# Internal Resistance and Circuits

• james_rich
Then do something similar with the first two to eliminate B. Finally, you'll have an equation only in terms of A, which you can solve.In summary, the individual currents for A, B, and C can be found by solving the system of equations formed by the relationships between the voltage and resistance of the circuit components. This can be done using Kirchhoff's laws and systematically eliminating variables until a single equation in terms of A is obtained.

#### james_rich

Hey, I'm revising for my summer exams, and gotten stuck on this particluar question.

_______ 6V ______2.0 ohms___
I I
A I I
I I
I_____4.0 ohms______________ I
I C I
I I
B I I
I I
I________1.5V_______________I

SORRY I HOPE THIS CIRCUIT IS READABLE

Question

The 6V cell has a 2ohm internal resistance...The 1.5V cell has a 1ohm internal resistance. Work out the currents A, B and C!

I have split the circuit into two series circuits to make it easier for me

okay...as V=IR

6V = (C x 4ohms) + (A x 2ohms) + (internal resistance - A x 2ohms)
6V = (C x 4) + (A x 4)

...and for the other series circuit

1.5V = (C x 4ohms) + (internal resistance - B x 1ohm)
1.5V = (c x 4) + B

errr...now I'm stuck ...have i made a mess of trying to work this out...or do i need to use simultanous equations to work it out. I haven't a clue!

Thanx James

okay it didn't come out that great...

current A is across the 2 ohm resistor, and 6V cell
current B is across the 1.5V cell
current C is the junction of A and C across the 4 ohm resistor.

I kno i may have to use Kirchhoff's laws! but not sure how

hope this helps if your as confused as i am! heehee :rofl:

james_rich said:
6V = (C x 4ohms) + (A x 2ohms) + (internal resistance - A x 2ohms)
6V = (C x 4) + (A x 4)

...and for the other series circuit

1.5V = (C x 4ohms) + (internal resistance - B x 1ohm)
1.5V = (c x 4) + B
Looks good to me. But you need one more equation relating the three currents: A + B = C.

Now you have three equations and three unknowns. Solve!

okay...

6V = 4C + 4A
6V = 4C + B

combining the equations together i get:

B = 4 x A

i can't seem to be able to get any values for either A or B, to be able to work out the current of C using Kirchoff's first law!

Is this problem actually solveable?

james_rich said:
okay...

6V = 4C + 4A
6V = 4C + B
First equation is OK. Second should be:
1.5 = 4C + B

But don't forget the third equation:
A + B = C.

Solve it systematically. Start by using the last equation to eliminate C from the first two.