Current flowing in a branch in a circuit

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

For calculation sake I will consider R = 1Ω . Final answer will be in mA instead of A .

The middle two resistors 3R and 4R are in parallel , so I consider their equivalent resistance 12R/7 in the middle branch .

In the left part of the circuit , suppose i1 current flows in R .

In the right part of the circuit , suppose i2 current flows in 2R .

Current flowing in the middle 12R/7 resistor will be i

Using KCL i = (i1+i2)

Writing KVL in the left loop ,
12 - i1 - 12i/7 = 0

Writing KVL in the right loop ,
24 - 2i2- 12i/7 = 0

Solving for i , i = 6.72

This is the sum of the currents that will flow in the middle 3R and 4R resistors .Since they are in parallel , current will be divided in the ratio 4:3 .This means (1/7)th of the current i =6.72 flows in the middle lower branch ( the one that is asked in the question )

This gives 0.96 which is incorrect .

What is the mistake ?

 

[cnh1995]

I believe node voltage method is an easier way to do this. Call the middle node voltage V and write KCL at that node. Assume the negatives of the batteries to be the ground node.

 

[Jahnavi]

I believe node voltage method is an easier way to do this. Call the middle node voltage V and write KCL at that node. Assume the negatives of the batteries to be the ground node.

I will do as you suggest . But for now , please help me identify the mistake in the OP .

 

[cnh1995]

This means (1/7)th of the current i =6.72 flows in the middle lower branch ( the one that is asked in the question )

How? I see you have assumed I=I_3Ω-I_4Ω.
Is that correct?

 

[Jahnavi]

How? I see you have assumed I=I_3Ω-I_4Ω.
Is that correct?

No .

It should be (4/7)i - i1 . But this gives 3.36

 

[Jahnavi]

I believe node voltage method is an easier way to do this. Call the middle node voltage V and write KCL at that node. Assume the negatives of the batteries to be the ground node.

I tried your approach . This also gives 3.36 .

 

[cnh1995]

I tried your approach . This also gives 3.36 A .

Right. I get the same answer.

 

[ehild]

No .

It should be (4/7)i - i1 . But this gives 3.36

The same as I got.

 

[Jahnavi]

Is it only me who is not able to see 3.36mA in any of the given options ?

 

[cnh1995]

Is it only me who is not able to see 3.36mA in any of the given options ?

It's not the first time that the options are incorrect.

 

[Jahnavi]

By the way , I agree that your approach is much simpler and most probably the paper setter expects us to apply the Node Voltage method .

 

[Jahnavi]

Assume the negatives of the batteries to be the ground node.

This is quite helpful . What if the polarity of one of the batteries is reversed ?

Do you still think that node voltage method will still be more efficient and simpler considering that now there will be two equations to deal with ?

 

[cnh1995]

Do you still think that node voltage method will be still more efficient and simpler considering that now there will be two equations to deal with ?

Yes. No matter what the polarities are, there are still only 4 nodes, one of which is the ground node and you know two node voltages. So you end up with a single equation with a single variable.
In general, you can assume any node to be the ground node. As long as you are consistent with signs in KCL, you are fine.

 

[Jahnavi]

Thank you so much !