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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