Current through shorted branch -- clarification

[palaphys]

Homework Statement

clarification in a part of a larger question

Relevant Equations

v=ir

View attachment 365575
In this seemingly simple circuit, applying KCL, I get the current in the shorted branch as
$3E/(2R) - 2E/(2R) = E/(2R)$ but some sources suggest that it is $5E/ (2R)$

which is right and why?

 

[nasu]

Are you sure that the polarity of the batteries in "some source" was the same as in your case?

 

[palaphys]

Are you sure that the polarity of the batteries in "some source" was the same as in your case?

okay let me post the larger part of the question here.

this was the original circuit. I am required to find the voltage across the inductor after S2 is closed at t=0+, given that initially S1 was closed for a long time, so as to establish an initial current $i_0= E/R$ through the inductor.
Here is my approach:
I can write $i(t)= i_0 e^{-t/\tau} + i_{s.s} (1-e^{-t/\tau})$ and for voltage at t=0
$L (di/dt) = \frac{L}{\tau}(i_{s.s} -i_0)$ (at t=0)

now, I have the time constant of the circuit, and the steady state current. the question also says that
$E = 20\text{V}, L = 0.5\text{H and } R = 10\Omega.$

I want to know if this approach is right

 

[nasu]

Is this diagram from the "some source"? Adding the inductor does not change the premise of your original question. "Some source" is not a very useful reference.

 

[palaphys]

Is this diagram from the "some source"? Adding the inductor does not change the premise of your original question. "Some source" is not a very useful reference.

I think there is some misunderstanding here. My original question posted here was a part of a larger question, which is the one described above in the previous message.

 

[palaphys]

okay let me post the larger part of the question here.View attachment 365577
this was the original circuit. I am required to find the voltage across the inductor after S2 is closed at t=0+, given that initially S1 was closed for a long time, so as to establish an initial current $i_0= E/R$ through the inductor.
Here is my approach:
I can write $i(t)= i_0 e^{-t/\tau} + i_{s.s} (1-e^{-t/\tau})$ and for voltage at t=0
$L (di/dt) = \frac{L}{\tau}(i_{s.s} -i_0)$ (at t=0)

now, I have the time constant of the circuit, and the steady state current. the question also says that
$E = 20\text{V}, L = 0.5\text{H and } R = 10\Omega.$

I want to know if this approach is right

I think I have found my mistake.
In this charging equation, I had to substitute $i_{s.s}$ as -E/(2R) and not just E/2R . (As the steady state current in the inductor is upward, as opposed to the initial current which is downward.) Have I identified my mistake correctly?(I'm getting the correct answer when I make this change) Or is it a coincidence, and something else is wrong?

 

[palaphys]

Are you sure that the polarity of the batteries in "some source" was the same as in your case?

also, to be precise the source is a Youtube video, which uses the incorrect polarity. thanks for pointing that out.

 

[cnh1995]

I think I have found my mistake.
In this charging equation, I had to substitute $i_{s.s}$ as -E/(2R) and not just E/2R . (As the steady state current in the inductor is upward, as opposed to the initial current which is downward.) Have I identified my mistake correctly?(I'm getting the correct answer when I make this change) Or is it a coincidence, and something else is wrong?

What is the correct answer that you are getting? Is it for the voltage across the inductor or for the current through the inductor?

As this problem is about the instant immediately after closing the switch, you really don't need the full time-dependent solution for the current or the voltage. You just need to remember how an inductor behaves at t=0+.

 

[nasu]

I think there is some misunderstanding here. My original question posted here was a part of a larger question, which is the one described above in the previous message.

The larger question does not matter for the OP question. As you already figured out, it was just a matter of comparing two different circuits so each version of the answer is right for the right circuit.

 

[palaphys]

What is the correct answer that you are getting? Is it for the voltage across the inductor or for the current through the inductor?

As this problem is about the instant immediately after closing the switch, you really don't need the full time-dependent solution for the current or the voltage. You just need to remember how an inductor behaves at t=0+.

the correct answer is for voltage at t=0+ is -E.