Question

Calculate the self-inductance of a coil using the following data obtained when an AC source of frequency `(200/pi)` Hz and a DC source are applied across the coil.

AC Source
S.No.	V (volts)	I (A)
1	3.0	0.5
2	6.0	1.0
3	9.0	1.5

DC Source
S.No.	V (volts)	I (A)
1	4.0	1.0
2	6.0	1.5
3	8.0	2.0

Answer 1

For the AC source given in the question, taking the first value as V = 3 volts, I = 0.5 A and calculating inductive reactance by the formula,

X_L = `V/I`

Where X_L is the inductive reactance, V is the voltage, I is the current.

`X_L = 3/0.5`

∴ `X_L = 6Omega`

and `f = 200/pi` (given)

`X_L = omegaL`

`X_L = 2pifL`

Where L is the inductance of the inductor, f is the frequency of the alternating current.

∴ `6 = 2pi xx (200/pi) xx L`

`L = 6/((2 xx 200))`

`L = 3/200` henry

L = 15 mH

Hence, the value of the self-inductance of a coil is 15 mH.