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Internal resistance of battery Some of the electrical energy is dissipated by Joule heating inside the battery. e.m.f. across the + & - terminal ξ of the battery is lower than the marked value when connected to r external components. 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 R of resistance R. ξ= V + I r =IR+Ir ξ I I => ξ = R + r r I => R = ξ - r V 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 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 I 3 V R1 V R2 V R3 1 V 1 R R R 1 2 1 3 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: P W V Q t t V Q t V I V2 2 P V I I 2R P I I 2R R 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. V2 V 2 12 2 P R 14 .4 R 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. V2 V 2 92 P R 1.62 R 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: I Rr Power output through R: 2 2R Pout I R R r 2 2r 2r 2 Max power occurs when R = r Pmax r r 2 4r 2 4r Proof of max. Power output 2R R 2 P R r 2 2 R 2 Rr r 2 R 2 2 2 R 2 Rr r 4 Rr R 2 d R Try 2 dR R r 2 ( R r ) 4 Rr Efficiency of Electric Circuit In general the power of the circuit. Power output I 2R R 100% Power input I (R r) R r 2 When max. power occurs, R = r, the efficiency is: r 100% 50% rr 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 1.52 Po I I 2 r 1.125W r 2 b.) Max. useful power output occur when R = r i.e. external resistance = 2Ω 21.52 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 92 Po I I 2 r 16 .2W r 5 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 ξ r1 Since any charge + or – cannot + - climb through the energy barrier set by the other cell, ξ r2 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 Net e.m.f. = 14 V – 12 V = 2 V 20Ω Total resistance = 15 + 5 + 20 = 40Ω 12V 15Ω Current = 2 V / 40 Ω = 0.05 = 50 mA 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 1.5V 5Ω Current = 7.5 V / 65 Ω = 0.115 A 2 parallel cell connected to a resistor I1 ξ r1 The terminal voltage of the cells are the same and equal to the voltage across R I2 ξ r2 ξ = I1 r1 + I R ξ = I2 r2 + I R I1 + I2 = I I R 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 V = I × 6Ω + I1 × 2Ω --- ( 1 ) I1 1.5 V 2Ω 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 I 6Ω 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. I1 1.5 V 1.5 V = I × 10Ω + I1 × 5Ω --- ( 1 ) 5Ω 1.5 V = I × 10Ω + I2 × 2Ω --- ( 2 ) I2 1.5 V ( 1 ) × 2, ( 2 ) × 5 2Ω 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 ) A diode allow current flow in 1 I / mA direction only. However, the voltage across the diode must reach a certain value to enable the charge 20 mA VD / V carriers to flow. 0.8 V For the diode to operate, I ≧20 mA, the largest I 0.8 V 2.2 V VR I min 20 mA Rmax R VR 2.2 Rmax 110 I max 20 mA 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 0.8 V 2.2 V 0.08 0.8V I max I max 0.1A VR 2 .2 Rmin 22 I max 0.1A R ∴ the diode can be operated with I ξ= 3V a resistor of 22Ω≦R ≦110 Ω 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. I / mA 1.2 V 7.8 V R 50 mA VD / V 1.2 V I ξ= 9V Rmax = 156 Ω 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 VD / V glass tube. 4V If a diode is saturated at I = 2 mA, VD 12V- VD the number electrons reaches a maximum value and nor more electrons can flow inside the glass - - R tube. I ξ= 12V I / mA VD 12V- VD 2 mA - - R VD / V 4V I 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 10Ω 10Ω 4V P Q I1 I2 R S VP= VR I1 P = I2 R P R VQ= VS I1 Q = I2 S 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

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posted: | 12/30/2010 |

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Internal resistance of battery

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