Internal Resistance and Circuits

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SUMMARY

The discussion revolves around solving a circuit problem involving two voltage sources (6V and 1.5V) and their respective internal resistances (2 ohms and 1 ohm). The user, James, attempts to apply Ohm's Law (V=IR) and Kirchhoff's laws to find the currents A, B, and C in the circuit. The key equations derived include 6V = 4C + 4A and 1.5V = 4C + B, along with the relationship A + B = C. The solution requires simultaneous equations to determine the values of the currents.

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james_rich
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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 :cry: ...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!

Please Help!

Thanx :smile:

James
 
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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 :smile:
 
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.
 

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