RL Circuit

Question

RL Circuit

anonymous · Answer

attachment

anonymous · Answer

A circuit is constructed with four resistors, one inductor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 25 Ω, R3 = 119 Ω and R4 = 146 Ω. The inductance is L = 281 mH and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign.

anonymous · Answer

) What is IL(∞), the magnitude of the current through the inductor after the switch has been closed for a very long time?

anonymous · Answer

An explanation more than answer would be greatly appreciated

ybarrap · Answer

The impedence across the inductor in steady-state is $j\omega L$, where $\omega=0$; therefore, the potential, using Ohms Law, is $I\times j\omega L=0$. This means that at steady-state, the inductor acts like a short-circuit. In this situation, you can ignore the current through $R_3$, since the potential across it is zero.

Now you can find the current through the inductor by treating $R_1,R_2 $ and $R_4$ in series: $I_L=\cfrac{V}{R_1+R_2+R_4}$.

That's it!.

Hope this makes sense.

anonymous · Answer

I tried that and got this response from the software. 
"It looks like you have calculated IL(∞) to be equal to the battery voltage divided by the sum of R1, R2 and R4. If this were true, the current through R3 would have to be zero. The current through R3 cannot be zero since the current through the inductor gives rise to a voltage drop across R2. To answer this question, you need to know the voltage across the inductor a very long time after the switch is closed." @ybarrap

ybarrap · Answer

Yes, that's right. I looked at R3 as not in the second loop but in the 1st. What happens in steady state IS that the voltage across the inductor is 0 but then the equivalent resistance (with $R_3 $ and $R_2$) will be $R_{eq}=R_3||R_2=\cfrac{R_3R_2}{R_3+R_2}$.

Then the current through $R_1$ and $R_4$ is $I=\cfrac{V}{R_1+R_{eq}+R_4}$.

Knowing this, then the current through the inductor, which is the same as the current through $R_2$ is $I_L=\cfrac{IR_3}{R_3+R_3}$, in steady state.

ybarrap · Answer

*I looked at R2 as not in the second loop ...

anonymous · Answer

@ybarrap Would you know how to approach this? After the switch has been closed for a very long time, it is then opened. What is I3(topen), the current through the resistor R3 at a time topen = 4.5 ms after the switch was opened? The positive direction for the current is indicated in the figure.

ybarrap · Answer

When the switch is closed for a long time then opened, there will not be a path from the battery to ground, so it immediately can be disregarded and treated as an open circuit. The only  energy available is that which is stored in the form of electromagnetic energy in the inductors. Therefore, there will only be current in the second loop, containing $R_3, R_2 $ and $L$.  $R_1$ and $R_4$ are completely out of of the picture since there is no current flowing in or out of the battery, due to lack of complete path from the positive terminal to negative terminal of the battery.

So we are just left with the second loop.

Since the question is asking us for the current after a specific time, we need to use the transient equation that describes what happens to the current in the inductor as its electromagnetic energy is released and converted to heat in resistors $R_3$ and $R_2$.

The equation we need is the one the that shows how the current through the inductor $decreases$  after the switch is opened and energy from the battery is no longer available. This decrease in current causes an EMF ($E$) that results from Faraday's Law of induction: $E=-\cfrac{d}{dt} \Phi_B=-L\cfrac{dI}{dt}$, where $\Phi_B$ is the magnetic flux through the inductor. the negative sign indicates that the voltage polarity will be opposing the change in current:

$IR_2+IR_3=-L\cfrac{dI}{dt}$.

This is a separable 1st order diffy-Q:

$$
-\cfrac{R_2+R_3}{L}dt=\cfrac{dI}{I}\
-\cfrac{R_2+R_3}{L}\int_0^{.0045} dt=\int_{I_0}^{I_f} \cfrac{dI}{I}\
-\cfrac{R_2+R_3}{L}(.0045)=\ln{I_f}-\ln{I_0}\
\exp[-\cfrac{R_2+R_3}{L}(.0045)]=\exp[\ln{I_f}-\ln{I_0}]=\cfrac{I_f}{I_o}\
\implies  I_f=I_0\times \exp[-\cfrac{R_3+R_2}{L}(.0045)]
$$
Where, $I_0=I_L$, calculated above, is the current that was going through $L$ when the switch was closed and in steady state and $I_f$ is the current flowing through the inductor 4.5 ms after the switch is opened. This is the same current flowing through resistor $R_3$ based on our discussion above.

anonymous · Answer

@ybarrap Awesome explanation! Thank you so much!