Current flowing in a branch in a circuit

In summary: 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.
  • #1
Jahnavi
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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
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
cnh1995 said:
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
Jahnavi said:
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?
 
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  • #5
cnh1995 said:
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
cnh1995 said:
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
Jahnavi said:
I tried your approach . This also gives 3.36 A .
Right. I get the same answer.
 
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  • #8
Jahnavi said:
No .

It should be (4/7)i - i1 . But this gives 3.36
The same as I got.
 
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  • #9
Is it only me who is not able to see 3.36mA in any of the given options o0) ?
 
  • #10
Jahnavi said:
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:
 
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  • #11
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
cnh1995 said:
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
Jahnavi said:
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.
 
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  • #14
Thank you so much !
 
Last edited:

Related to Current flowing in a branch in a circuit

1. What is current flowing in a branch in a circuit?

Current flowing in a branch in a circuit refers to the movement of electric charge through a specific path or branch in a circuit. This current is measured in amperes (A) and is caused by the flow of electrons from a higher potential to a lower potential.

2. How is current flowing in a branch in a circuit calculated?

To calculate the current flowing in a branch in a circuit, you can use Ohm's Law which states that current (I) is equal to voltage (V) divided by resistance (R). This can be represented as I = V/R. Alternatively, you can use Kirchhoff's Current Law which states that the sum of all currents entering and leaving a node in a circuit must equal zero.

3. What factors affect the amount of current flowing in a branch in a circuit?

The amount of current flowing in a branch in a circuit is affected by the voltage applied, the resistance of the branch, and the type of circuit (series or parallel). Higher voltage and lower resistance will result in a higher current, while a series circuit will have the same current in all branches and a parallel circuit will have different currents in each branch.

4. How does current flowing in a branch in a circuit affect the components in the branch?

The amount of current flowing in a branch in a circuit can affect the components in that branch in different ways. If the current exceeds the component's rated capacity, it can cause damage or failure. On the other hand, if the current is too low, the component may not function properly. Additionally, the direction of current flow can also affect certain components, such as diodes and transistors.

5. Can current flowing in a branch in a circuit be controlled?

Yes, the amount of current flowing in a branch in a circuit can be controlled by using components such as resistors, capacitors, and inductors. These components can be used to limit or regulate the current in a circuit by changing the voltage or resistance. Additionally, using a variable power supply or adjusting the circuit's resistance can also control the amount of current flowing in a branch.

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