Internal resistance circuit problem

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Homework Help Overview

The discussion revolves around a circuit problem involving two identical batteries connected in parallel, each with an EMF of 22 V and an internal resistance of 7 Ohms, supplying power to a load resistance of 35 Ohms. The focus is on understanding the power transferred to the load resistor.

Discussion Character

  • Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between load resistance and total resistance, with some questioning the interpretation of the load resistance in the context of the circuit. There are attempts to clarify the use of Thevenin's Theorem and Superposition, although one participant later retracts their suggestion.

Discussion Status

The discussion is ongoing, with participants providing insights and clarifications regarding the circuit setup and resistance definitions. There is no explicit consensus yet, as different interpretations of the circuit components are being explored.

Contextual Notes

Some participants express uncertainty about the definitions of load resistance versus total resistance and the implications of the circuit configuration. There is also a mention of potential confusion regarding the application of circuit analysis techniques.

Sarlizwx
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Homework Statement



In a circuit, two identical batteries with EMF of 22 V and internal resistance of 7 Ohm are placed in parallel to provide power to a load resistance of RL = 35 Ohms. What is the power transferred from the current to RL to heat the resistor?


Homework Equations


Is R(L) (Resistance load) mean the same thing as R(total)? I know EMF = Isys * Rsys as well.



The Attempt at a Solution



Only drew the diagram, labeled areas and put the RL = [1/R(1) + 1/R(2)] ^ -1I also know that Itot = I1 + I2. I'm assuming here that EMF of 22V is for each battery?
 
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The load resistance is the resistance of the resistor that is added to the circuit. In this case, the total resistance is not the same as the load resistance.
 
So RL = ((1/R1) + (1/R2))^-1 is an incorrect statement?
 
Your equation RL = ((1/R1) + (1/R2))^-1 only equates the total (internal) resistance of the batteries. The load resistance is a separate load (resistor) in the circuit which is in series with the two batteries. R(total)=(total internal resistance)+(load resistance RL).
 
Circuit/Power

Howdy,
If I've got the right circuit in mind, you could use Thevenin's Theorem to find the thevenin equivalent circuit then determine I and P.

Peace out.

Fez07
 
Circuit/Power

Me again,
Forget my last comment. I was thinking of the wrong circuit. Use Superposition.
 
Don't recognize either of those, Fez07
 
Circuit/Power

Firstly, is this the circuit you have?
 

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