Answer :
After the switch is opened in an RL circuit, the current decreases, and the inductor induces an emf to sustain it. The current is given by I(t) = I0 * e^(-(R/L)t), and the voltage across the inductor is ε_L = -(I0 * e^(-(R/L)t)) * R.
When the switch S in an RL circuit is opened after a long time, the current begins to decrease. As the current changes, the inductor induces an electromotive force (emf) in the same direction as the original emf to keep the current from dying out immediately. We can analyze this situation using Kirchhoff's loop rule for the RL circuit.
Kirchhoff's loop rule for the RL circuit is given by:
ε - L(dI/dt) - IR = 0
where:
ε is the electromotive force (emf) of the voltage source,
L is the inductance of the inductor,
(dI/dt) is the rate of change of current with respect to time,
I is the current flowing through the circuit, and
R is the resistance in the circuit.
Since we are interested in what happens when ε (voltage source emf) is set to 0 (switch opened), the equation becomes:
0 - L(dI/dt) - IR = 0
Now, we can rearrange the equation to solve for the current (I) as a function of time (t):
L(dI/dt) + IR = 0
L(dI/dt) = -IR
Now, we'll solve this first-order ordinary differential equation. To do this, we'll separate variables and integrate both sides:
∫(dI/I) = -∫(R/L)dt
ln|I| = -(R/L)t + C
where C is the constant of integration.
To find the constant of integration (C), we can use the initial condition when the switch was closed for a long time (t = 0) and the current (I0) was at its maximum value. At t = 0, I = I0, so:
ln|I0| = C
Now, substitute the value of C back into the equation:
ln|I| = -(R/L)t + ln|I0|
Using logarithm properties, we can rewrite the equation as:
ln(|I|/|I0|) = -(R/L)t
Now, remove the logarithm by taking the exponent of both sides:
|I|/|I0| = e^(-(R/L)t)
Now, since I and I0 are positive values, we can remove the absolute value:
I/I0 = e^(-(R/L)t)
Now, we want to solve for I as a function of time, so:
I(t) = I0 * e^(-(R/L)t)
Finally, to obtain the voltage across the inductor (ε_L), we can use Ohm's law for an inductor:
ε_L = L * (dI/dt)
We have already calculated dI/dt from our initial differential equation:
L(dI/dt) = -IR
So:
ε_L = -IR
Now, using our expression for the current (I(t)) from earlier:
ε_L = -I(t) * R
Substitute the value of I(t):
ε_L = -(I0 * e^(-(R/L)t)) * R
Thus, the current across the inductor as a function of time is given by:
I(t) = I0 * e^(-(R/L)t)
And the voltage across the inductor as a function of time is given by:
ε_L = -(I0 * e^(-(R/L)t)) * R
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