10 joules of work is done by the object
I believe the answer is free electrons
Answer:
Explanation:
Given an RL circuit
A voltage source of.
V = 108V
A resistor of resistance
R = 1.1-kΩ = 1100 Ω
And inductor of inductance
L = 34 H
After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor
A. Time the inductor current will reduce to 12% of it's initial current
Let the initial charge current be Io
Then, final current is
I = 12% of Io
I = 0.12Io
I / Io = 0.12
The current in an inductor RL circuit is given as
I = Io ( 1—exp(-t/τ)
Where τ is time constant and it is given as
τ = L/R = 34/1100 = 0.03091A
So,
I = Io ( 1—exp(-t/τ))
I / Io = ( 1—exp(-t/τ))
Where I/Io = 0.12
0.12 = 1—exp(-t/τ)
0.12 — 1 = —exp(-t/τ)
-0.88 = -exp(-t/0.03091)
0.88 = exp(-t/0.03091)
Take In of both sides
In(0.88) = In(exp(-t/0.03091)
-0.12783 = -t/0.030901
t = -0.12783 × 0.030901
t = 3.95 × 10^-3 seconds
t = 3.95 ms
B. Energy stored in inductor is given as
U = ½Li²
So, the current at this time t = 3.95ms
I = Io ( 1—exp(-t/τ))
Where Io = V/R
Io = 108/1100 = 0.0982 A
Now,
I = Io ( 1—exp(-t/τ))
I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))
I = 0.0982(1—exp(-0.12783)
I = 0.0982 × 0.12
I = 0.01178
I = 11.78mA
Therefore,
U = ½Li²
U = ½ × 34 × 0.01178²
U = 2.36 × 10^-3 J
U = 2.36 mJ
