Terminal Voltage Battery 12V - 1.90A 6W Resistor

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The discussion revolves around calculating the terminal voltage and internal resistance of a 12V battery supplying 1.90 A to a 6.00 W resistor. Participants clarify that the power rating of the resistor is crucial for determining its resistance using the formula R = P/I², yielding a load resistance of 1.66 ohms. The internal resistance is calculated based on the total voltage drop and current, resulting in an internal resistance of 4.65 ohms. There is confusion regarding the values used, particularly the power rating, which some participants suggest may have been a typo. Ultimately, the correct terminal voltage is determined to be 11.4 V.
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"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
 
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Note that the resistor value is given in watts, which is a measure of power.

~H
 
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 .
 
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...:confused:
 
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?:frown:
 
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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
 
so i guess it was a typo...
 
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