Adding resistance and finding current in each branch

In summary, In this conversation, the student is struggling with a homework problem and is looking for help from an expert. The student has completed Kirchoff's laws and superposition theorem, but is struggling to find a solution to the problem. The student calculates the current and voltage drop through the circuit using these laws, then removes one of the voltage sources. They then use current division to calculate the current in each branch. They then use a matrix table to calculate the total current in each branch.
  • #1
nothing909
170
4

Homework Statement


I have to calculate the current through each branch - I1, I2 and I3.

Homework Equations

The Attempt at a Solution


The only reason I'm having trouble with this problem is because the voltage source is in the middle and I'm a little confused. I am struggling even to add up the resistances.

Are the 2, 5, and 4Ω resistors in series with each other? Do I just add them all up to get the total resistance?

After I have the total resistance, I calculated the supply current (I2) by just using ohms law.

How is it I calculate I1 and I3? Is it current division?
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  • #2
You can use KVL in the two loops..
 
  • #3
cnh1995 said:
You can use KVL in the two loops..
I forgot to mention that I'm doing superposition, I can't use KVL.
 
  • #4
nothing909 said:
I forgot to mention that I'm doing superposition, I can't use KVL.
 
  • #5
The resistors are not in series. In fact you should be able to confirm quite easily that the 2 and 4 Ohm resistors are in parallel (they share the same potential difference as their leads are connected to the same two nodes). In fact, there are three parallel branches comprising the circuit. What does that tell you about the potential difference across all three? What relationship exists between the three branch currents (think KCL).

Draw in the currents:
upload_2016-3-1_14-9-55.png


Can you write KCL and KVL equations for the circuit?
 
  • #6
nothing909 said:
I forgot to mention that I'm doing superposition, I can't use KVL.
For superposition, you need more than one sources.
 
  • #7
Yes, I had a 32V on the right, which I already dealt with. It's removed, now I'm left with the 18V source.
 
  • #8
nothing909 said:
I forgot to mention that I'm doing superposition, I can't use KVL.
Sure you can. Superposition does not exclude the use of KVL or KCL. They are the basic circuit laws that you need to apply.

Superposition involves suppressing all but one source at a time and determining the circuit conditions, then summing the contributions from each of the individual sources. You can still invoke all the usual tools to find the individual contributions.
 
  • #9
nothing909 said:
Yes, I had a 32V on the right, which I already dealt with. It's removed, now I'm left with the 18V source.
Use KCL and KVL. It is allowed in superposition( and in Thevenin and Norton too)..
 
  • #10
gneill said:
Sure you can. Superposition does not exclude the use of KVL or KCL. They are the basic circuit laws that you need to apply.

Superposition involves suppressing all but one source at a time and determining the circuit conditions, then summing the contributions from each of the individual sources. You can still invoke all the usual tools to find the individual contributions.
Is the total resistance for that circuit 6.3 ohms?
 
  • #11
nothing909 said:
Is the total resistance for that circuit 6.3 ohms?
For KVL, it doesn't matter. But looking at the circuit, no two resistors are in series or in parallel.
Edit: 2 and 4 are in parallel..sorry! I drew it wrong..
 
  • #12
nothing909 said:
Is the total resistance for that circuit 6.3 ohms?
6.33Ω, yes.
 
  • #13
cnh1995 said:
For KVL, it doesn't matter. But looking at the circuit, no two resistors are in series or in parallel.
Edit: 2 and 4 are in parallel..sorry! I drew it wrong..
The first question I have on my question paper is:

Determine the current flowing through the 4 ohm resistor and the voltage drop across it using Kirchoff's laws, which I completed fine.

The next question is:

Determine the current flowing through the 4 ohm resistor and the voltage drop across it using superposition theorem.

Usually what I have is a circuit with two voltage sources. I remove one, calculate the total resistance and then the current in each branch.
I then remove the second voltage source and do the same thing - calculate the total resistance and then the current in each branch.

I then draw a matrix table thing to calculate the total current in each branch.
 
  • #14
nothing909 said:
Usually what I have is a circuit with two voltage sources. I remove one, calculate the total resistance and then the current in each branch.
I then remove the second voltage source and do the same thing - calculate the total resistance and then the current in each branch.
That would be possible if there are series and parallel combinations of resistors(like this circuit). For complicated circuits, KVL and KCL work better.
 
  • #15
cnh1995 said:
That would be possible if there are series and parallel combinations of resistors(like this circuit). For complicated circuits, KVL and KCL work better.
I calculated that the supply current (I2) is 2.86A. Is this correct? If so, how do I now go from that to calculating I1 and I3? Can I do current division or something?
 
  • #16
nothing909 said:
I calculated that the supply current (I2) is 2.86A. Is this correct? If so, how do I now go from that to calculating I1 and I3? Can I do current division or something?
You can use current division since 2 ohm and 4 ohm are in parallel.
 
  • #17
cnh1995 said:
You can use current division since 2 ohm and 4 ohm are in parallel.
So I3 = 0.953A and I1 = 1.906

Is that correct?
 
  • #18
nothing909 said:
So I3 = 0.953A and I1 = 1.906

Is that correct?
Yes.
 
  • #19
cnh1995 said:
Yes.
Okay, now I can do the matrix table thing and calculate the total current in each branch. I know where to go from here.

Thanks for your help.
 

Related to Adding resistance and finding current in each branch

1. What is resistance?

Resistance is a measure of how much a material or component impedes the flow of electric current. It is measured in ohms (Ω) and is represented by the symbol "R". The higher the resistance, the more difficult it is for current to flow through the material or component.

2. How do you add resistance in a circuit?

To add resistance in a circuit, you can either add resistors in series or in parallel. In series, the resistors are connected end-to-end, so the total resistance is the sum of the individual resistances. In parallel, the resistors are connected side-by-side, so the total resistance is calculated using the formula 1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ...

3. What is the formula for calculating current in each branch?

The formula for calculating current in each branch of a circuit is I = V/R, where I is the current in amperes (A), V is the voltage in volts (V), and R is the resistance in ohms (Ω). This formula is known as Ohm's law and is used to determine the amount of current flowing through a specific branch in a circuit.

4. How do you measure the current in each branch of a circuit?

To measure the current in each branch of a circuit, you will need a multimeter. Set the multimeter to the current (amps) setting and connect it in series with the branch you want to measure. The multimeter will then display the amount of current flowing through that particular branch.

5. Why is it important to find the current in each branch of a circuit?

Finding the current in each branch of a circuit is important because it allows you to understand how much current is flowing through each component. This information is crucial for troubleshooting and designing circuits, as it helps determine the appropriate size and type of components needed for a circuit to function properly.

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