# Terminal voltage

1. Jun 1, 2006

### endeavor

"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

Last edited: Jun 1, 2006
2. Jun 1, 2006

### Hootenanny

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

~H

3. Jun 1, 2006

### arunbg

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.

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

4. Jun 1, 2006

### endeavor

P = I2R
R = 6/1.92 = 1.66 ohms
now what?
not really...

5. Jun 1, 2006

### 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?

Last edited: Jun 2, 2006
6. Jun 2, 2006

### 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

7. Jun 2, 2006

### endeavor

so i guess it was a typo...