Calculating Terminal Voltage: Battery with 3.0V EMF and 0.70 Ohm Resistance

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Homework Help Overview

The discussion revolves around calculating the terminal voltage of a battery with a specified electromotive force (EMF) and internal resistance, connected to an external circuit with its own resistance. The problem involves applying Ohm's law and understanding the relationship between current, voltage, and resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore how to calculate the current in the circuit using Ohm's law, questioning the need for the current value to determine terminal voltage. There is discussion about the relationship between internal resistance and current, and how to derive terminal voltage from the given EMF.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on how to approach the calculation of current and voltage. There is recognition of the two-step nature of the problem, involving first finding the current and then using it to determine the voltage drop across the internal resistor.

Contextual Notes

There is a noted confusion regarding the terminology used for current and resistance, as well as the absence of the current value in the initial problem statement. Participants are navigating these conceptual clarifications while adhering to the constraints of the homework problem.

sp1974
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I have another homework problem involving termal voltage that I am stuck on since an Ampere wasn't given any help?


A battery whose emf is 3.0 V. and whose internal resistance is 0.70 ohms is connected to a circuit whose net resistance is 14.7 ohms. What is the terminal voltage of the battery?
 
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It's true, the current was not given to you, because you can calculate the current in this circuit using Ohm's law. That is the whole point of the question.
 
cepheid said:
It's true, the current was not given to you, because you can calculate the current in this circuit using Ohm's law. That is the whole point of the question.

Yeah but don't you use intermal resistance * ampere to figure a number for voltage and than take that and subtract from emf to figure terminal voltage?
 
sp1974 said:
Yeah but don't you use intermal resistance * ampere to figure a number for voltage and than take that and subtract from emf to figure terminal voltage?

First of all, it's not "ampere." The name of the physical quantity you are measuring is 'current.' An 'ampere' is the name of the unit by which current is measured. What you are saying is an equivalent mistake to saying that force = "kilogram" * acceleration. Try not to confuse the names of quantities in physics with the names of the units used to measure those quantities.

Secondly, it is true that the voltage across the internal resistor will be the current through it multiplied by its resistance. The point I was trying to make was that it's okay that you haven't been given this current, because you can use Ohm's law for the whole circuit in order to calculate what the current is. (Hint: you have two resistors in series).
 
So 3.0V = I(R1 + R2)

3V = I (15.4) and than solve for I which would be .1948A?
 
sp1974 said:
So 3.0V = I(R1 + R2)

3V = I (15.4) and than solve for I which would be .1948A?

Yeah, that's right.
 
But the question is asking for terminal voltage of the battery and this is in Ampere not volts.
 
sp1974 said:
But the question is asking for terminal voltage of the battery and this is in Ampere not volts.

Yeah, but now that you have the current, you can do what you yourself suggested in post #3 in order to find the voltage across the internal resistor. It was a two-step problem. Do you understand? You needed to find the current through the circuit in order to find the voltage drop across the internal resistor.
 
sweet! 2.86V thanks!
 

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