Question

At t = 0, a battery is connected to a series arrangement of a resistor and an inductor. If the inductive time constant is 43.4 ms, at what time is the rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in the inductor's magnetic field?

Answer 1

Answer:30.08 ms

Step-by-step explanation:

Given

time Constant
`\tau =43.4 ms =(L)/(R)`

Also rate at which energy is dissipated in the resistor equal to the rate at which energy is stored in inductor's magnetic Field

Energy stored in Inductor is
`U_L=(1)/(2)Li^2`

rate of Energy storing
`(dU_L)/(dt)=(1)/(2)L\cdot 2i* (di)/(dt)----1`

Rate of Energy dissipation from resistor i.e. Power is given by

`(dU_R)/(dt)=i^2R-----2`

Equating 1 and 2

`Li\cdot (di)/(dt)=i^2R`

`L((di)/(dt))=R(i)-----3`

i is given by
`i=(V)/(R)(1-e^{-(t)/(\tau )})`

`(di)/(dt)=(V)/(L)e^{-(t)/(\tau )}`

substitute the value of
`(di)/(dt)` in 3

`L((V)/(L)e^{-(t)/(\tau )})=R\cdot (V)/(R)(1-e^{-(t)/(\tau )})`

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

`2e^{-(t)/(\tau )}=1`

`e^{-(t)/(\tau )}=0.5`

`e^{-(t)/(43.4* 10^(-3))}=0.5`

`(t)/(43.4* 10^(-3))=0.693`

`t=30.08 ms`