C = 80 μF = 80 x 10-6 F
Resistor R = 40 Ω
Voltage V = 230 V
Impedeance (z)
then
1/z = √ 1/R2 + (1/ωL - ωC)2
Where, Angular frequency ω
At resonance, 1/ωL - ωC = 0
ω = 1/√ LC = 1/ √ 5 x 80 x 10-6 = 50 rad/s
rms current flowing through Inductor L
IL = V/ ωL = 230/ 50 x 5 = 0.92 A
rms current flowing through capacitor C
Ic = V/1/ ωC = ωCV = 5 x 80 x 10-6 = 0.92 A
Rms current flowing through resistor R
IR = V/R = 230/40 = 5.75 A
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