Termal voltage question

1. Mar 14, 2010

sp1974

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?

2. Mar 14, 2010

cepheid

Staff Emeritus
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.

3. Mar 14, 2010

sp1974

Yeah but dont you use intermal resistance * ampere to figure a number for voltage and than take that and subtract from emf to figure terminal voltage?

4. Mar 14, 2010

cepheid

Staff Emeritus
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).

5. Mar 14, 2010

sp1974

So 3.0V = I(R1 + R2)

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

6. Mar 14, 2010

cepheid

Staff Emeritus
Yeah, that's right.

7. Mar 14, 2010

sp1974

But the question is asking for terminal voltage of the battery and this is in Ampere not volts.

8. Mar 14, 2010

cepheid

Staff Emeritus
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.

9. Mar 14, 2010

sp1974

sweet! 2.86V thanks!