How Do You Calculate Current Through a Resistor in a Parallel Circuit?

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Homework Help Overview

The discussion revolves around calculating the current through a resistor in a parallel circuit, given the e.m.f. and internal resistance of the power source. Participants are exploring the relationships between total resistance, current, and voltage drops in the context of parallel resistors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating total resistance and current, with some questioning whether the e.m.f. should be considered the supply voltage. Others suggest using the internal resistance in conjunction with the e.m.f. to find terminal voltage and current through the resistors.

Discussion Status

The discussion is active, with participants offering various approaches to the problem. Some guidance has been provided regarding the use of e.m.f. and internal resistance, but there is no explicit consensus on the best method to proceed.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available or the methods they can use to solve the problem.

QueenFisher
i've been given this circuit, i know the e.m.f. and internal resistance, it has 2 resistors connected in parallel so i can work out Rtotal, but i need to work out the current through one of the resistors. can anyone point me in the right direction?
thanks :biggrin:
 
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Well, once you know R you can calculate the total current. With the current, you can calculate the volt-drops certain resistances, etc. That's generally the approach to take.
 
so do i take the e.m.f. as being the supply voltage in this case? cos i can't think of any other way to find it
 
The internal resistance [of the battery] acts as if it were a resistor in series with the battery. You could find the ternimal voltage.
[tex]V_{ab} = \varepsilon - Ir[/tex]

[tex]\varepsilon[/tex] = Emf
[tex]r[/tex] = internal resistance

USing ohms law we get.
[tex]IR = \varepsilon -Ir[/tex]

Where [tex]R[/tex] is the 2 resistors in parallel.

I'm sure you can take it from here.
 
Last edited:

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