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- 1. flipperworks.comDirectCurrentCircuits
- 2. flipperworks.com Learning Outcomesa. Recall and use appropriate circuit symbols as set out in SI Units, Signs, Symbols and Abbreviations (ASE, 1981) and Signs, Symbols and Systematics (ASE, 1995).b. Draw and interpret circuit diagrams containing sources, switches, resistors, ammeters, voltmeters, and/or any other type of component referred to in the syllabus.c. Solve problems using the formula for the combined resistance of two or more resistors in series.d. Solve problems using the formula for the combined resistance of two or more resistors in parallel.e. Solve problems involving series and parallel circuits for one source of e.m.f.f.* Show an understanding of the use of a potential divider circuit as a source of variable p.d.g.* Explain the use of thermistors and light-dependent resistors in potential dividers to provide a potential difference which is dependent on temperature and illumination respectively.h.* Recall and solve problems using the principle of the potentiometer as a means of comparing potential differences.
- 3. flipperworks.comThe Electrical Circuit Symbols Symbol Meaning Symbol Meaning Cell / battery thermistor power supply diode switch potential divider A ammeter earth V voltmeter aerial/antenna galvanometer wires crossing with no connection filament lamp wires crossing with resistor connection variable resistor Loudspeaker light dependent resistor
- 4. flipperworks.com Principles Involved in Solving Circuits ProblemsConservation of charges states that charges cannot be created nor destroyed.Hence, the total charge that enters a junction per unit time (i.e. current entering)must be equal to the total charge that leaves the same junction per unit time (i.e.current leaving).Principle 1: The sum of currents entering any junction in an electric circuit is alwaysequal to the sum of currents leaving that junction (aka. Kirchhoff’s First Law) V I = I 1 + I2 I1 I R1 I I2 R2
- 5. flipperworks.com Principles Involved in Solving Circuits Problems Resistors In Parallel Resistors In Series V E I1 I R1 I R1 R2 I I2 R2 V1 V2From Principle 1: I = I1 + I2 From Principle 2: E = V1 + V2 V V VWhich implies that: = + Which implies that: I Rc = I R1 + I R2 Rc R1 R2 1 1 1Thus: = + Thus: Rc = R1 + R2 Rc R1 R2
- 6. flipperworks.comPrinciples Involved in Solving CircuitsProblemsE.g. 1 The diagrams show connected wires which carry currents I1, I2, I3 and I4.The currents are related by the equation I1 + I2 = I3 + I4.To which diagram does this equation apply? [N98/P1/Q16] A B C D I1 I3 I4 I4 I3 I2 I1 I2 I4 I1 I1 I2 I2 I3 I4 I3 C
- 7. flipperworks.com Principles Involved in Solving Circuits ProblemsExtra: Calculate the currents IA, IB, IC, ID, IE, and IF as shown in the diagram below. IF IA = 8.4 – 3.7 = 4.7 mA IB = 8.4 mA 1.3 mA IC = 1.3 mA 2.4 mA IE ID IC ID = 8.4 – 1.3 = 7.1 mA IB IE = 7.1 – 2.4 = 4.7 mA IA IF = 8.4 mA 3.7 mA 8.4 mA
- 8. flipperworks.comPrinciples Involved in Solving CircuitsProblemsE.g. 2 Referring to the circuit drawn, determine the value of I1, I and R, thecombined resistance in the circuit. 2V Applying Ohm’s Law, I1 160 Ω I1 = 2 ÷ 160 = 0.0125 A I2 = 2 ÷ 4000 = 5 × 10-4 A I I3 = 2 ÷ 32000 = 6.25 × 10-5 A I2 4000 Ω Since I = I1 + I2 + I3, I3 32000 Ω I = 13.1 mA Applying Ohm’s Law, Note: Rc of resistors in // is always R = 2 ÷ (13.1 × 10-3) smaller than smallest R in the // network = 153 Ω
