Turn on thread page Beta

EMF and internal resistance question help watch

    • Thread Starter
    Offline

    0
    ReputationRep:
    A car battery has an emf of 12V and internal resistance of 5.0 * 10-3 ohms. The battery is used to provide the starting motor of a car with a current of 800A.

    Calculate the potential difference across the terminals of the battery.
    I got this to be 4V (IR = 800*5*10-3)

    Calculate the rate of dissipation of energy due to it's internal resistance stating an appropriate unit.
    Not sure how to work this one out.

    State and explain the effect of attempting to use a battery with a much higher i.r. to start the car.
    Offline

    10
    (Original post by TSR561)
    A car battery has an emf of 12V and internal resistance of 5.0 * 10-3 ohms. The battery is used to provide the starting motor of a car with a current of 800A.

    Calculate the potential difference across the terminals of the battery.
    I got this to be 4V (IR = 800*5*10-3)

    Calculate the rate of dissipation of energy due to it's internal resistance stating an appropriate unit.
    Not sure how to work this one out.

    State and explain the effect of attempting to use a battery with a much higher i.r. to start the car.
    In the first part you have calculated the pd across the internal resistance, not the terminals.


    What formula do you know for the power dissipated in a resistor?
    The resistor is the internal resistance in this case. (You know the current and the resistance.)
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Stonebridge)
    In the first part you have calculated the pd across the internal resistance, not the terminals.


    What formula do you know for the power dissipated in a resistor?
    The resistor is the internal resistance in this case. (You know the current and the resistance.)
    What's the difference? My teacher hasn't been great in explaining it to be honest. Power dissipated in a resistor... IV/ I2R / V2/R?
    Offline

    10
    (Original post by TSR561)
    What's the difference?
    The pd across the terminals is the emf minus the pd across the internal resistance.
    My teacher hasn't been great in explaining it to be honest. Power dissipated in a resistor... IV/ I2R / V2/R?
    Yes, so use the formula that contains I and R as you know both of these for the internal resistance.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Stonebridge)
    The pd across the terminals is the emf minus the pd across the internal resistance.


    Yes, so use the formula that contains I and R as you know both of these for the internal resistance.
    Do I know R? I've learnt internal resistance to be r. I thought emf = IR + Ir .. so IR = emf - Ir.. thank you. How about the second question though? I have no idea how to calculate rate.
    Offline

    10
    The formulas with R in are general formulas that apply to any resistor. It's only when you have two, as here, that you call internal r and the rest of the circuit R so as the distinguish between the two.


    Power is rate of dissipation of energy. Use the formula for power with I and R in with R (or r) as the internal resistance as I said in my other post.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 27, 2013
The home of Results and Clearing

2,569

people online now

1,567,000

students helped last year

University open days

  1. Bournemouth University
    Clearing Open Day Undergraduate
    Wed, 22 Aug '18
  2. University of Buckingham
    Postgraduate Open Evening Postgraduate
    Thu, 23 Aug '18
  3. University of Glasgow
    All Subjects Undergraduate
    Tue, 28 Aug '18
Poll
How are you feeling about GCSE results day?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.