Question

An LCR circuit contains resistance of $100\Omega$ and a supply of $200\text{V}$ at $300\text{rad/s}$ angular frequency. If only capacitor 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. $\text{1 A}$
• B. $\text{1.5 A}$
• C. $\text{2 A}$
• D. $\text{2.5 A}$

Answer 1

The correct option is C $\text{2 A}$


For the given LCR circuit the phasor diagram would be

In this the voltage of inductor leads current by ${90}^{\circ }$ and that of capacitor lags by the same.
So if capacitor is removed first then

We know that current is same
$\therefore$ dividing both voltages of inductor and resistance we get
$\frac{V_L}{i}=X_L$ $\frac{V_R}{i}=R$

$\frac{V_C}{i}=X_C$
As we remove capacitor


So resultant will give impedence(Z)
And from the triangle we can say that
$tan\varphi =\frac{X_L}{R}$

Now we remove inductor

From this we get
$tan\varphi =\frac{X_C}{R}$
$\therefore X_L=X_C$
which implies that circuit is in resonance.
So effective impedence $Z=\sqrt{R^2+(X_L-X_C{)}^2}\Rightarrow Z=R$
So to get current
$i=\frac{V_{rms}}{R}\Rightarrow i=\frac{200}{100}=2A$