Results are out! Find what you need...fast. Get quick advice or join the chat
Hey there Sign in to join this conversationNew here? Join for free

Internal resitance and efficiency

Announcements Posted on
Applying to Uni? Let Universities come to you. Click here to get your perfect place 20-10-2014
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    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?
    Attached Thumbnails
    Click image for larger version. 

Name:	Untitled.png 
Views:	17 
Size:	4.1 KB 
ID:	263448  
    • 43 followers
    Offline

    ReputationRep:
    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.
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    (Original post by 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.
    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?
    • 43 followers
    Offline

    ReputationRep:
    (Original post by You-know-who)
    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?
    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
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    (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 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
    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"...
    Attached Thumbnails
    Click image for larger version. 

Name:	Untitled.png 
Views:	18 
Size:	24.4 KB 
ID:	263488  
    • 43 followers
    Offline

    ReputationRep:
    (Original post by You-know-who)
    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"...
    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...
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    (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!

    This is what I'm solving: http://www.scribd.com/doc/47429136/U...hysics-edexcel

    This question is number 55. Here's the mark scheme: http://www.scribd.com/doc/47429134/U...hysics-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.
    • Thread Starter
    • 0 followers
    Offline

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

    Sorry, but I'm getting confused...
    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!
    • 43 followers
    Offline

    ReputationRep:
    (Original post by You-know-who)
    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!
    That's right.

    We got there in the end.

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: January 19, 2014
New on TSR
Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.