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?

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

i'm on my way to history not a mistory.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.