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

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To calculate the terminal voltage of a battery with a 3.0 V EMF and 0.70 ohm internal resistance connected to a 14.7 ohm circuit, first determine the current using Ohm's law. The total resistance in the circuit is 15.4 ohms, leading to a calculated current of approximately 0.1948 A. The voltage drop across the internal resistor is then found by multiplying the current by the internal resistance. Subtract this voltage drop from the EMF to find the terminal voltage, which is approximately 2.86 V. This problem illustrates the importance of calculating current to determine terminal voltage accurately.
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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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