Internal Resistance and Circuits

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Homework Help Overview

The discussion revolves around a circuit problem involving internal resistance in batteries and the calculation of currents A, B, and C. The circuit includes a 6V cell with a 2-ohm internal resistance and a 1.5V cell with a 1-ohm internal resistance, with participants attempting to analyze the circuit using Kirchhoff's laws and Ohm's law.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of Kirchhoff's laws and Ohm's law to set up equations based on the circuit configuration. There are discussions about splitting the circuit into series circuits and forming equations to relate the currents. Some participants express uncertainty about whether they have set up the equations correctly and question if simultaneous equations are necessary.

Discussion Status

There is an ongoing exploration of the relationships between the currents and the equations derived from the circuit analysis. Some participants have provided guidance on forming additional equations to relate the currents, while others express confusion about how to proceed with solving the equations.

Contextual Notes

Participants note potential issues with the readability of the circuit diagram and the complexity of the problem, indicating that they are working under the constraints of a homework assignment.

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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