Question

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?

Answer 1

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