Terminal Voltage Battery 12V - 1.90A 6W Resistor

[endeavor]

http://img413.imageshack.us/img413/2593/physics26is.th.png
"A battery labeled as 12.0-V is measured to supply 1.90 A to a 6.00-W resistor (see figure). (a) What is the terminal voltage of the battery? (b) What is its internal resistance?"
I'm not sure how to solve this problem...

I originally did R = V/I = 6.32 Ohms, and E = 12V, so V = E - IR = 0... but that is obviously wrong

 

[Hootenanny]

Note that the resistor value is given in watts, which is a measure of power.

~H

 

[arunbg]

Your method would work had the cell had negligible internal resistance .
Note that the battery is only marked 12V. This is not the actual voltage drop across the load resistance .
You can start off by finding the value of the resistor . It is given that current through the resistor is 1.90 A .
What is the relation connecting power and current ?

Then you must find what the value of a resistance connected in series should be , so that from the 12 V source, a current of 1.90 A is drawn. This is the internal resistance .
Now find the voltage drop across load resistor, this will be terminal voltage.
Can you follow ?

Edit : 6W may be the power consumed by the resistor when 1.9 A flows theough it .

 

[endeavor]

P = I²R
R = 6/1.9² = 1.66 ohms
now what?

Then you must find what the value of a resistance connected in series should be , so that from the 12 V source, a current of 1.90 A is drawn. This is the internal resistance .
Now find the voltage drop across load resistor, this will be terminal voltage.
Can you follow ?

not really...

 

[endeavor]

so...
I = 1.90A, P = 6.00W, V = 12V

R_{load} = \frac{P}{I^2} = 1.66\Omega

V_{emf} = 12V = I \cdot R_{total} = I \cdot R_{load} + I \times R_{internal}

R_{internal} = 4.65\Omega

but can't I just skip all that and do

V_{terminal} = I \cdot R_{load} = 3.16 V

The answer is supposed to be 11.4 V...what am I doing wrong?

 

[arunbg]

What you have done is completely correct . In your original question, it was asked to calculate the internal resistance as well as the terminal potential difference .And of course you should skip and use
V_{terminal} = I \cdot R_{load}.
However, from the answer that you have provided , it seems that in the question, it is given that the value of the resistance is indeed 6 ohms .
So you need not consider power at all and I am sure you can do the rest.
It is much simpler as compared to the question that we have been discussing .

Arun

 

[endeavor]

so i guess it was a typo...