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
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,
Amplitude of the current at the resonating frequency is given as:
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,
(VL)rms = I × ωRL
I = rms current
Potential drop across the capacitor,
Potential drop across the resistor,
(VR)rms = IR
= × 40 = 230 V
Potential drop across the LC combination,
Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.