Question

Answer the following question.
When an inductor is connected to a 200 V dc voltage, a current at 1A flows through it. When the same inductor is connected to a 200 V, 50 Hz ac source, only 0.5 A current flows. Explain, why? Also, calculate the self-inductance of the inductor. 

Answer 1

When the inductor is connected across the 200 V DC circuit 1 A current flows because in this case the inductor simply acts as a resistor and there is no inductive reactance. Whereas, when we connect the same inductor across 200 V AC, due to inductive reactance the overall impedance is changed and hence the value of current also changes.  

When the inductor is connected across 200 V DC  

The resistance of the coil 

`R = 200/1 = 200 Ω`

Let the self-inductance of the inductor be L,
When the inductor is connected across 200 V A.C.

Net impedance, Z = `sqrt((2pifL)^2 + R^2) = 200/0.5 = 400 Omega`

⇒ `(2pi xx 50 xx L)^2 + 200^2 = 400^2`

⇒ `(100piL)^2 = 160000 - 40000`

⇒ `10000pi^2L^2 = 120000`

⇒ `L^2 = 12/pi^2`

`therefore L = sqrt(12)/pi H`