Electromotive Force v.s Potential Difference

Electromotive Force (e.m.f) v.s Potential Differences (p.d)

Electromotive Force (e.m.f) Potential Difference (p.d)
Definition Work done by the source to drive a unit charge round a complete circuit. Work done to drive a unit charge through a component or across two points.
Example Cell, battery Light bulb
Energy Conversion Other forms of energy to electrical energy Electrical energy to other forms of energy
Formula       ε = W/Q
where ε = e.m.f (in V)
         W = work done (in J)
         Q = charge (in C)		       V = W/Q
where V = p.d (in V)
         W = work done (in J)
         Q = charge (in C)
SI Unit Volt (V) or J/C or JC-1
Instrument Voltmeter
                     

Symbol:
Note
  • Voltmeter must be connected in parallel to the component.
  • After connecting the voltmeter to the circuit, if the needle deflects to the left side of zero mark, simply interchange the connections on the terminal of the voltmeter.
  • An ideal voltmeter has very large resistance so no current passing through it.
  • (Very Important!!) We cannot say e.m.f of a bulb or p.d of a cell.

Example 1:
What does voltmeter reading show?
(A)     the e.m.f of cell
(B)     the e.m.f of bulb
(C)     the p.d of cell
(D)     the p.d of bulb

Answer: A and D

Potential at a point

Potential difference can be the difference of potential at any two points. It is not necessarily must be across an electric component like light bulb or resistor.

p.d  at two points = potential at the higher point – potential at the lower point

Example 2:
Potential at A = Potential at D = 9V as they are on the same wire that connected to positive terminal of battery (9V).
Potential at C = Potential at F = 0V as they are on the same wire that connected to negative terminal of battery (0V).

Looking at the circuit diagram, we should know that 
Potential at A > B > C;
Potential at D > E > F.

By doing calculation, 
Potential difference across 2Ω is 1.8V so Potential at B = 9 – 1.8 = 7.2V or
Potential difference across 8Ω is 7.2V so Potential at B = 0 + 7.2 = 7.2V

Potential difference across 3Ω is 3.0V so Potential at E = 9 – 3 = 6.0V or
Potential difference across 6Ω is 6.0V so Potential at E = 0 + 6 = 6.0V

So potential difference between B and E = 7.2 – 6 = 1.2V
Example 3:
(1) When switch is open,
V1 = 3.0 V as it measures the e.m.f of cell.

Since it is an open circuit, no current flowing in the circuit.
V2 = 0V as V = IR

Since V2 is 0V, the potential at B = potential at A = 3.0V
V3 = 3.0 – 0 = 3.0V as potential at C is 0V

(2) When switch is closed,
V1 = 3.0 V as it measures the e.m.f of cell.

Since it is a complete circuit, B and C are actually same potential point.
V2 = 3.0V and
V3 = 0V
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