Power efficiency of Battery with Internal Ressitance

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papie
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#1
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Hi, I have a question i want to ask that i got in the physics AS examination. I will draw a diagram of the circuit.
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papie
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Not sure if I have the right values. The question was something like that : what is the power efficiency of the battery. How would you do it ?
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papie
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anyone ?
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Callicious
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(Original post by papie)
anyone ?
I mean, you can calculate the ratio of \frac{\textrm{useful}}{\textrm{total}} if you had another component in the circuit... but with just an internal resistance I can't see how you'd be able to get an "efficiency" per-say from this... someone else might know more?

Edit: Didn't see that XY was actually a component- there's nothing there to indicate it on the diagram (you should draw in a resistor between X and Y- I thought that thing on the top right was just "X1" or something else unrelated to the problem. In the textbook I got afforded for circuits they usually use a "dummy resistor" to represent wire resistance as opposed to drawing two nodes and slapping a label to them.

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P_{total} = P_{int} + P_{ext}

P_{useful} = P_{ext}

\mu = \frac{P_{useful}}{P_{total}} = \frac{P_{ext}}{P_{int} + P_{ext}}

The power is just given by I^2R and since it's a series, this will reduce to

\mu = \frac{R_{ext}}{R_{int} + R_{ext}}

Solve for internal resistance R_{int} and you'll have your problem solved. Note that \mu is fractional and not a percentage.
Last edited by Callicious; 2 months ago
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Itsmikeysfault
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Find the total resistance by doing R = V/I. Then find the useful power and total power using the total resistance and the load resistance (the resistance between x and y). Finally, find efficiency using efficiency = useful power/total power
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papie
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#6
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(Original post by Callicious)
I mean, you can calculate the ratio of \frac{\textrm{useful}}{\textrm{total}} if you had another component in the circuit... but with just an internal resistance I can't see how you'd be able to get an "efficiency" per-say from this... someone else might know more?

Edit: Didn't see that XY was actually a component- there's nothing there to indicate it on the diagram (you should draw in a resistor between X and Y- I thought that thing on the top right was just "X1" or something else unrelated to the problem. In the textbook I got afforded for circuits they usually use a "dummy resistor" to represent wire resistance as opposed to drawing two nodes and slapping a label to them.

Name:  res.PNG
Views: 10
Size:  29.8 KB

P_{total} = P_{int} + P_{ext}

P_{useful} = P_{ext}

\mu = \frac{P_{useful}}{P_{total}} = \frac{P_{ext}}{P_{int} + P_{ext}}

The power is just given by I^2R and since it's a series, this will reduce to

\mu = \frac{R_{ext}}{R_{int} + R_{ext}}

Solve for internal resistance R_{int} and you'll have your problem solved. Note that \mu is fractional and not a percentage.
It's very different than what i did . It is probably wrong but i will write down what i did: I first calculated the Internal resistance then did P=VI to find max power of battery w/o internal resistance using the same value for current(probably sure value of current is different w/o internal resistance). Then i did P=I^2r to find the power loss in the internal resistance. To get %efficiency i did (Power loss in internal Resistance/ VI ). That was a 3 mark question BTW in AS physics CIE exam this year. Also the question was something like this: What is the efficiency of the battery in delivering power to resistance XY(It wasn't phrased like this but something like that ) .
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