Internal resistance of batterySome of the electrical energy is dissipated by Joule heating inside the battery.

r

ξ

Terminal voltage V

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.

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.

I I

R

V

ξ= V + I r ξr

= I R + I r

=> ξ = R + r

I

=> R = ξ - r

I

∴ 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.

Combination of Resistors1. 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.

321 IIII 321 R

V

R

V

R

V

321

111

RRRV

321

1111

RRRR

Power and heating effect

The charge pass through the resistor:

tIQ The electrical energy converted into other form of energy in Δt:

QVW

∴ Power of the resistor:

t

WP

t

QV

t

QV

IV

R

VRIIVP

22

RRIIP

22

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.

4.1410

12222

P

VR

R

VP

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

WRIP 5.54.14619.0 22

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.

62.150

9222

P

VR

R

VP

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

WRIP 97.662.11775.0 22

The current flow through the above circuit:rR

I

Power output through R:RIPout

2 22

rR

R

Power output and Resistance

Max power occurs when R = r rr

r

rr

rP

44

2

2

2

2

2

max

22

rR

RP

2

rR

R

dR

dTry

Proof of max. Power output

22

2

2 rRrR

R

RrrRrR

R

42 22

2

RrrR

R

4)( 2

2

Efficiency of Electric Circuit

%100)(input

output 2

2

rR

R

rRI

RI

Power

Power

In general the power of the circuit.

When max. power occurs, R = r, the efficiency is:

%50%100

rr

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.

Wr

rIIPo 125.12

5.1 222

b.) Max. useful power output occur when R = ri.e. external resistance = 2Ω

Wr

P 28.024

5.1

4

22

max

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.

Wr

rIIPo 2.165

9222

b.) Max. useful power output occur when R = ri.e. external resistance = 5Ω

Wr

P 05.454

9

4

22

max

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

RIP Lloss 2

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

Typical Example ---- Combination of 2 cells

ξ r1

ξ r2

+ -Since any charge + or – cannot climb through the energy barrier set by the other cell,

NO current flow through the cells.

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

12V 15Ω

Net e.m.f. = 14 V – 12 V = 2 V

Total resistance = 15 + 5 + 20 = 40Ω

Current = 2 V / 40 Ω = 0.05 = 50 mA

20Ω

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.

Net e.m.f. = 9 V – 1.5 V = 7.5 V

Total resistance = 10 + 5 + 50 = 65Ω

Current = 7.5 V / 65 Ω = 0.115 A

9V 10Ω

1.5V 5Ω

50Ω

2 parallel cell connected to a resistor

ξ = I1 r1 + I R

ξ r1

ξ r2

I1

I2

I R

The terminal voltage of the cells are the same and equal to the voltage across R

ξ = I2 r2 + I R

I1 + I2 = I

This is known as Kirchoff’s 1st law of current

Example 6 The light bulb in the diagram shown below has a resistance of 6 Ω. Find the power output.

1.5 V2Ω

1.5 V

4Ω

I1

I2

I 6Ω

1.5 V = I × 6Ω + I1 × 2Ω --- ( 1 )

1.5 V = I × 6Ω + I2 × 4Ω --- ( 2 )

I = I1 + I2 --- ( 3 )

1.5 V = (I1 + I2 ) × 6Ω + I1 × 2Ω

1.5 V = (I1 + I2 ) × 6Ω + I2 × 4Ω

Solve, I1 = 0.06818 A and I2 = 0.136 A

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.

1.5 V 5Ω

1.5 V

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 = 0.13125 A

∴ Power output of the bulb = 0.131252 × 10 = 0.17226 W

I-V characteristics of junction diode ( semiconductor diode )

ξ= 3V

R

I

0.8 V 2.2 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.

I / mA

VD / V20 mA

0.8 VFor the diode to operate, I 20 mA, ≧the largest I

mAR

VI R 20

maxmin

11020

2.2

maxmax mAI

VR R

ξ= 3V

R

I

0.8 V 2.2 V

For the diode to operate safely,

maxmax IVP D

221.0

2.2

maxmin AI

VR R

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:

AIIV 1.08.008.0 maxmax

∴ the diode can be operated with a resistor of 22Ω R 110 Ω≦ ≦

ξ= 9V

R

I

1.2 V 7.8 VI / mA

VD / V50 mA

1.2 V

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.

Rmin = 9.36 ΩRmax = 156 Ω

Thermionic diode 熱放電二極管

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.

I / mA

VD / V

2 mA

4 V

ξ= 12V

R

I

VD 12V- VD

- -

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

- -

I / mA

VD / V

2 mA

4 V

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

10Ω10Ω10Ω10Ω

No current same p.d.∵

4 V

P Q

SR

VP= VR I1 P = I2 R

VQ= VS I1 Q = I2 S S

R

Q

P

I1

I2

Classwork 6 The ammeter in the diagram below shows no reading. Calculate the resistance of the unknown resistor P

P 25Ω

40Ω64Ω

P = 40 Ω