How Do You Solve a Basic KVL Circuit Problem?

[nchin]

http://imageshack.us/a/img585/9687/kvlf.jpg

the work i did on that sheet is completely wrong.

I know that Ic = Ia + Ib

but can someone help me set up the equations for the two loops please! I can't seem to figure it out and its frustrating me!

the answer is supposed to be Ic = (RbE1 + RaE2) / RaRb + RaRc + RbRc

 

[SammyS]

http://imageshack.us/a/img585/9687/kvlf.jpg

the work i did on that sheet is completely wrong.

I know that Ic = Ia + Ib

but can someone help me set up the equations for the two loops please! I can't seem to figure it out and its frustrating me!

the answer is supposed to be Ic = (RbE1 + RaE2) / RaRb + RaRc + RbRc

Write the KVL equation for the "left loop".

Write the KVL equation for the "right loop".

Write the KCL equation for either junction.

 

[nchin]

thanks i just figured it out. this is just an intense algebra problem!

 

[SammyS]

thanks i just figured it out. this is just an intense algebra problem!

Yes. Algebra is involved.

 

[Nickie]

I would be happy to assist with your KVL circuit problem. It seems like you are on the right track by using the equation Ic = Ia + Ib. To set up the equations for the two loops, you can start by labeling the currents flowing through each resistor and then applying KVL to each loop.

For the first loop, starting at the top left corner and moving clockwise, we have:
-IRa - IRc + E1 = 0

For the second loop, starting at the top right corner and moving clockwise, we have:
-IRb - IRc + E2 = 0

From these two equations, we can solve for IRc by substituting in the equation Ic = Ia + Ib. This will give us:
-IRc + (Ic - Ib) + E1 = 0
-IRc + Ic - Ib + E1 = 0
Ic = Ib + IRc - E1

Next, we can substitute this expression for Ic into the second loop equation to solve for IRb:
-IRb - IRc + E2 = 0
-IRb - (Ib + IRc - E1) + E2 = 0
-IRb - Ib - IRc + E1 + E2 = 0
IRb = Ib + IRc - E1 - E2

Finally, we can substitute both expressions for IRc and IRb into the equation Ic = Ia + Ib to get the final answer:
Ic = Ib + (Ib + IRc - E1 - E2) - E1
Ic = 2Ib + IRc - 2E1 - E2

Using the equation Ic = Ia + Ib, we can also solve for IRa:
IRa = Ic - Ib
IRa = (2Ib + IRc - 2E1 - E2) - Ib
IRa = Ib + IRc - 2E1 - E2

Finally, we can substitute these expressions into the original equation Ic = (RbE1 + RaE2) / (RaRb + RaRc + RbRc) to get:
Ic = (RbE1 + RaE2) / (RaRb + RaRc + RbRc)
Ic = (Rb(I