Question

An $LCR$ circuit contains resistance of $100\Omega$ and a supply of $200V$ at $300rad/s$ 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

• A. $1A$
• B. $1.5A$
• C. $2A$
• D. $2.5A$

Answer 1

The correct option is C $2A$

if only capacitance is remove,
$Z=R+jL\omega \Rightarrow \left|Z\right|cos\left(60\right)=R=100$

$\Rightarrow \left|Z\right|=200\Omega$

$\Rightarrow L\omega =\left|Z\right|sin\left(60\right)=173.205$
if only inductor is taken out,
$Z=R-j\frac{1}{C\omega }\Rightarrow \left|Z\right|cos(-60)=R=100$

$\Rightarrow \left|Z\right|=200\Omega$

$\Rightarrow \frac{1}{C\omega }=\left|Z\right|sin(-60)=-173.205$
therefore actual impedance= $R+jL\omega -j\frac{1}{C\omega }=100\Omega$
therefore current in circuit=$\frac{200V}{100\Omega }=2A$