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

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SUMMARY

The discussion centers on calculating the terminal voltage of a battery with an electromotive force (EMF) of 3.0V and an internal resistance of 0.70 ohms, connected to a circuit with a net resistance of 14.7 ohms. Participants clarify that the current can be determined using Ohm's Law, leading to the calculation of terminal voltage. The correct approach involves finding the total resistance, calculating the current, and then using that current to find the voltage drop across the internal resistance. The final terminal voltage is determined to be 2.86V.

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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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