Question

A solenoid having an inductance of 5.41 μH is connected in series with a 0.949 kΩ resistor. (a) If a 16.0 V battery is connected across the pair, how long will it take in seconds for the current through the resistor to reach 77.9% of its final value? (b) What is the current through the resistor at a time t = 1.00τL?

Answer 1

(A)
`9.14* 10^(-9)sec`

(B)
`6.20* 10^(-3)A`

Step-by-step explanation:

We have given inductance
`L=5.41\mu H=5.41* 10^(-6)H`

Resistance
`R=0.949kohm=0.949* 10^3ohm`

Time constant of RL circuit is equal to
`\tau =(L)/(R)`

`\tau =(5.41* 10^(-6))/(0.949* 10^3)=5.70* 10^(-9)sec`

Battery voltage e = 16 volt

(a) It is given current becomes 79.9% of its final value

Current in RL circuit is given by

`i=i_0(1-e^{(-t)/(\tau )})`

According to question

`0.799i_0=i_0(1-e^{(-t)/(\tau )})`

`e^{(-t)/(\tau )}=0.201`

`{(-t)/(\tau )}=ln0.201`

`{(-t)/(5.7* 10^(-9) )}=-1.6044`

`t=9.14* 10^(-9)sec`

(b) Current at
`t=\tau sec`

`i=i_0(1-e^{(-t)/(\tau )})`

`i=(16)/(0.949* 10^3)(1-e^{(-\tau )/(\tau )})`

`i=6.20* 10^(-3)A`