Electricity - Lesson 8 - EMF and Internal Resistance

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  • 7/25/2019 Electricity - Lesson 8 - EMF and Internal Resistance

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    EMF and Internal

    Resistance

    Electricity Lesson 8

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    Learning Objectives

    To know why the pd of a cell in useis less than its emf.

    To know how to measure the internalresistance of a cell.

    To know how to calculate how muchpower is wasted in a cell.

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    Question

    What is the dierence between emfand potential dierence?

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    Answer

    Emf is the total energy supplied tothe circuit per unit charge by thesource.

    pd is the energy per unit chargeconerted to other energies by thecomponents.

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    Electromotive Force

    The energy supplied to a circuit by a battery isgien by!

    Wis the energy in "# Qis the charge in $# % is theemf in olts &'(T 'ewtons).

    'o circuit at all is *++ , eicient. -ome energyis dissipated in the wires# or een in the batteryitself.

    Q

    E=

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    Internal Resistance

    The internal resistance of a source is theloss of potential dierence per unitcurrent in the source when current

    passes through the wire.

    t is caused by the opposition to the /owof charge through the source.

    We will use the symbol r to representinternal resistance.

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    Terminal pd

    The electrical energy per unit chargedeliered by the source when it is in acircuit.

    The terminal pd# 0# is less than the emf# %#wheneer current passes through thesource.

    The dierence is the lost pd# # due to theinternal resistance of the source.

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    Circuit iagram

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    A simple circuit!

    n this circuit theoltmeter reads &erynearly) the emf.

    &1 per"ect voltmeterhas in#niteresistance. 1 digitalmultimeter has aver$

    %ig%resistance# soneeds a tiny current2it is almost perfect. )

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    Add a resistor!

    This time we 3nd thatthe terminal oltagegoes down to V.

    -ince 0 is lessthanE#

    this tells us that not allof the oltage is beingtransferred to theoutside circuit2 some islost due to the internal

    resistance which heatsthe battery up. Emf 4 5seful olts 6

    Lost olts

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    Including internalresistance

    The resistors areconnected in series.

    -o total resistanceis!7

    6 r

    $urrent throughthe cell!7

    rRI

    +

    =

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    Lost pd

    -o the cell

    n other words!7

    n energy terms the lost pd is the energyper coulomb dissipated or wasted insidethe cell due to the internal resistance.

    IrIRrRI +=+= )(

    pdlost''thepdterminaltheemfcell +=

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    Lost pd

    The terminal pdcan be calculatedusing!7

    The e9uationbecomes!7

    The lost pd can becalculated using!7

    Irv =IRV =

    vVIrIR +=+=

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    &or'ed E(ample

    1 battery of emf *: olts andinternal resistance +.; ohms isconnected to a *+ ohm resistor.

    What is the current and what is theterminal oltage of the battery underload?

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    &or'ed E(ampleiagram

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    )tep *

    Treat the circuit as a perfect battery inseries with an internal resistor. Thecircuit becomes!

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    &or'ed )olution

    )tep +, Work out the total resistance

    tot 4 * 6 : 4 *+ ohms 6 +.; ohms 4*+.; ohms

    )tep -,'ow work out the current! 40 1

    )tep .,work out the oltage across theinternal resistor &lost oltage)! 4 r 4*.*> amps +.; ohms 4 +.;@ olts

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    &or'ed )olution

    -tep ;! work out the terminal oltage!Terminal oltage 4 emf 7 lost oltage4 *: 7 +.;@ 4 **/.- volts

    We can of course work out theterminal oltage by working the

    oltage across the *+ ohm resistor#assuming there are no losses.