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प्रश्न
In a given coil of self-inductance of 5 mH, current changes from 4 A to 1 A in 30 ms. Calculate the emf induced in the coil.
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उत्तर
Self inductance, L=5 mH=5×10−3 HChange in current,
dI=(1−4)=−3 A
Change in time, `dt=30 ms=30×10−3 s`
The emf induced in the coil is given by ` e=−L(dI)/(dt)`
`⇒e=((−5xx10^(−3) )xx(−3))/(30×10^(−3))`
`⇒e=0.5 V`
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संबंधित प्रश्न
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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 |
