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प्रश्न
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
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उत्तर
Initial current, I1 = 5.0 A
Final current, I2 = 0.0 A
Change in current, dI = I1 − I2 = 5 A
Time taken for the change, t = 0.1 s
Average emf, e = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as:
e = `"L""dl"/"dt"`
`"L" = "e"/((("dl")/("dt")))`
= `200/(5/0.1)`
= 4 H
Hence, the self-induction of the coil is 4 H.
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संबंधित प्रश्न
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| 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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