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Question
A coil of self-inductance 2.5H and resistance 20Ω is connected to a battery of emf 120V having the internal resistance of 5 n. Find:
1) The time constant of the circuit.
2) The current in the circuit in steady state
Solution
Given
l = 2.5h
R = 20Ω
E = 120V
r = 5Ω
1) Tme constant , `t = L/R = 2.5/20 = 1/8 = 0.125 s`
2) Current, `I = E/(R + r) =- 120/25 = 4.8 A`
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