Question

An $LCR$ series circuit with $R=100\Omega$ is connected to a $200V,50Hz$ a.c source. When only the capacitance is removed, the current leads the voltage by ${60}^o$. When only the inductance is removed, the current leads the voltage by ${60}^o$. The current in the circuit is :

• A. $2A$
• B. $1A$
• C. $\frac{\sqrt{3}}{2}A$
• D. $\frac{2}{\sqrt{3}}A$

Answer 1

The correct option is A $2A$

Given,

$R=100\Omega$

$V=200V$

$f=50Hz$

$\varphi ={60}^0$

When a capacitance is removed, the current leads the voltage by $\varphi$

$tan\varphi =\frac{X_L}{R}$

$X_L=Rtan60=\sqrt{3}R$. . . .(1)

When the inductance is removed current leads the voltage by $\varphi$

$tan\varphi =\frac{X_C}{R}$

$X_C=\sqrt{3}R$. . . . . .(2)

From equation (1) and (2), we get

$X_C=X_L$ (this is the condition of resonance)

And at the resonance, $Z=R$

The current in the circuit is

$I=\frac{V}{R}$

$I=\frac{200}{100}$

$I=2A$

The correct option is A.