EMF and Internal Resistance

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EMF and Internal Resistance

• Electricity Lesson 9

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Learning Objectives

• To know why the pd of a cell in use is less than
  its emf.
• To know how to measure the internal resistance of
  a cell.
• To know how to calculate how much power is wasted
  in a cell.

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Question

• What is the difference between emf and potential
  difference?

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Answer

• Emf is the total energy supplied to the
  circuit per unit charge by the cell.
• pd is the energy per unit charge converted to
  other energies by the components. 

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Electromotive Force

• The energy supplied to a circuit by a cell is
  given by
• E is the energy in J, Q is the charge in C, e is
  the emf in volts (NOT Newtons).
• No circuit at all is 100 efficient.  Some
  energy is dissipated in the wires, or even in the
  cell itself.

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Internal Resistance

• The internal resistance of a source is the loss
  of potential difference per unit current in the
  source when current passes through the wire.
• It is caused by the opposition to the flow of
  charge through the source.
• We will use the symbol r to represent internal
  resistance.

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A simple circuit

• In this circuit the voltmeter reads (very nearly)
  the emf.
• (A perfect voltmeter has infinite resistance.  A
  digital multimeter has a very high resistance, so
  needs a tiny current it is almost perfect.  )

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Add a resistor

• This time we find that the terminal voltage goes
  down to V. 
• Since V is less than E, this tells us that not
  all of the voltage is being transferred to the
  outside circuit some is lost due to the internal
  resistance which heats the battery up. 
• Emf Useful volts  Lost volts

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Terminal pd

• The electrical energy per unit charge delivered
  by the source when it is in a circuit.
• The terminal pd, V, is less than the emf, e,
  whenever current passes through the source.
• The difference is the lost pd, v, due to the
  internal resistance of the source.

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Including internal resistance

• The resistors are connected in series.
• So total resistance is-
• R r
• Current through the cell-

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Lost pd

• So the cell
• In other words-
• In energy terms the lost pd is the energy per
  coulomb dissipated or wasted inside the cell due
  to the internal resistance.

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Circuit Diagram
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Lost pd

• The terminal pd can be calculated using-
• The equation becomes-

• The lost pd can be calculated using-

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Worked Example

• A battery of emf 12 volts and internal resistance
  0.5 ohms is connected to a 10 ohm resistor.  What
  is the current and what is the terminal voltage
  of the battery under load?

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Worked Example Diagram
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Step 1

• Treat the circuit as a perfect battery in series
  with an internal resistor.  The circuit becomes

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Worked Solution

• Step 2  Work out the total resistance
• R tot R1 R2 10 ohms 0.5 ohms 10.5 ohms
• Step 3 Now work out the currentI V/R 12
  10.5 1.14 A
• Step 4 work out the voltage across the internal
  resistor (lost voltage) v Ir 1.14 amps 0.5
  ohms 0.57 volts

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Worked Solution

• Step 5 work out the terminal voltage Terminal
  voltage emf - lost voltage 12 - 0.57 11.43
  volts
• We can of course work out the terminal voltage by
  working the voltage across the 10 ohm resistor,
  assuming there are no losses.

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Power Wasted in a Cell

• Recall that power can be calculated using-
• So we can multiply our equation by I to get-
• Or in words-

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Measuring Internal Resistance
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Measuring Internal Resistance

• Terminal pd and current can be plotted on a
  graph.
• This equation is in the form of a straight line