Internal resistance of battery
Some of the electrical energy is
dissipated by Joule heating inside the
battery.
e.m.f. across the + & - terminal
of the battery is lower than the
marked value when connected to
external components.
∴ the voltage across the terminal of a
cell is called the terminal voltage and
is usually less than the e.m.f. of the
cell.
ξ
r
Terminal voltage V
Finding the internal resistance of the battery
The current I through the circuit is varied
by a resistance box which has known value
of resistance R.
ξ= V + I r
=> ξ = R + r
=IR+Ir
I
R
ξ
r
I
=> R = ξ
I
- r
V
∴ by plotting a graph of R against 1 / I, a straight line can
be obtained. The slope of the graph is ξ and the intercept on
the y-axis is the internal resistance r.
I
Combination of Resistors
1. In series
All the resistors carry
same current I
V = V1 + V2 + V3
= I R1 + I R2 + I R3
= I ( R1 + R2 + R3 )
∴ R = R1 + R2 + R3
Combination of resistors
2. In parallel
The p.d. V across each
resistors is the same., but
the current branches into
I1, I2 and I3.
I  I1  I 2  I3
V
V
V
 1
1
1 





V



R1
R2
R3
R
R
R
2
3
 1
1
1
1
1



R
R1 R2
R3
Power and heating effect
The charge pass through the resistor:
Q  I  t
The electrical energy converted into
other form of energy in Δt:
W  V  Q
∴ Power of the resistor:
W V  Q
Q
P

V 
t
t
t
2
V
P V  I  I R 
R
2
V I
P    I  I R 
2

2
R
Example 5 A light bulb labelled 12 V 10W is connected
across a 12 V cell with internal resistance 5Ω.
Find the power output.
∵ the cell has internal resistance, the p.d. across the light
bulb ≠12 V. But the resistance of the bulb is fixed.
2
V
P 
R
2
2
V
12
R

 14.4
P
10
The total resistance of the circuit is 5Ω + 14.4 Ω = 19.4 Ω
The current flow through the circuit and the bulb is
= 12 V / 19.4 Ω = 0.619 A
2
2
P  I  R  0.619 14.4  5.5W
Classwork 1 A student connect a toy motor labelled
9 V 50 W across a 9 V battery with internal
resistance 10 Ω. Find the power output by the
toy motor.
2
V
P 
R
2
2
V
9
R

 1.62
P 50
The total resistance of the circuit is 1.62Ω + 10 Ω = 11.62 Ω
The current flow through the circuit and the toy motor is
= 9 V /11.62 Ω = 0.775 A
2
2
P  I  R  0.775 11.62  6.97W
Power output and Resistance
The current flow through the above circuit:
Power output through R:
2
I
Pout  I R
Max power occurs when R = r
Pmax 


Rr
 2R
R  r 2
 2r
r  r 2

 2r
4r 2

2
4r
Proof of max. Power output
P


 2R

R  r 
2
 R
2
2
R  2Rr  r
2
 R
2
2
2
R  2Rr  r  4Rr
 R
2
2
( R  r )  4 Rr
d  R 
Try


2
dR  R  r  
Efficiency of Electric Circuit
In general the power of the circuit.
Power output
I 2R
R



100%
2
Power input
I (R  r) R  r
When max. power occurs, R = r, the efficiency is:
r

100 %  50%
rr
Example 6 A 1.5 V cell has an internal resistance of 2 Ω.
Find the condition for
a.) Max. Power.
b.) Max. Efficiency.
a.)
Max. Power consumption occur when the cell
is short circuit.

2
2
1.5
 Po    I  I r 

 1.125W
r
2
2
b.)
Max. useful power output occur when R = r
i.e. external resistance = 2Ω
2
1.52
 Pmax 

 0.28W
4r 4  2
Classwork2 A 9 V cell has an internal resistance of 5 Ω.
Find the power out put for
a.) Max. Power.
b.) Max. Efficiency.
a.)
Max. Power consumption occur when the cell
is short circuit.

2
2
9
 Po    I  I r 

 16.2W
r
5
2
b.)
Max. useful power output occur when R = r
i.e. external resistance = 5Ω
2
92
 Pmax 

