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

Answer:

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

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

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

$I(t)=\dfrac{V}{R}\times(1-e^{-t\times\dfrac{R}{L}})$

Put the value into the formula

$I(t)=\dfrac{8.2}{150}\times(1-e^{-0.110\times10^{-3}\times\dfrac{150}{38\times10^{-3}}})$

$I(t)=0.01925\ A$

$I(t) = 19.25\ mA$

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

Using formula of energy

$E=\dfrac{1}{2}LI^2$

Put the value into the formula

$E=\dfrac{1}{2}\times38\times10^{-3}\times(0.01925)^2$

$E=7.04\times10^{-6}\ J$

{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.