A variable frequency 230V ac voltage source is connected across a series combination of L=5H,R=40ohm,C=80uF.Calculate
(i)Angular frequency of the source which drives the circuit in resonance,
(ii)Amplitude of current at resonant frequency
(iii)r.m.s voltage across inductor at resonance

Dear Student,
 

Inductance of the inductor, L = 5.0 H

Capacitance of the capacitor, C = 80 μH = 80 × 10⁻⁶ F

Resistance of the resistor, R = 40 Ω

Potential of the variable voltage source, V = 230 V

(a) Resonance angular frequency is given as:

Hence, the circuit will come in resonance for a source frequency of 50 rad/s.

(b) Impedance of the circuit is given by the relation,

At resonance,

Amplitude of the current at the resonating frequency is given as: 

Where,

V₀ = Peak voltage

Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.

(c) Rms potential drop across the inductor,

(V_L)_rms = I × ω_RL

Where,

I = rms current

Potential drop across the capacitor,

Potential drop across the resistor,

(V_R)_rms = IR

= × 40 = 230 V

Potential drop across the LC combination,

At resonance,

∴V_LC= 0

Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.