Answer:
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
Answer:
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
