Q. The internal resistance of the cells increases as they are used. Explain what effect this will have on the efficiency of the system.
I've attached what the mark scheme says (can't seem to paste it here).
Here's how I understand it: Efficiency = Useful power output/Power input. There are two equations for power and resistance, P = I^{2}R and P = V^{2}/R. The battery with the internal resistance is connected in series with the circuit, thus since current is the same in series, using P = I^{2}R , a constant current shows that P is directly proportional to R. Thus, if R increases, P increases? However, if we consider the other equation with V^{2}, an increase in resistance decreases power. How do I end up with two conclusions using an equation with the same derivations?
Internal resitance and efficiency


You say "using P = I^{2}R , a constant current shows that P is directly proportional to R. Thus, if R increases, P increases.
Yes but if R increases I decreases (V=IR), reducing the power.
And as it depends on I^{2} then current "wins". The decrease in current outweighs the increase in R.
What the mark scheme is pointing to is
The power (heat) developed in the cell (internal resistance) is lost.
The power developed by the cell is emf x current delivered.
Power lost is I^{2}r in the cell.
Power used is I^{2}R in the external circuit. 
(Original post by Stonebridge)
You say "using P = I^{2}R , a constant current shows that P is directly proportional to R. Thus, if R increases, P increases.
Yes but if R increases I decreases (V=IR), reducing the power.
And as it depends on I^{2} then current "wins". The decrease in current outweighs the increase in R.
What the mark scheme is pointing to is
The power (heat) developed in the cell (internal resistance) is lost.
The power developed by the cell is emf x current delivered.
Power lost is I^{2}r in the cell.
Power used is I^{2}R in the external circuit. 
(Original post by Youknowwho)
When we say V = IR means that as resistance increase, current decreases, aren't we assuming that the voltage is constant? In a series circuit, voltage is not the same everywhere. Besides, we have already initially assumed that the current is constant, so shouldn't that mean that resistance would cause a change in voltage according to V = IR and not current?
You have only provided a fragment of this question so we can only guess what's going on.
You need to post the whole question.
However, guessing...
We can say
The emf in the circuit is assumed to be constant and the current will vary according to the total resistance.
If you increase the internal resistance of the circuit (in the cell) you will reduce the total current in the circuit. This will reduce the power in the external resistor (from I^{2}R)
The power delivered by the cell will always be emf x current. Emf is assumed constant.
This will also decrease with I but less than the power in the external resistor which decreases with I^{2} 
(Original post by Stonebridge)
It depends on the circuit.
You have only provided a fragment of this question so we can only guess what's going on.
You need to post the whole question.
However, guessing...
We can say
The emf in the circuit is assumed to be constant and the current will vary according to the total resistance.
If you increase the internal resistance of the circuit (in the cell) you will reduce the total current in the circuit. This will reduce the power in the external resistor (from I^{2}R)
The power delivered by the cell will always be emf x current. Emf is assumed constant.
This will also decrease with I but less than the power in the external resistor which decreases with I^{2}
I found similar questions in other year papers, and they all have the same explanations.
Also, from what you said about emf being constant, the mark scheme says "V decreases"... 
(Original post by Youknowwho)
Right, sorry, attached the first part of the question. I can't find part b of the question, but judging from the mark scheme, it just asked us to find the time taken from a charge of 14400 C and a current of 0.31 A. Which was simple enough, Q = It...
I found similar questions in other year papers, and they all have the same explanations.
Also, from what you said about emf being constant, the mark scheme says "V decreases"...
I thought we were talking about the efficiency of a system when considering internal resistance and power developed using I^{2}R? This doesn't look anything like the questions you have just mentioned.
Sorry, but I'm getting confused... 
(Original post by Stonebridge)
The mark scheme for what? What V? V is not the emf of the cell, it would the pd across the torch bulb in the attached question.
I thought we were talking about the efficiency of a system when considering internal resistance and power developed using I^{2}R? This doesn't look anything like the questions you have just mentioned.
Sorry, but I'm getting confused...
This is what I'm solving: http://www.scribd.com/doc/47429136/U...hysicsedexcel
This question is number 55. Here's the mark scheme: http://www.scribd.com/doc/47429134/U...hysicsedexcel
What I'm basically having problems with is, understanding which formula to use to explain the effect of internal resistance and power, P = I^{2}R or P = V^{2}/R and why. 
(Original post by Stonebridge)
The mark scheme for what? What V? Decreases when? V is not the emf of the cell, it would the pd across the torch bulb in the attached question.
I thought we were talking about the efficiency of a system when considering internal resistance and power developed using I^{2}R? This doesn't look anything like the questions you have just mentioned.
Sorry, but I'm getting confused...
Emf = Terminal voltage + Lost voltage.
Increase in internal resistance increases the voltage lost. Thus the terminal voltage across the component decreases. According to P = V^{2}R, this decrease in voltage decreases power.
OR
Increase in internal resistance increases the total resistance in the circuit. Thus, according to V = IR, this decreases the total current in the circuit. According to P = I^{2}R, the decrease in total current decreases power.
Does that sound right?
Sorry for all the trouble, and thank you for all your help! 
(Original post by Youknowwho)
Wait, I think I finally got it.
Emf = Terminal voltage + Lost voltage.
Increase in internal resistance increases the voltage lost. Thus the terminal voltage across the component decreases. According to P = V^{2}R, this decrease in voltage decreases power.
OR
Increase in internal resistance increases the total resistance in the circuit. Thus, according to V = IR, this decreases the total current in the circuit. According to P = I^{2}R, the decrease in total current decreases power.
Does that sound right?
Sorry for all the trouble, and thank you for all your help!
We got there in the end.
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