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प्रश्न
A coil of self-inductance 3 H carries a steady current of 2 A. What is the energy stored in the magnetic field of the coil?
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उत्तर
Energy stored in the magnetic field, UB = `1/2 "LI"^2 = 1/2 xx 3 xx 2^2 = 6` J
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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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