Question

A coil of inductance 0.1H is connected to 50V, 100Hz generator and current is found to be 0.5A. The potential difference across resistance of the coil is

• A. 15V
• B. 20V
• C. 25V
• D. 39V

Answer 1

Answer: (D)

39V

$Z=\frac{V}{i}$

$=\frac{50}{0.5}$

$=100\Omega$

$Z^2=X_L^{+}{R}^2$

$(100{)}^2=(2\pi fL{)}^2+R^2$

$(100{)}^2-[2\pi 100\left(0.1\right){]}^2=R^2$

$6052=R^2$

$R=78\Omega$

$V_R=iR$

$=0.5\times 78$

$=39V$