 4.05W
4r 4  5
Power transmission
In power transmission, the
voltage across the power
cable is VL – V’L. or ILR,
where L is the current
through the cable.
The power loss by the
cable is
2
Ploss  I L  R
∵ Power transfer to the user equal to the
power generate, it usually remains constant
Typical Example ---- Combination of 2 cells
Since any charge + or – cannot
climb through the energy
barrier set by the other cell,
NO current flow through the
cells.
+
ξ
ξ
r1
r2
-
Example 6 A 12V battery of internal resistance 15 Ω is
recharged by a 14 V d.c. supply with internal resistance
5Ω via a 20 Ω. Find the current through the battery.
14V
5Ω
Net e.m.f. = 14 V – 12 V = 2 V
20Ω
Total resistance = 15 + 5 + 20 = 40Ω
Current = 2 V / 40 Ω = 0.05 = 50 mA
12V
15Ω
Classwork3 A student uses a 9V battery of internal
resistance 10 Ω to charged a 1.5 V
‘AA’battery with internal resistance
5Ω via
a 50 Ω resistor.
Find the current through the battery.
9V
10Ω
Net e.m.f. = 9 V – 1.5 V = 7.5 V
50Ω
Total resistance = 10 + 5 + 50 = 65Ω
Current = 7.5 V / 65 Ω = 0.115 A
1.5V 5Ω
2 parallel cell connected to a resistor
The terminal voltage of the cells
are the same and equal to the
voltage across R
I1 ξ
r1
I2 ξ
r2
ξ = I1 r1 + I R
ξ = I2 r2 + I R
I1 + I2 = I
This is known as Kirchoff’s
1st law of current
I
R
Example 6 The light bulb in the diagram shown below has
a resistance of 6 Ω. Find the power output.
1.5 V = I × 6Ω + I1 × 2Ω --- ( 1 )
I1 1.5 V
1.5 V = I × 6Ω + I2 × 4Ω --- ( 2 )
I = I1 + I2 --- ( 3 )
I2 1.5 V
1.5 V = (I1 + I2 ) × 6Ω + I1 × 2Ω
4Ω
1.5 V = (I1 + I2 ) × 6Ω + I2 × 4Ω
Solve, I1 = 0.06818 A and I2 = 0.136 A
2Ω
I
6Ω
i.e. I = 0.06818 + 0.136 = 0.2042 A
∴ Power output of the bulb = 0.20422 × 6 = 0.25 W
Classwork 4 The light bulb in the diagram shown
below has a resistance of 10 Ω.
Find the power output.
I1 1.5 V
5Ω
1.5 V = I × 10Ω + I1 × 5Ω --- ( 1 )
1.5 V = I × 10Ω + I2 × 2Ω --- ( 2 )
I2 1.5 V
2Ω
( 1 ) × 2, ( 2 ) × 5
I = 0.13125 A
I
10Ω
∴ Power output of the bulb = 0.131252 × 10 = 0.17226 W
I-V characteristics of junction diode ( semiconductor diode )
I / mA
A diode allow current flow in 1
direction only. However, the
voltage across the diode must reach
a certain value to enable the charge 20 mA
carriers to flow.
For the diode to operate, I ≧20 mA,
the largest I
I min 
Rmax 
0.8 V
0.8 V
VR
 20m A
Rmax
VR
2.2

 110
I max 20m A
VD / V
2.2 V
R
I
ξ= 3V
For a diode to operate safely, prevent it burn out, a maximum
current must be noticed. This current is limited by the power
rating of the diode.
Assume a diode has a maximum rating of 0.08 W, the smallest
R is given by:
For the diode to operate safely,
Pmax  VD  I max
0.08  0.8V  I max  I max  0.1A
0.8 V
VR
2.2
Rmin 

 22
I max 0.1A
∴ the diode can be operated with
a resistor of 22Ω≦R ≦110 Ω
2.2 V
R
I
ξ= 3V
Classwork 5 The I-V characteristic of a diode is shown below.
The diode is operated with a resistor R. Given that
the maximum power rating of the diode is 1 W.
Calculate the maximum and minimum value of the
resistor.
1.2 V
I / mA
7.8 V
R
50 mA
VD / V
1.2 V
Rmax = 156 Ω
I
ξ= 9V
Rmin = 9.36 Ω
Thermionic diode 熱放電二極管
I / mA
A thermionic diode is a evacuated
glass tube with a heated cathode.
2 mA
Electrons is excited in the cathode
and ‘jump’ to the anode inside the
glass tube.
4V
If a diode is saturated at I = 2 mA,
the number electrons reaches a
maximum value and nor more
electrons can flow inside the glass
tube.
VD
-I
VD / V
12V- VD
R
ξ= 12V
I / mA
VD
2 mA
VD / V
4V
-I
12V- VD
R
ξ= 12V
If the diode is just saturated, the voltage across the diode
VD is 4 V and the current I = 2 mA.
12 V = VD + I R = VD + 2 mA × R  VD = 12 V - 2 mA × R
Since I must be larger than 2 mA, the resistor must be
smaller than a certain value.
VD > 4V  12V – 2 mA × R > 4V
∴ R < 4000 Ω
Bridge circuit
1Ω
1Ω
1Ω
1Ω
No current
∵ same p.d.
10Ω
10Ω
4V
10Ω
10Ω
P
Q
R
S
I1
I2
VP= VR
 I1 P = I2 R
VQ= VS
 I1 Q = I2 S
P R

Q S
Classwork 6 The ammeter in the diagram below shows
no reading. Calculate the resistance of the unknown
resistor P
P
25Ω
64Ω
40Ω
P = 40 Ω