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?

Respuesta :

Answer:

18.26 mA

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 [tex]i=\frac{V}{R}(1-e^\frac{-t}{\tau })[/tex] here [tex]\tau[/tex] is time constant which is given by [tex]\tau =\frac{L}{R}=\frac{43\times 10^{-3}}{180}=0.238ms[/tex]

So the current [tex]i=\frac{V}{R}(1-e^\frac{-t}{\tau })=\frac{8.3}{180}(1-e^\frac{-0.120}{0.238})=18.26mA[/tex]