The document discusses electromotive force (EMF), internal resistance of power supplies, and how they relate to current, voltage, and circuits. It introduces the Wheatstone bridge circuit for measuring unknown resistances. The bridge balances voltages when the ratio of resistances in each arm are equal, allowing calculation of an unknown value. Slight imbalances produce readings proportional to resistance changes, making it useful for measurement near the balance point.

1. Electricity
6D: Monday 29th November

2. Electromotive Force (e.m.f.)
• The energy given to a coulomb of charge by a
source of electrical energy is called the e.m.f.
of the source
• This is measured in J C-1
or volts

3. Internal resistance, r
• When a power supply is part of a closed circuit, it must
itself be a conductor. All conductors have some
resistance. A power supply has internal resistance, r,
measured in Ohms
• Energy will be wasted in getting the charges through the
supply (heat from supply is noticeable) and so energy at
the output (the terminal potential difference) will fall
r
E

4. • When supply of e.m.f., E, and internal resistance,
r, is connected to a circuit of resistance, R, it
causes a current, I, to flow through circuit:
rE
R
I
I
Applying Ohm’s law to this circuit gives:
Voltage = current x resistance
e.m.f. = I (R + r)
E = IR + Ir
Or
E = V + Ir
Ir = lost volts: the voltage dropped across the
internal resistance, r
V = terminal voltage: the voltage across the
terminals of the power supply

5. Open circuit
• When no current is taken from the power
supply, no energy is wasted
• The terminal potential difference is therefore
the maximum available and equals the e.m.f.

6. General circuit
• With switch open,
voltmeter gives the e.m.f.
(an open circuit)
• With switch closed, the
voltmeter reading will fall
(lost volts)
– Voltmeter now giving the
output voltage, the terminal
potential difference, t.p.d.
rE
R
V

7. Experiment
• Alter variable resistor and
take readings of terminal
voltage, V, and current, I
• Plot V versus I
rE
V
A
V
I
the e.m.f.
the gradient = - internal resistance

8. Short circuit current
• The greater the current, the more energy will be dissipated in the
power supply until eventually all the available energy (e.m.f.) is
wasted and none is available outside the power supply. This
maximum current is the short circuit current – the current which
would flow if the terminals were connected with a short piece of
thick wire (zero resistance: R=0).
• I shortcircuit= e.m.f / internal resistance = E/r

9. Wheatstone bridge circuit
• Wheatstone bridge consists of 4 resistors:
V
R3
R4
R1
R2
A B
If VOA = VOB there is no p.d. between A and B so
no current flows.
The potentials at A and B depend on the ratio
of the resistors that make up each of the two
voltage dividers.
The voltmeter forms a ‘bridge’ between the two
voltage dividers to make up a bridge circuit.
The bridge is said to be balanced when the
resistors in the two potential dividers are in the
same ratio as each other and voltmeter reads
zero:
O
R1
R2
=
R3
R4

10. Finding an unknown resistance
• Circuit is set up with three resistors of known values – one needs
to be a variable resistor
• The unknown resistor is connected into circuit as fourth resistor
• Variable resistor is adjusted until the voltmeter reads zero, i.e. the
bridge is balanced
• Substitute known values into equation to calculate unknown
resistance
• Note: Balance condition does not depend on supply voltage
• The meter must be sensitive to detect zero precisely
• Circuit is sometimes drawn in a diamond shape or rotated through
90 degrees
R1
R2
=
R3
R4

11. Wheatstone bridge – slightly out of balance
• When one of the resistors is varied slightly above and below
its balance value, the voltmeter shows a reading which is
directly proportional to the change in resistance
• This direct proportionality (or linear) relationship allows
Wheatstone bridge to be used as as type of measuring device
when operated near to its balance point
Out of balance voltage
Out of balance resistance
+
+
-
-
0