Answer :

Given: L = 5.0 H

capacitor C = 80 × 10^{-6}F

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^{-6}F × 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

Rate this question :

Figure shows a tyHC Verma - Concepts of Physics Part 2

In a series LCR cHC Verma - Concepts of Physics Part 2

A circuit containNCERT - Physics Part-I

Obtain the answerNCERT - Physics Part-I

A lamp is connectPhysics - Exemplar

(a) State tPhysics - Board Papers