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Current flowing in a branch in a circuit

  • Thread starter Jahnavi
  • Start date
  • #1
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102

Homework Statement


circuit3.jpg


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 ?
 

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  • #2
cnh1995
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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.
 
  • #3
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102
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 .
 
  • #4
cnh1995
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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-I.
Is that correct?
 
  • #5
848
102
How? I see you have assumed I=I-I.
Is that correct?
No .

It should be (4/7)i - i1 . But this gives 3.36
 
  • #6
848
102
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 :wideeyed: .
 
  • #7
cnh1995
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I tried your approach . This also gives 3.36 A .
Right. I get the same answer.
 
  • #8
ehild
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No .

It should be (4/7)i - i1 . But this gives 3.36
The same as I got.
 
  • #9
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102
Is it only me who is not able to see 3.36mA in any of the given options o0) ?
 
  • #10
cnh1995
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Is it only me who is not able to see 3.36mA in any of the given options o0) ?
It's not the first time that the options are incorrect.:wink:
 
  • #11
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102
By the way , I agree that your approach is much simpler :smile: and most probably the paper setter expects us to apply the Node Voltage method .
 
  • #12
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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 ?
 
  • #13
cnh1995
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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.
 
  • #14
848
102
Thank you so much !
 
Last edited:

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