Question

When a coil is connected to a 200 V dc supply, the current flowing through it is found to be 1A. However, when it is connected to the 200 V, 50 Hz ac supply, the current is found to be 0.5 A. 

1. Explain why current flowing is less when the coil is connected to an ac supply.
2. Calculate coefficient of self-inductance (L) of the coil.

Answer 1

(1) The current flowing is less when the coil is connected to an AC supply.

1. When the coil is connected to an AC supply, it offers inductive reactance in addition to resistance.
2. Due to self-inductance, an induced emf opposes the change in current (Lenz’s law).
3. This opposition increases the coil’s effective impedance in AC.
4. Hence, the current flowing in AC is less than in DC.

(2) Coefficient of self-inductance (L) of the coil.

In DC supply:

`R = V/I`

= `200/1`

= 200

In AC supply:

Z = `V/I`

= `200/0.5`

= 400

The relationship between impedance, resistance, and reactance in a series RL circuit is:

`Z = sqrt(R^2 + (X_L)^2)`

`400 = sqrt(200^2 + X_L^2)`

400² = 200² + X_L²

160000 = 40000 + X_L²

X_L² = 160000 − 40000 

X_L² = 120000

`X_L = sqrt(120000)`

= 346.4 Ω

The formula for inductive reactance is:

X_L = 2πfL

L = `(X_L)/(2pif)`

= `346.4/(2pi xx 50)`

= `346.4/(2 xx 3.1416 xx 50)`

= `346.4/314.16`

= 1.10 H