Question

In a $L-R$ circuit, the value of $L$ is $(0.4/\pi )H$ sar the value of $R$ is $30\Omega$. If in circuit, an alternating emf of $200V$ at $50cycle/s$ is connected, the impedance of the circuit and current will be

• A. $11.\Omega ,17.5A$
• B. $30.\Omega ,6.5A$
• C. $40.\Omega ,5A$
• D. $50\Omega ,4A$

Answer 1

Answer: (D)

Impedance of $L-R$ circuit is
$Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+(\omega L{)}^2}$
$=\sqrt{R^2+(2\pi fL{)}^2}$
Given, $L=\frac{0.4}{\pi }H,R=30\Omega ,f=50$ cycle/s
So, $Z=\sqrt{(30{)}^2+{\left(2\pi \times 50\times \frac{0.4}{\pi }\right)}^2}$
$=\sqrt{(30{)}^2+(40{)}^2}=\sqrt{900+1600}=50\Omega$
and current $I=\frac{V}{Z}=\frac{200}{50}=4A$