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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 |
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Solution
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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