Question

 A circuit is constructed with four resistors, one inductor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 48 Ω, R3 = 100 Ω and R4 = 130 Ω. The inductance is L = 330 mH and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign.

1)

The switch has been open for a long time when at time t = 0, the switch is closed. What is I₄(0), the magnitude of the current through the resistor R₄ just after the switch is closed?

2)

What is I₄(?), the magnitude of the current through the resistor R₄ after the switch has been closed for a very long time?

3)

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

4)

After the switch has been closed for a very long time, it is then opened. What is I₃(t_open), the current through the resistor R₃ at a time t_open = 5.6 ms after the switch was opened? The positive direction for the current is indicated in the figure.

5)

What is V_L,max(closed), the magnitude of the maximum voltage across the inductor during the time when the switch is closed?

6)

What is V_L,max(open), the magnitude of the maximum voltage across the inductor during the time when the switch is open?

Answer 1

Something we need to understand before solving the problem is that when we have an inductor and the problem have a switch that will be open/closed for long/short periods of time the inductor acts like a huge resistor that basically drains all the energy and dont let it flow when the switch is recently opened (t=0) and will act like a wire (as if it was not even on the picture) when the switch was opened a long time ago (t=infinity). This will help us solve some of the questions.

Question 1

To calculate the current in resistor 4 we have

$I_4=\frac{V}{R_T}$

And R_T is the sum of all the resistors, but, since t=0 no current will flow after the inductor so we ignore R₂

Question 2

Same as last question but this time we do consider R₂ since the inductor is acting as just a wire (we ignore it)

Question 3

Again like last question the inductor acts like a wire so the current throught it will be the same as the current on R₂ (because they are in series) so we basically need I₂ as if there was no inductor there. In order to find it we need the voltage that is going on the second loop that splits (parallel) between R₃ and R₂ and the current for this case is the same as last problem since the conditions are the same

Now we calculate I₂ since is the same current as the current in the inductor and remember that we calculated the voltage that goes to both R₃ and R₂ so V₂₃=V₂=V₃

Question 4

Now to find current with a certain specific time we go by:

I made it bigger so you can see it better. Now we need since I_max is the current we calculated before since is the current with the switch open for the maximum time and t is the given time (in seconds). We have then

And R is the sum of the resistors in the loop which are R₂ and R₃ and this time they are in series (only for this loop not the circuit) so we have

This is a complex number so i will keep it for the formula to avoid having a serious difference in the result due to using less significant numbers.

Is negative because it goes in the opposite direction as of what the problem says is positive

Question 5

For this question we have that

And since is max closed we take as if it were t=0 (same current as question 1) and the resistor R₃ is parallel to the inductor the voltage in both of them will be the same and like i said in question 1 we ignore R₂, so we have.

Question 6

This time we follow the same procedure as last question but this time we have max open, so, we take the I from question 3 (same conditions) and the resistance now includes R₂ so we need the total resistance R₂₃ that are in parallel for this loop so we have

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This was a really long question, i hope you could understand everything.

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