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.

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.

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

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.

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:

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 + 14.4 Ω = 19.4 Ω The current flow through the circuit and the bulb is = 12 V / 19.4 Ω = 0.619 A

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 = 11.62 Ω The current flow through the circuit and the toy motor is = 9 V /11.62 Ω = 0.775 A

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

Proof of max. Power output

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

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

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

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

Typical Example ---- Combination of 2 cells ξ r1 - + Since any charge + or – cannot climb through the energy barrier set by the other cell, NO current flow through the cells. ξ 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 20Ω Net e.m.f. = 14 V – 12 V = 2 V 12V 15Ω Total resistance = 15 + 5 + 20 = 40Ω Current = 2 V / 40 Ω = 0.05 = 50 mA

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 = 10 + 5 + 50 = 65Ω Current = 7.5 V / 65 Ω = 0.115 A

2 parallel cell connected to a resistor ξ 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

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 + 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 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 = 0.13125 A ∴ Power output of the bulb = 0.131252 × 10 = 0.17226 W

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

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

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

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 -

- 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

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

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

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