What is the current in a circuit with two cells?

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Homework Help Overview

The discussion revolves around determining the current in a circuit with two cells, focusing on concepts related to electromotive force (EMF), terminal voltage, and internal resistance.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between EMF, terminal voltage, and internal resistance, questioning how these factors influence the current in the circuit. There is an emphasis on understanding the definitions and implications of potential rises and drops within the circuit.

Discussion Status

Some participants have provided insights into the nature of EMF and its role in driving current, while others express confusion about the concepts involved. The discussion is ongoing, with various interpretations of the circuit's behavior being explored.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the depth of exploration into the problem. There is a noted lack of explicit consensus on the implications of the EMF values in relation to the current.

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Homework Statement


What is the current in this circuit:
http://img802.imageshack.us/img802/4646/8ekh.png

Homework Equations


All potential differences in a closed portion of a circuit must add to 0.
Terminal Voltage = EMF - I x internal resistance.

The Attempt at a Solution


I do not know how to do this, the answer says the current is 0A. I have no idea why.
 
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You forgot the Relevant Equations portion of the template. What circuit laws have you learned about?
 
gneill said:
You forgot the Relevant Equations portion of the template. What circuit laws have you learned about?

Sorry, I have added them in.
 
Okay, so potential changes come in two flavors: Rises and drops. Typically sources are responsible for potential rises and resistors for potential drops.

If you "walk" around an isolated loop in a given direction and sum up just the changes in potential due to the sources you'll arrive at the total amount of EMF that can drive current in that loop. What do you find when you do this "walk" around your loop?
 
gneill said:
Okay, so potential changes come in two flavors: Rises and drops. Typically sources are responsible for potential rises and resistors for potential drops.

If you "walk" around an isolated loop in a given direction and sum up just the changes in potential due to the sources you'll arrive at the total amount of EMF that can drive current in that loop. What do you find when you do this "walk" around your loop?

I'm getting confused with EMF, terminal voltage and internal resistance etc... isn't the 1.5V the EMF, which means that the actual potential rise (terminal voltage) will be less that 1.5V?

If I were to add up the EMFs, I would be assuming no current is flowing.
 
BMW said:
I'm getting confused with EMF, terminal voltage and internal resistance etc... isn't the 1.5V the EMF, which means that the actual potential rise (terminal voltage) will be less that 1.5V?

The EMF, or "Electromotive force", refers to the potential across the thing inside the battery that produces the potential difference, and ignored any internal losses due to internal resistance. It's the potential that you would measure across the cell if you were to employ a perfect voltmeter that draws no current (so there would be no drop across any internal resistances).

In summing up the EMFs that are driving current in the circuit, ignore all resistances no matter where they are located and concentrate on the raw EMF values. You want to find out what the total EMF available to drive current is, not the effects of the current (not yet, anyways!).
 
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gneill said:
The EMF, or "Electromotive force", refers to the potential across the thing inside the battery that produces the potential difference, and ignored any internal losses due to internal resistance. It's the potential that you would measure across the cell if you were to employ a perfect voltmeter that draws no current (so there would be no drop across any internal resistances).

In summing up the EMFs that are driving current in the circuit, ignore all resistances no matter where they are located and concentrate on the raw EMF values. You want to find out what the total EMF available to drive current is, not the effects of the current (not yet, anyways!).

Aha! So the net EMF is what creates a current in a closed circuit or part of a circuit? And the net EMF in this circuit is zero so the current will also be zero?

:D
 
BMW said:
Aha! So the net EMF is what creates a current in a closed circuit or part of a circuit? And the net EMF in this circuit is zero so the current will also be zero?

:D

Exactly! :smile:
 
gneill said:
Exactly! :smile:

Thank you for your time :biggrin:
 

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