Question

A 8.3-V battery is connected in series with a 43-mH inductor, a 180-Ω resistor, and an open switch. What is the current in the circuit 0.120 ms after the switch is closed? How much energy is stored in the inductor at this time?

Answer 1

18.26 mA

Step-by-step explanation:

We have given voltage V = 8.3 Volt

Resistance R=180 ohm

Inductance L=43 mH

Time at which we have to find the current t=0.120 ms

The current in the RL circuit is given by
`i=(V)/(R)(1-e^(-t)/(\tau ))` here
`\tau` is time constant which is given by
`\tau =(L)/(R)=(43* 10^(-3))/(180)=0.238ms`

So the current
`i=(V)/(R)(1-e^(-t)/(\tau ))=(8.3)/(180)(1-e^(-0.120)/(0.238))=18.26mA`