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.
Solution
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`
