What is the current in this circuit?

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Homework Help Overview

The discussion revolves around a circuit problem involving two batteries with different EMFs and internal resistances connected in a loop with a resistor. The original poster expresses difficulty in applying Kirchhoff's rule to determine the current flowing in the circuit.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Kirchhoff's rule and Ohm's law, questioning how to account for internal resistance and opposing current directions from the batteries.

Discussion Status

Some participants have offered guidance on using Ohm's law and considering the effects of internal resistance. There is an ongoing exploration of how to set up the equations correctly, with no explicit consensus reached yet.

Contextual Notes

The original poster mentions that they have already lost points on this problem, indicating a time constraint and the importance of understanding the topic for an upcoming test.

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Ok, I already got all 6 points off on this homework problem:( but it might be on the test tomorrow. I know what the answers should be. He says you apply Kirchoff's rule, I've tried, and I don't get the right answers. Help! if you can get (a) I can get the rest.


Two batteries and a 50-ohm resistor are connected in a single, simple loop. The first battery has an EMF of 15 V and an internal resistance of 9 ohms. The second battery has an EMF of 9 V and an internal resistance of 4 ohms. The circuit is connected in such a way that first battery is trying to drive current clockwise in the circuit while the second battery is trying to drive current counterclockwise.

For questions (a) through (f) below, enter each answer as a positive number.

(a) How much current flows in the circuit?

0.0952 (real answer)
 
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Draw a picture and use Ohm's law. The sum of the voltages around the loop will equal 0.
 
hmmm

Let's say I don't know what that is... I drew a picture. Ohm's law is v=ir but what do I do about the internal resistance, or that one's pushing current one way, and one's pushing current the other way...
 
That's the sweet thing about V=IR. Internal resitance, and the Source Voltage tells you how much Current the source is developing, does it not?
 

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