Kanha wants to calculate the current and power dissipated in an L-C-R series circuit. He connected $100 Ω$ resistance to an AC source of peak value 200V and angular frequency 300 rad/s. When he removed only the capacitance, the current was found to be lagging behind the voltage by $60^{0}$ . While on removing the inductance he found, the current leading the voltage by $60^{0}$ . The value of peak current and the power dissipated obtained by him will be

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3A,120W

2A,200W

4A,100W

None of these

answer is A.

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Detailed Solution

$\tan 60^{\circ} = \frac{ωL}{R}, \tan 60^{\circ} = \frac{1 / ωC}{R} ωL = \frac{1}{ωC}$

$∴$ Impedance of the circuit, $Z = {[R^{2} + {(ωL - \frac{1}{ωC})}^{2}]}^{1 / 2} = R$

Current in the circuit, $i_{0} = \frac{V_{0}}{Z} = \frac{V_{0}}{R} = \frac{200}{100} = 2 A$

Average power, $\bar{P} = \frac{1}{2} V_{0} i_{0} cosϕ$

But, $\tan ϕ = \frac{ωL - \frac{1}{ωC}}{R} = 0 ∴ \cos ϕ = 1 ∴ \bar{P} = \frac{1}{2} \times 200 \times 2 \times 1 = 200 W$

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