# Branch current analysis

Discussion in 'Homework Help' started by Gregg, Dec 2, 2009.

1. ### Gregg Thread Starter New Member

Dec 2, 2009
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0
Please anyone help me with this circuit. i keep getting to nothing after creating the simultaneous equations (3) and finding the current relationship at two nodes. i am required to solve it only by the method of branch current.
Thanks!

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2. ### studiot AAC Fanatic!

Nov 9, 2007
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I don't think this can be solved by the branch current method alone as you only have 2 nodes, but three unknowns.

1) Two node current equations plus a voltage loop equation

2) Superposition

3) Three loop current equations as you have three loops.

Is this homework? If so it should be in the homework section.

3. ### Gregg Thread Starter New Member

Dec 2, 2009
5
0
hi!
i have tried the three loop equations using the KCL at the nodes to eliminate and then solve but i keep getting the same equations that gives me zero.

4. ### zgozvrm Member

Oct 24, 2009
115
2

I'm not sure if I'm doing this right, but here's what I came up with:

1st, calculate the currents with V2 shorted out
2nd, calculate the currents with V1 shorted out
3rd, combine the calculated currents from the 1st two steps.

Assuming current flow from positive battery terminal to negative terminal:

Step 1 currents:
R1 - 120.6 mA (to the right)
R2 - 120.6 mA (to the left)
R3 - 24 mA (down)
R4 - 96.6 mA (right)
R5 - 96.6 mA (left)

Step 2 currents:
R1 - 96.6 mA (right)
R2 - 96.6 mA (left)
R3 - 13.5 (up)
R4 - 110.1 mA (right)
R5 - 110.1 mA (left)

Step 3 currents:
R1 - 217.2 mA (right)
R2 - 217.2 mA (left)
R3 - 10.5 mA (down)
R4 - 206.7 mA (right)
R5 - 206.7 mA (left)

5. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
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Can't agree with the working of zgozvrm's superposition solution. I should check the signs (directions) at your nodes.

Using Maxwells Mesh Current method is the most economical solution. I said 3 equations before, but it's only two required. Maxwells methods always need one less equation.

Assume clockwise fictitious mesh current I1 in left hand loop and I2 in right hand loop leads to the following mesh equations

36 = +1500I1 - 1200I2
20 = -1200I1 + 1480I2

Solving these yields I1 = 99.08mA & I2 = 93.85mA

Since actual current in R1 and R2 = I1 ; actual current in R4 and R5 = I2 ; actual current in R3 = I1 - I2

thus voltages
VR1 = 9.908
VR2 = 19.816
VR3 = 6.276
VR4 = 9.385
VR5 = 16.893

As a check using KVL

left hand loop resistor voltages add up to 36
right hand loop resistor voltages add up to 20

Last edited: Dec 3, 2009
6. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,281
326
Using the branch current method there's only one unknown--the voltage at node A.

You sum the currents into node A to zero; you only have one equation in one unknown.

(Va-36/300) + Va/1200 + (Va+20)/280 = 0

The solution is Va = 408/65 = 6.2769

From this, you can derive everything else.

7. ### Gregg Thread Starter New Member

Dec 2, 2009
5
0
Hi!
Thanks for your input. Could you please explain more what you did as i am required to find the currents.

Thanks!

8. ### studiot AAC Fanatic!

Nov 9, 2007
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513
Which method are you choosing?

The Electricians is actually a mixture of KVL and KCL.

zgozvrms is superposition.
Set each battery to zero in turn and solve the resulting circuit, with all the other elements in place. Then add the two solutions, with due regard to sign.

Maxwells mesh current is a true current only method. What do you not understand about it?

9. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,281
326
Have a look at the eBook description of the branch current method:

The circuit they show is almost identical to yours. Just replace the two resistors in the left hand branch with a single one equal to the sum of R1 and R2, and similarly in the right hand branch, and it will be identical.

However, for this simple circuit you don't need to go to the trouble to solve any loop equations.

If you know the voltage at node A, finding the currents is trivial. For example, if the voltage at node A is Va, then the current in R3 is just Va/R3, using Ohm's law. That is the second term in the equation I gave in post #6.

The current in the left hand branch is (Va-36)/(R1+R2); this is a voltage divided by a resistance, an application of Ohm's law. That is the first term I gave in post #6.

The current in the right hand branch is (Va+20)/(R4+R5). That is the third term I gave.

The three terms are just an application of Ohm's law. Setting the sum of the 3 branch currents to zero is an application of KCL.

This circuit is simple enough that you don't need to use KVL to find the currents after you have used KCL to find the voltage at node A; you only need to use Ohm's law to find the currents. This is why I said that there is only one unknown in this circuit; once you have the voltage at node A, the rest is trivial. If you use the node method to solve a circuit, you haven't yet found the currents, but once you have the node voltages, finding the currents is usually a trivial exercise.

You can use KVL, and set up some loop equations as shown in the eBook description of the branch current method, but as I said, you don't need to for this circuit. However, if that is what your instructor expects, then do so.

Last edited: Dec 3, 2009
10. ### zgozvrm Member

Oct 24, 2009
115
2
I'm not sure everyone has realized that the batteries are in series, not parallel. Therefore, there is essentially a 56 volt circuit around the outside (through resistors, 1, 2, 4, & 5)

11. ### Gregg Thread Starter New Member

Dec 2, 2009
5
0
thanks guys!

The link the Electrician sent i already have read and understand but if i replace say R1 and R4 with equivalent resistor terms so that i only have R1, R3,R4, and the two batteries in series, will using KVL and KCL still give me the right current values when i decompose the resistors. Sorry guys, instructor wants only branch current method.

Cheers!

12. ### studiot AAC Fanatic!

Nov 9, 2007
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?? What do you mean ??

13. ### studiot AAC Fanatic!

Nov 9, 2007
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If you are prepared to work through for yourself, in the method of your choice, you can compare your answers with the ones I posted. I have already listed the currents for you.

14. ### zgozvrm Member

Oct 24, 2009
115
2
Look at the diagram...

The left-hand battery has its cathode (the positive terminal) on top. The right-hand battery has its cathode on the bottom. Therefore, if you follow the circuit around the outside loop (through both batteries), the voltage is additive.

Oct 9, 2007
2,281
326
Yes. Try it.

16. ### zgozvrm Member

Oct 24, 2009
115
2
After re-working my numbers, I came up with the same voltage drops as studiot (you did noticed that the batteries are in series)!

17. ### studiot AAC Fanatic!

Nov 9, 2007
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But then if you follow a figure of 8 you can show a connnection path for the batteries the other way round.

As soon as you get loops or meshes the concept of series or parallel starts to break down and disappears altogether when you have lots of loops.

Dec 2, 2009
5
0
Thanks guys!