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

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प्रश्न

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
योग
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उत्तर

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,

XL = `V/I`

Where XL 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.

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