Keeping the sourc

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:

I_L = V/ωL

Substituting the values, we get

I_L = 230(V)/(50 (rad/s) × 5 (H))

On calculating, we get

I_L = 0.92 A

Root mean square value of the current through capacitor is given by the following relation:

I_c = ωCV

I_c = 50rad/s × 80 × 10^-6F × 230V = 0.92 A

Root mean square value of the current through resistor is given by the following relation:

I_R = V/R

I_R = 230V/40Ω

I_R = 5.75A