- 9. flipperworks.com Principles Involved in Solving Circuits ProblemsConservation of energy states that energy cannot be created nor destroyed.Therefore, the electrical energy produced by the source is equal to the sum ofelectrical energy consumed by all the components.From the conservation of energy,Power supplied by = Power dissipated by Ethe source the resistors I R1 R2 IE = I2 R1 + I2 R2 E = I R1 + I R2i.e. E = V1 + V 2 V1 V2Principle 2: In any closed loop in an electric circuit, the total potential rise equalsto the total potential drop along that loop.i.e., in the circuit shown above the EMF of the battery, E = V1 + V2
- 10. flipperworks.com Principles Involved in Solving Circuits Problems Resistors In Parallel Resistors In Series V E I1 I R1 I R1 R2 I I2 R2 V1 V2From Principle 1: I = I1 + I2 From Principle 2: E = V1 + V2 V V VWhich implies that: = + Which implies that: I Rc = I R1 + I R2 Rc R1 R2 1 1 1Thus: = + Thus: Rc = R1 + R2 Rc R1 R2
- 11. flipperworks.com Principles Involved in Solving Circuits ProblemsE.g. 3 Referring to the circuit drawn, determine the value of the current I and thepotential difference across each resistor. 2V Total resistance = 100 + 600 + 300 = 1000 Ω I I = 2 ÷ 1000 100 Ω 600 Ω 300 Ω = 0.002 A Applying Ohm’s law, Potential difference across the 100 Ω resistor = 0.002 x 100 = 0.2 V Potential difference across the 600 Ω resistor = 0.002 x 600 = 1.2 V Potential difference across the 300 Ω resistor = 0.002 x 300
- 12. flipperworks.comPrinciples Involved in Solving CircuitsProblemsExtra: A battery with an EMF of 20 V and an internal resistance of 2.0 Ω isconnected to resistors R1 and R2 as shown in the diagram. A total current of 4.0A is supplied by the battery and R2 has a resistance of 12 Ω. Calculate theresistance of R1 and the power supplied to each circuit component. E - I r = I 2 R2 2Ω 20 – 4 (2) = I2 (12) I2 = 1 A 20 V Therefore, I1 = 4 – 1 = 3 A 4A R1 E – I r = I 1 R1 I1 12 = 3 R1 R2 Therefore, R1 = 4 Ω I2 Power supplied to R1 = (I1)2 R1 = 36 W Power supplied to R2 = (I2)2 R2 = 12 W
- 13. flipperworks.comPrinciples Involved in Solving CircuitsProblemsE.g. 4 The diagram shows a network of three resistors. Two of which have aresistance of 2.5 Ω and one of which has a resistance of 5.0 Ω, as shown.Determine the resistance between X and Y. [J91/P1/Q14] Y Ω 5.0 2.5 Ω Resistance between X and Y Z 1 1 = ( + ) −1 2.5 2.5 + 5 Ω = 1.875 Ω 2.5 X
- 14. flipperworks.comPrinciples Involved in Solving CircuitsProblemsE.g. 5 Three resistors of resistance 2 Ω, 3 Ω and 4 Ω respectively are used tomake the combinations X, Y and Z shown in the diagrams. List the combinationsin order of increasing resistance. X Y Z 2Ω 4Ω 3Ω 2Ω 2Ω 4Ω 3Ω 3Ω 4Ω 1 1 −1 Resistance for X = ( + ) = 1.56 Ω 2 4+3 1 1 −1 Therefore, the combination of Resistance for Y = 2 + ( + ) = 3.71 Ω resistors in order of increasing 4 3 resistance is Z X Y. 1 1 1 −1 Resistance for Z = ( + + ) = 0.923 Ω 3 2 4
- 15. flipperworks.comPrinciples Involved in Solving CircuitsProblemsE.g. 6 Calculate the resistance across XY for the network shown. 2Ω 3Ω 5Ω 4Ω 6Ω X Y −1 1 −1 1 −1 1 Resistance = + + 5 + = 3.277 Ω 2 + 3 4 6
- 16. flipperworks.com The Potential DividerA potential divider is a circuit in which two (or more) resistors are connected inseries with a supply voltage (EMF).The resistors “divide” the EMF proportionally according to their resistance. R1 = 6 kΩ V1 E = 12 V (negligible internal R2 = 4 kΩ V2 resistance) R1 R2 V1 = ×E Similarly, V2 = ×E R1 + R 2 R1 + R 2 6 4 = ×12 = ×12 4+6 4+6 = 7.2 V = 4 .8 V
- 17. flipperworks.comThe Potential DividerE.g. 7 Two resistors, of resistance 300 kΩ and 500 kΩ respectively, form apotential divider with outer junctions maintained at potentials of +3 V and –15 V. +3 V 300 kΩ X 500 kΩ -15 VDetermine the potential at the junction X between the resistors. 300 The potential difference across the 300 kΩ resistor = [ 3 − ( − 15) ] 300 + 500 = 6.75 V The potential at X = 3 - 6.75 = - 3.75 V
