Question

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.

Answer 1

a.) E = 30.8J

b.) E = 12.3J

Step-by-step 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

Answer 2

A) 30.82 J

B) 12.31 J

Step-by-step 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