Question

A 8.2-V battery is connected in series with a 38-mH inductor, a 150-Ω resistor, and an open switch.A 8.2-V battery is connected in series with a 38-mH inductor, a 150-Ω resistor, and an open switch. Part A

What is the current in the circuit 0.110 ms after the switch is closed? unit (mA)
Part B
How much energy is stored in the inductor at this time? unit(micro J)

Answer 1

(A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

Step-by-step explanation:

Given that,

Voltage = 8.2 V

Inductor = 38 mH

Resistance = 150 Ω

Time t = 0.110 ms

The battery has negligible internal resistance, so that the total resistance in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance

We need to calculate the current

Using formula of current

`I(t)=(V)/(R)*(1-e^{-t*(R)/(L)})`

Put the value into the formula

`I(t)=(8.2)/(150)*(1-e^{-0.110*10^(-3)*(150)/(38*10^(-3))})`

`I(t)=0.01925\ A`

`I(t) = 19.25\ mA`

(B). We need to calculate the store energy in the inductor

Using formula of energy

`E=(1)/(2)LI^2`

Put the value into the formula

`E=(1)/(2)*38*10^(-3)*(0.01925)^2`

`E=7.04*10^(-6)\ J`

{tex]E=7.04\ \mu J[/tex]

Hence, (A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.