- 18. flipperworks.com The Potential DividerE.g. 8 The diagram below shows a potential divider circuit which by adjustmentof the position of the contact X, can be used to provide a variable potentialdifference between the terminals P and Q.Determine the limits of this potential difference. X 4 kΩ P 25 V Q 6 kΩ When X is at the upper end of the resistor, the potential difference across PQ 4 = × 25 = 10 V 4+6 When X is at the lower end of the resistor, the potential difference across PQ 0 = × 25 = 0 V 0 + 10
- 19. flipperworks.com The Potential DividerE.g. 9 A 20 V battery with an internal resistance of 2.5 kΩ is connected to a 50 kΩdevice. A voltmeter with an internal resistance of 120 kΩ and an ammeter withnegligible internal resistance are used to measure the potential difference across thedevice and the current passing through it respectively.(a) Sketch the circuit diagram for this setup.(b) Determine the reading on the voltmeter and the ammeter. −1 Rc (parallel to the voltmeter) = 1 1 + 50 k 120 k 20 V 2.5 kΩ = 35 kΩ 35.3 kI Voltmeter reading, V = × 20 = 19 V 50 kΩ 35.3 k + 2.5 k A 20 I= = 0.53 mA 35.3 k + 2.5k V 120 120 kΩ Ammeter reading = × 0.529 = 0.37 mA 120 + 50
- 20. flipperworks.comThe Use of a Thermistor in a PotentialDividerA thermistor is a resistor whose resistance varies greatly with temperature. Itsresistance decreases with increasing temperature. It can be used in temperaturepotential divider circuits to monitor and control temperatures.E.g. 10 In the figure below, the thermistor has a resistance of 800 Ω when hot,and a resistance of 5000 Ω when cold. Determine the potential at W when thetemperature is hot. +7 V When thermistor is hot, potential difference across it 800 = × ( 7 − 2) 1.7 kΩ 800 + 1700 W = 1.6 V The potential at W = 2 + 1.6 V = 3.6 V 2V
- 21. flipperworks.com The Use of a Light Dependent Resistor (LDR) in a Potential DividerAn LDR is a resistor whose resistance varies with the intensity of light falling on it.Its resistance decreases with increasing light intensity. It can be used in a intensitypotential divider circuit to monitor light intensity.E.g. 13 In the figure below, the resistance of the LDR is 6.0 MΩ in the dark butthen drops to 2.0 kΩ in the light. Determine the potential at point P when theLDR is in the light. +18 In the light the potential difference across the LDR V 2k = × (18 − 3) 3k +2k P =6V 3.0 kΩ The potential at P =18 – 6 3V = 12 V
- 22. flipperworks.comThe PotentiometerThe potentiometer is a device used to compare PD or EMF. It can measure anunknown EMF by comparing this EMF with a known standard EMF. Driver cell, E Note: 1) E > V L2 2) Same polarity L1 B A Potentiometer wire (Usually, 1 m) Sliding Contact p.d. source, VPotential difference along wire ∝ length of the wire.Sliding contact will move along AB until galvanometer shows 0 (i.e. current = 0)potentiometer is said to be “balanced”.If the cell has negligible internal resistance, and if the potentiometer is balanced, L1EMF / PD of the unknown source, V = ×E L1 + L 2
- 23. flipperworks.comThe Potentiometer 10 V 2V 2Ω 2Ω 2Ω 2Ω 2Ω 2V 2V No current flowing
- 24. flipperworks.comThe Potentiometer 10 V 4V 2Ω 2Ω 2Ω 2Ω 2Ω 2V 2V 10 V 1V 1Ω 1Ω 2Ω 2Ω 2Ω 2Ω 2V 2V
- 25. flipperworks.comThe PotentiometerE.g. 12 In the circuit shown, the potentiometer wire has a resistance of 60 Ω.Determine the EMF of the unknown cell if the balanced point is at B. 20 Ω 12 V Resistance of wire AB 0.65 0.35 m B 0.65 m = × 60 0.65 + 0.35 C A = 39 Ω 500 Ω 30 Ω EMF of the test cell 39 = × 12 60 + 20 = 5.85 V

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