Q. 174.7( 6 Votes )

# Keeping the sourc

Answer :

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

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