Current flowing in the middle branch -- Why 2 different answers?

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

The two batteries are in series in both the left and right branch .This means the potential difference applied across the two middle resistors is 20 V .Hence the current flowing through them is 20/10= 2 A . This is indeed the correct answer .

Now , my doubt is that why does superposition method fails in this case ?

If I assume current I1 flowing clockwise in the left loop and I2 flowing anti clockwise in the right loop . I1 = 2A and I2 = 2A . Superimposing the two currents in the middle branch gives 4A . But this is incorrect .

 

[gneill]

When you apply superposition and you select a source to find its contribution to circuit conditions, what must you do to other sources in the circuit?

 

[Jahnavi]

When you apply superposition and you select a source to find its contribution to circuit conditions, what must you do to other sources in the circuit?

Replace the other voltage sources by a short circuit .

 

[gneill]

Replace the other voltage sources by a short circuit .

Right! And with the batteries being ideal, what are the consequences for the resulting circuit configuration?

 

[Jahnavi]

Does that mean Superposition method cannot be applied in this circuit ?

 

[gneill]

Does that mean Superposition method cannot be applied in this circuit ?

Yup!

If the batteries were not ideal and had some internal resistance (as all real components do), then you would leave the internal resistances in place when you suppress them and everything would be fine and you could go ahead and use superposition.

 

[Jahnavi]

Thanks !

 

[Jahnavi]

@gneill ,

I am really sorry . Actually in the OP , I tried the "Mesh Analysis" and not the Superposition method .

Why does "Mesh Current" method fails in this case ?

If the batteries were not ideal OR if there were resistors in the left top and left right branch , it works nicely .

Why does it fail this time ?

 

[gneill]

Why does it fail this time ?

It fails because you end up with two identical equations for the loops, which mathematically means that they are not independent. Hence you have only one equation and two unknowns.

 

[gneill]

You have two variables, $i_1$ and $i_2$, yes, but they should both appear in both equations when you perform mesh analysis on this circuit. Both currents run through the central branch.

 

[Jahnavi]

Sorry . I had deleted my post . I realize my mistake .

Thanks !