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Question
The figure shows a series LCR circuit with L = 10.0 H, C = 40 μF, R = 60 Ω connected to a variable frequency 240 V source, calculate
(i) the angular frequency of the source which drives the circuit at resonance,
(ii) the current at the resonating frequency,
(iii) the rms potential drop across the inductor at resonance.
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Solution
(i) Resonant angular frequency
`omega_0 = 1/(sqrt(LC)) = 1/sqrt(10 xx 40 xx 10^6)`
`= 1/(sqrt (400 xx 10^-6 ))= 1/(20 xx 10^-3)`
`=1000/20`
`= 50 \text {rads}^-1`
(ii) At resonant frequency, we know that the inductive reactance cancels out the capacitive reactance.
The impedance = Z = 60Ω the value of resistance
The current amplitude at resonant frequency
`I_0 = E_0/Z = sqrt(2E_v)/R = (sqrt2 xx 240)/60`
` = 339.36/60 = 5.66A`
(iii) The R.M.S. value of current
`I_v = I_0/sqrt2 = 5.66/sqrt2 = 4A`
For R.M.S potential drop across inductor
`V_1 = I_VX_L`
`= I_V xx omega L`
`= 4 xx 50 xx 10`
`= 200 xx 10`
`= 2000 V`
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