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A coil of self inductance 10 mH and resistance 0.1 Ω is connected through a switch to a battery of internal resistance 0.9 Ω. After the switch is closed, the time taken for the current to attain 80% of the saturation value is [take ln 5 = 1.6]
- 0.324 s
- 0.002 s
- 0.103 s
- 0.016 s
Correct answer: 0.016 s
Solution
The time constant for an RL circuit is given by τ = (L)/(Rₜₒₜₐₗ), where Rₜₒₜₐₗ includes both the coil's resistance and the internal resistance of the battery. In this case, the total resistance is 1.0 Ω, leading to a time constant of 0.01 s, and the time to reach 80% of the saturation current is approximately 0.2 times the time constant, resulting in 0.016 s.
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