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# Can some one please help me solve for currents in this problem?

I am trying to solve this problem below, but it is not working out. I would be able to solve this it it would only have one power source, but when there are multiple I get confused.

The answers are I1=3.5a, I2=2.5a, I3=1a.

I struggling to get the right numbers.

Maybe on a side note, if any one knows a good book or a website where I could practice these would be great.

Thank you very much

## Comments

Best Answer 9 years ago

You need Kirchoff !!!

Consider the upper loop

24=I1 x 9 - I2 x 3

Consider the lower loop

12 = i2 x 9 - i1 x 3

Solve for I1 and I2

You can also do this by nodal analysis, instead of loops, but I got good at loops to the exclusion of nodes....

Steve

Answer 9 years ago

When I am considering the upper loop

24=I1 x 9 - I2 x 3

9 is the total resistance of the upper loop, right?

But why are you subtracting i2*3?

same question for the lower loop

Answer 9 years ago

To clarify...The two virtual currents through R3 oppose one another, that's the actual "why" behind it. You can think of loop 1 as draining OOmPh (electromotive force) from loop 2 and visa versa in the shared leg, or in another way if you can think of one ;-)

Draw a loop indicating the direction of flow for each loop. You'll notice that in the shared leg they are going in opposite directions. So they are represented by a difference, where loop 2's opposition to I1 is accounted in it's equation and loop 1's opposition to I2 is accounted in its.

Answer 9 years ago

I am in <3, thank you so much.

steveastrouk thank you as well

Answer 9 years ago

This yields the answers stated.

Steve

9 years ago

How do you do it ?

1.) Draw loops inside the fewest number of closed paths in the circuit. Make each loop circle CLOCKwise, for consistency.

2.) Label each loop consistently. I1, I2, I3, I4 etc etc etc.

3.) Start with the voltage sources, place them, summed algebraically on the left of the equation for the loop.

4.) Take your loop current, and sum algenraically, all the impedances that it passes through in the loop ( 9 x I1 here)

5.) Look for OTHEr current loops appearing in your loop equation ( I2) I2's effect is to oppose the current in I1, so its effect is -ve (-3 x I2)

Repeat for each loop.

Solve for each loop current - it helps to know how to do determinants if there are more than 2 loops.

If there ISN'T a voltage source in a loop, then the left hand side is zero.