Question

An inductor is connected to the terminals of a battery that has an emf of 12.0V and negligible internal resistance. The current is 4.86mA at 0.940ms after the connection is completed. After a long time the current is 6.45mA

Part A) What is the resistance R of the inductor?

R=?

Part B) What is the inductance L of the inductor?

L=?

Answer 1

PartA) resistance R=1.86Ω

PartB) inductnce L=1.26mH

Step-by-step explanation:

In an RL circuit,
`i(t)=(E)/(R)(1-e^{(-t)/((L)/(R))})`

where E is the emf of battery

L is inductance of inductor

R is the internal resistance of inductor

given
`i(\infty)=`
`i`max=
`(E)/(R)`

⇒
`6.45=(E)/(R)`

⇒
`6.45=(12)/(R)`

⇒
`R=(6.45)/(12)=1.86`Ω

given
`i(0.94ms)=4.86`

⇒
`4.86=6.45(1-e^{(-0.94*10^(-3))/((L)/(R))})`

⇒
`0.75=1-e^{(-0.94*10^(-3))/((L)/(R))}`

⇒
`0.25=e^{(-0.94*10^(-3))/((L)/(R))}`

applying logarithm on both sides,

⇒
`log(0.25)=(-0.94*10^(-3))/((L)/(R))`

⇒
`log(4)=(0.94*10^(-3))/((L)/(1.86))`

⇒
`L=(0.94*10^(-3)*1.86)/(log4)=1.26mH`