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Internal resitance and efficiency

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 = I²R and P = V²/R. The battery with the internal resistance is connected in series with the circuit, thus since current is the same in series, using P = I²R , a constant current shows that P is directly proportional to R. Thus, if R increases, P increases? :confused:

However, if we consider the other equation with V², an increase in resistance decreases power. How do I end up with two conclusions using an equation with the same derivations?

(edited 10 years ago)

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Reply 1

Stonebridge

You say "using P = I²R , 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 I² 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 I²r in the cell.
Power used is I²R in the external circuit.

Reply 2

You-know-who

Original post by Stonebridge

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?

Reply 3

Stonebridge

Original post by You-know-who

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 I²R)

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 I²

Reply 4

You-know-who

Original post by Stonebridge

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"...

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Reply 5

Stonebridge

Original post by You-know-who

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 I²R? This doesn't look anything like the questions you have just mentioned.

Sorry, but I'm getting confused...

(edited 10 years ago)

Reply 6

You-know-who

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 I²R? This doesn't look anything like the questions you have just mentioned.

Sorry, but I'm getting confused...

The part about efficiency is the c) part of the question I attached later (since you asked me to post the whole question). Sorry this is so confusing! :tongue:

This is what I'm solving: http://www.scribd.com/doc/47429136/Unit-2-loads-of-questions-Physics-edexcel

This question is number 55. Here's the mark scheme: http://www.scribd.com/doc/47429134/Unit-2-ans-loads-of-questions-Physics-edexcel

What I'm basically having problems with is, understanding which formula to use to explain the effect of internal resistance and power, P = I²R or P = V²/R and why.

Reply 7

You-know-who

Original post by Stonebridge

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 = V²R, 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 = I²R, the decrease in total current decreases power.

Does that sound right?
Sorry for all the trouble, and thank you for all your help! :colondollar: