An LCR series circuit with 100 Ω resistance is connected to an a.c. source of 200 V and angular frequency 300 radians per second. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. Calculate the current and power dissipated in LCR circuit.

Solution

Given, an LCR series circuit.
Resistance, R = 100 Ω
Rms voltage, V = 200 V
Angular frequency = 300 radians per second.
Current lags behind the voltage by 60^o

Using the formula,

$\tan 60 ° = \frac{ωL}{R}$

or,    $\tan 60 ° = \frac{1 / ωC}{R}$
$∵$    $ωL = \frac{1}{ωC}$

Impedance of circuit,

$Z = \sqrt{[R^{2} + {(ωL - \frac{1}{ωC})}^{2}]} = R$

Current in the circuit,

$I_{0} = \frac{V_{0}}{Z} = \frac{V_{0}}{R} = \frac{200}{100} = 2 amp .$

Average power, $P = \frac{1}{2} V_{0} I_{0} \cos ϕ$

But,
$\tan ϕ = \frac{ωL - (1 / ωC)}{R} = 0$ $(\cos ϕ = 1)$

Now, $P = \frac{1}{2} \times 200 \times 2 \times 1 = 200 watt .$

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