A 24-V battery is connected in series with a resistor and an inductor, with R = 5.8 Ω and L = 3.6 H, respectively. (a) Find the energy stored in the inductor when the current reaches its maximum value. J (b) Find the energy stored in the inductor one time constant after the switch is closed.

a.) E = 30.8J

b.) E = 12.3J

Explanation:

Given that the

Voltage V = 24v

Resistance R = 5.8 Ω

Inductor L = 3.6 H

The relationship between the energy and inductor is

E = 1/2LI^2

Where

E = energy stored

I = current in the circuit

But V = IR + Ldi/dt

Where

Current I = a (1-e^-kt)

for large t, current will be

i = 24/5.8 = a

so

a = 4.14

i = 4.14 (1-e^-kt)

di/dt = 4.14 k e^-kt

24 = 24-24e^-kt + 2 (4.14) k e^-kt

24 = 2 (4.14) k

k = 24 / (2 * 4.14) = R/L

so

i = 4.14 (1-e^ - (Rt/L))

current is max at large t

i max = 4.14 - 0

Substitute the values of current and inductor into the formula

Energy E = (1/2) L i^2 = (1/2) (3.6) 4.14^2

= 30.8Joules

For one time constant T = L/R and

e^ - (Rt/L) = 1/e = 0.368

i = 4.14 (1-.368) = 4.14 * 0.632 = 2.62A

Substitute the values of current and inductor into the formula

Energy E = (1/2) (3.6) 2.62^2 = 12.32 Joules

A) 30.82 J

B) 12.31 J

Explanation:

We are given;

Voltage; E = 24V

Resistance; R = 5.8 Ω

Inductance; L = 3.6 H

A) The formula for the maximum current is: I_max = E/R

Where E is voltage and R is resistance

So, I_max = 24/5.8 A

Now, the formula for the Energy stored in the inductor at maximum current is;

P_max = ½L (I_max) ²

P_max = ½ * 3.6 * (24/5.8) ²

P_max = 30.82 J

B) The formula for the current after the switch is closed is;

I = I_max (1 - e^ (-tr))

Since, one time constant, then tr = 1

So, I = (24/5.8) (1 - e^ (-1))

I = (24/5.8) (1 - 0.3679)

I = 2.6156 A

Energy stored in the inductor is;

P = ½LI²

P = ½ * 3.6 * 2.6156²

P = 12.31 J