When 100 V DC is applied across a solenoid, a current of 1 A flows in it. When 100 VAC is applied across the same coil, the current drop is 0.5 A . If the frequency of the A C source is 50 Hz , then inductance of solenoid is found to be x mH , then find the value of x .

Step-by-step solution

When DC is applied across a solenoid, then resistance, i.e. R= V I R= 100 1 =100 When A C is applied, then Z= 100 0.5 =200 Net impedance, aligned Z & = R^2+X_L^2 200 & = (100)^2+X_L^2 (200)^2 & =(100)^2+X_L^2 30000 & =X_L^2 X_L=173.205 aligned Now, X_L= L aligned & 173.205=2 f L 173.205 2 f =L & 173.205 2 3.14 50 =L 0.55 =L aligned L=0.55 H =550 mH