Question

A resistor 30 Ω, inductor of reactance 10 Ω and the capacitor of reactance 10 Ω are connected in series to an ac voltage source e = 300√2sin(ωt). The current in the circuit is 

(a) 10√2A 

(b) 10A 

(c) 30√11A 

(d) (30/√11)A

Answer 1

The correct option (b) 10A

Explanation:

R = 30 Ω

X_L = 10 Ω

X_C = 10 Ω

e = 300√2 sin ωt i.e. e_m = 300√2

Hence e_rms = [{300√2}/{√2}] = 300 V ------ [e_rms = (e_m/√2)]

as X_L = X_C

Hence resonance takes place

∴ circuit is purely resistive

∴ Z = R

current in the circuit = I_rms = [{e_rms} / {|Z|}]

∴ I_rms = [{e_rms}/R]

∴ I_rms = [{300}/{30}]

I_rms = 10A