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2 November, 05:30

A resistor and an inductor are connected in series to a battery with emf 240 V and negligible internal resistance. The circuit is completed at time t=0. At a later time t=T the current is 3.00 A and is increasing at a rate of 20.0 A/s. After a long time the current in the circuit is 25.0 A. What is the value of T?

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  1. 2 November, 06:21
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    Answer: 0.141s

    Explanation: given

    Battery EMF = 240

    Current = 25A

    Rate of Current = 20A/s

    Time = ?

    From Ohms law, V=IR

    R = V/I

    R = 240/25

    R = 9.6Ω

    Voltage that pass across the inductor at 3A = V

    V = 240 - [ (3*9.6) remember, v=ir]

    V = 240 - 28.8

    V = 211.2

    Also, V = L[d (i) / d (t) ]

    211.2 = L * 20

    L = 10.56H

    To get the time constant, then,

    τ = L/R

    τ = 10.56/9.6

    τ = 1.1s

    i = i•[1 - e^ (-T/τ) ]

    3 = 25[1 - e^ (-T/1.1) ]

    0.12 = 1 - e^ (-T/1.1)

    e^ (-T/1.1) = 1 - 0.12

    e^ (-T/1.1) = 0.88

    Taking log of both sides

    -T/1.1 = - 0.128

    -T = - 0.1408

    T = 0.141s
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