An $L C R$ circuit contains resistance of $100 Ω$ and a supply of $200 V$ at $300 r a d / s$ angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by $60^{\circ}$ . If on the other hand, only inductor is taken out, the current leads by $60^{\circ}$ with the applied voltage. The current flowing in the circuit is

1 A

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1.5 A

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2 A

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2.5 A

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Solution

The correct option is C $2 A$
if only capacitance is remove,
$Z = R + j L ω \Rightarrow | Z | c o s (60) = R = 100$
$\Rightarrow | Z | = 200 Ω$
$\Rightarrow L ω = | Z | s i n (60) = 173.205$
if only inductor is taken out,
$Z = R - j \frac{1}{C ω} \Rightarrow | Z | c o s (- 60) = R = 100$
$\Rightarrow | Z | = 200 Ω$
$\Rightarrow \frac{1}{C ω} = | Z | s i n (- 60) = - 173.205$
therefore actual impedance= $R + j L ω - j \frac{1}{C ω} = 100 Ω$
therefore current in circuit= $\frac{200 V}{100 Ω} = 2 A$

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