1. 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. ξ r
Terminal voltage V
∴ 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.

2. 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. R
ξ= V + I r
= I R + I r ξ r I
=> ξ = R + r I
=> R = ξ r I 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.

3. Combination of Resistors
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

4. 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.

5. Power and heating effect
The charge pass through the resistor:
The electrical energy converted into other form of energy in Δt:
∴ Power of the resistor:

6. Example 5. A light bulb labelled 12 V 10W is connected
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.
The total resistance of the circuit is 5Ω Ω = 19.4 Ω
The current flow through the circuit and the bulb is = 12 V / 19.4 Ω = A

7. Classwork 1. A student connect a toy motor labelled
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.
The total resistance of the circuit is 1.62Ω + 10 Ω = Ω
The current flow through the circuit and the toy motor is = 9 V /11.62 Ω = A

8. Power output and Resistance
The current flow through the above circuit:
Power output through R:
Max power occurs when R = r

9. Proof of max. Power output

10. Efficiency of Electric Circuit
In general the power of the circuit.
When max. power occurs, R = r, the efficiency is:

11. Example 6. A 1. 5 V cell has an internal resistance of 2 Ω
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.
b.) Max. useful power output occur when R = r i.e. external resistance = 2Ω

12. Classwork2. A 9 V cell has an internal resistance of 5 Ω
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.
b.) Max. useful power output occur when R = r i.e. external resistance = 5Ω

13. 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
∵ Power transfer to the user equal to the power generate, it usually remains constant

14. Typical Example ---- Combination of 2 cellsξ r1 - +
Since any charge + or – cannot climb through the energy barrier set by the other cell,
NO current flow through the cells. ξ r2

15. 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
20Ω
Net e.m.f. = 14 V – 12 V = 2 V
12V
15Ω
Total resistance = = 40Ω
Current = 2 V / 40 Ω = 0.05 = 50 mA

16. Classwork3. A student uses a 9V battery of internal
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
1.5V 5Ω
50Ω
Net e.m.f. = 9 V – 1.5 V = 7.5 V
Total resistance = = 65Ω
Current = 7.5 V / 65 Ω = A

17. 2 parallel cell connected to a resistorξ r1 r2 I1 I2
The terminal voltage of the cells are the same and equal to the voltage across R I R
ξ = I1 r1 + I R
ξ = I2 r2 + I R
I1 + I2 = I
This is known as Kirchoff’s 1st law of current

18. Example 6. The light bulb in the diagram shown below has
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 )
1.5 V 2Ω 4Ω I1 I2 I 6Ω
1.5 V = I × 6Ω + I2 × 4Ω ( 2 )
I = I1 + I ( 3 )
1.5 V = (I1 + I2 ) × 6Ω + I1 × 2Ω
1.5 V = (I1 + I2 ) × 6Ω + I2 × 4Ω
Solve, I1 = A and I2 = A
i.e. I = = A
∴ Power output of the bulb = × 6 = 0.25 W

19. Classwork 4 The light bulb in the diagram shown
Classwork 4 The light bulb in the diagram shown below has a resistance of 10 Ω. Find the power output.
1.5 V 5Ω 2Ω I1 I2 I
10Ω
1.5 V = I × 10Ω + I1 × 5Ω ( 1 )
1.5 V = I × 10Ω + I2 × 2Ω ( 2 )
( 1 ) × 2, ( 2 ) × 5
I = A
∴ Power output of the bulb = × 10 = W

20. I-V characteristics of junction diode ( semiconductor diode )
I / mA
VD / V
20 mA
0.8 V
A diode allow current flow in 1 direction only. However, the voltage across the diode must reach a certain value to enable the charge carriers to flow.
For the diode to operate, I ≧20 mA, the largest I
ξ= 3V R I
0.8 V
2.2 V

21. 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,
ξ= 3V R I
0.8 V
2.2 V
∴ the diode can be operated with a resistor of 22Ω≦R ≦110 Ω

22. Classwork 5. The I-V characteristic of a diode is shown below
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.
ξ= 9V R I
1.2 V
7.8 V
I / mA
VD / V
50 mA
1.2 V
Rmax = 156 Ω
Rmin = 9.36 Ω

23. Thermionic diode 熱放電二極管
I / mA
VD / V
2 mA
4 V
A thermionic diode is a evacuated glass tube with a heated cathode. Electrons is excited in the cathode and ‘jump’ to the anode inside the glass tube.
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.
ξ= 12V R I VD
12V- VD -

24. - I / mA VD / V 2 mA 4 V ξ= 12V R I VD 12V- VD
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

25. Bridge circuit 1Ω
10Ω
No current ∵ same p.d.
4 V

26. P Q I1 S R I2
VP= VR
 I1 P = I2 R
VQ= VS
 I1 Q = I2 S

27. Classwork 6 The ammeter in the diagram below shows no reading
Classwork 6 The ammeter in the diagram below shows no reading. Calculate the resistance of the unknown resistor P P
25Ω
40Ω
64Ω
P = 40 Ω