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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,

At resonance,

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

Where,

*V*_{0} = 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*_{R}*L*

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.

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