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प्रश्न
Explain why the inductance of two coils connected in parallel is less than the inductance of either coil.
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उत्तर १
- For a parallel combination of two coils, the current through each parallel inductor is a fraction of the total current and the voltage across each parallel inductor is the same.
- As a result, a change in total current will result in less voltage dropped across the parallel array than for any one of the individual inductors.
- There will be less voltage drop across parallel inductors for a given rate of change in current than for any of the individual inductors.
- Less voltage for the same rate of change in current results in less inductance.
- Thus, the total inductance of two coils is less than the inductance of either coil.
उत्तर २
When two inductors with inductances L1 and L2 are connected in parallel, the equivalent inductance is given by
`1/"L"_"equivalent" = 1/"L"_1 + 1/"L"_2`
which is less than the individual inductance value L1 and L2.
As a result, the inductance of two parallel coils is smaller than the inductance of either coil.
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संबंधित प्रश्न
Explain the phenomenon of self induction
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Choose the correct option
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Energy needed to establish an alternating current I in a coil of self-inductance L is
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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 |
An air-cored solenoid, 40 cm long and of cross-sectional area 5 cm2, is tightly wound with 400 turns of copper wire and carries a steady current of 10 A. (a) Calculate the self-inductance of the solenoid. (b) Find the emf induced if the current in the solenoid decreases to zero in 0.2 s.
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Obtain an expression for the self inductance of a solenoid.
