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 11 for this frequency.
Given: L = 5.0 H
capacitor C = 80 × 10-6F
Resistance = 40 Ω
Voltage = 230 V
Impedance is expressed as:
At resonance condition, 
ω = 1/ √LC
substituting the values, we get
ω = 1/ √(5(H) × 80 × 10-6(F))
On calculating, we get
ω = 50 rad/s
Since, the magnitude of impedance is maximum at 50 rad/s due to which the total current is minimum.
The root mean square current flowing through the inductor is given by the relation:
IL = V/ωL
Substituting the values, we get
IL = 230(V)/(50 (rad/s) × 5 (H))
On calculating, we get
IL = 0.92 A
Root mean square value of the current through capacitor is given by the following relation:
Ic = ωCV
Ic = 50rad/s × 80 × 10-6F × 230V = 0.92 A
Root mean square value of the current through resistor is given by the following relation:
IR = V/R
IR = 230V/40Ω
IR = 5.75A