Not a typical RL circuit...

[rickyw2777]

Homework Statement

The circuit contains an ideal battery with emf E = 30 V, a resistor R1 = 10 Ω, and a parallel
branch consisting of a resistor R2 = 20 Ω and an ideal inductor L (the inductor’s internal resistance is negligible). The branch is connected to the battery through a switch K (see Fig.)
The switch K is left open for a long time, then at t = 0 it is suddenly closed, and after a long
time it is opened again.
Find the current I20 through resistor R2:
1. immediately after the switch is closed;
2. after a sufficiently long time has passed with the switch closed;
3. immediately after the switch is opened again

Relevant Equations

I(t)=I of infinite*(1-e^(-t/tau))
I of infinite=epsilon/R
tau=L/R

I tried to use the formula of I(t), but I can not determine the R to substitute. Also, it is the circuit of both parallel and series, so I am not sure how the current goes.

 

[kuruman]

You don't need any formula to answer this. You need to understand what kind of circuit element the inductor behaves as in each of the limiting cases. Before t = 0 there is no current anywhere.

1. Immediately after the switch is closed, the inductor will oppose any current that tries to go through it so it behaves as a _________________ .
2. After a long time with the switch closed, there is a steady current everywhere, so the inductor behaves as a _________________ .
3. Immediately after the switch is opened again, the inductor behaves as a _________________ . (Don't forget there is a current in it up to that point.)

I suggest that you draw three separate circuit diagrams with the inductor replaced by what you put in the blanks and analyze each separately to find the current in the 20 Ω resistor.

 

[Gordianus]

What formula did you use for I(t)?

 

[rickyw2777]

You don't need any formula to answer this. You need to understand what kind of circuit element the inductor behaves as in each of the limiting cases. Before t = 0 there is no current anywhere.

1. Immediately after the switch is closed, the inductor will oppose any current that tries to go through it so it behaves as a _________________ .
2. After a long time with the switch closed, there is a steady current everywhere, so the inductor behaves as a _________________ .
3. Immediately after the switch is opened again, the inductor behaves as a _________________ . (Don't forget there is a current in it up to that point.)

I suggest that you draw three separate circuit diagrams with the inductor replaced by what you put in the blanks and analyze each separately to find the current in the 20 Ω resistor.

I understand now. Thank you!

 

[Gavran]

I tried to use the formula of I(t), but I can not determine the R to substitute. Also, it is the circuit of both parallel and series, so I am not sure how the current goes.

The formula can be obtained by using the Thevenin equivalent circuit at the terminals of the inductor.