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Name: ________________ Class: _________________ Index: ________________

D.C. Circuit. Name: ________________ Class: _________________ Index: ________________. Objectives

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Name: ________________ Class: _________________ Index: ________________

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  1. D.C. Circuit Name: ________________ Class: _________________ Index: ________________

  2. Objectives -- draw circuit diagrams with power sources (cell or battery), switches, lamps, resistors (fixed and variable), fuses, ammeters and voltmeters, bells, light-dependent resistors, thermistors and light-emitting diodes --state that the current at every point in a series circuit is the same and apply the principle to new situations or to solve related problems --state that the sum of the p.d.'s in a series circuit is equal to the p.d. across the whole circuit and apply the principle to new situations or to solve related problems --state that the current from the source is the sum of the currents in the separate branches of the parallel circuit and apply the principle to new situations or to solve related problems -- state that the potential differences across the separate branches of a parallel circuit is the same and apply the principle to new situations or to solve related problems

  3. -- recall and apply the relevant relations, including R = V/I and those for potential differences in series and in parallel circuits, resistors in series and in parallel, in calculations involving a whole circuit --describe the action of a variable potential divider (potentiometer) --describe the action of thermistors and light-dependent-resistors and explain their use as input transducers in potential dividers -- solve simple circuit problems involving thermistors and light-dependent resistors -- describe the use of a CRO to display waveforms and to measure p.d.’s and short time intervals of time (detailed circuits, structure and operation of the CRO are not required -- interpret CRO displays of waveforms, p.d.’s and time intervals to solve related problems

  4. Electrical Circuit Symbols Electrical Circuit symbols are used in circuit diagrams which show how a circuit is connected together. LED Bell

  5. Cells in Series and Parallel • Cell : In series (connected side by side) In parallel (connected in parallel) (b) For (b), If a cell has V = 2V, then total emf = 8V (b) A cell = 2V; emf = 2V

  6. Examples of cell arrangement 2V -2V 2V 2V 2V 2V 6 V 2 V 2V 2V   2V 2V

  7. Resistors Arrangement

  8. Series and Parallel Circuits • Resistance: In series(connected side by side) In parallel (connected in parallel)

  9. Resistors in series • When resistors are connected in series, the total resistance (effective resistance or resultant resistance) is equal to the sum of of the individual resistance. Thus, Rtotal = R1 + R2 + R3

  10. Resistors in parallel • When resistor connected in parallel: The total resistance (or effective resistance or resultant resistance) is

  11. Two resistors connected in parallel • If two resistors of resistance R1 and R2 are connected in parallel, Then Hence

  12. Current through resistors in series I Current flow through R1, R2 andR3 with the same current, I.

  13. Current through resistors in parallel I1 I I2 I3 R1 = R2 = R3: I = I1 + I2 + I3 R1 > R2 > R3: I3 > I2 > I1

  14. Voltage – Series/parallel circuit V1 V2 E = V1 + V2 E E V1 V2 E = V1 = V2

  15. Example The diagram shows the magnitude and directions of the electric currents entering and leaving junction P. What will be the magnitude and direction of the current in the wire PQ ? Q 3A Solution 5A P Resultant current = (7 + 3) - (5) = 5 A (in the direction QP) 7A

  16. Example A voltage of 4V is supplied to two resistors of (6  and 2  ) connected in series. Calculate (a) the combined resistance, (b) the current flowing, (c) the p.d. across the 6  resistor. Solution (a) combined resistor = 6 + 2 = 8  (b) since V= RI 4 = 8 I, I = 0.5 A (c) V6 = 6 x 0.5 = 3 V I 6 2 4V

  17. Example A voltage of 12 V is applied to two resistors of 3  and 6  connected in parallel. Calculate (a) the combined resistance, (b) the current flowing in the main circuit, (c) the current in the 3  resistor. Solution (a) the combined resistance = (3 x 6) / (3 + 6) = 2  (b) since V = RI I = V / R = 12 / 2 = 6 A (c) current through 3  = 12 / 3 = 4 A 3  6  12 V

  18. Example The battery in the circuit illustrated has an e.m.f. of 16 V and negligible internal resistance. Calculate (a) the combined resistance of the system. (b) the current flowing through the 8  resistor. 16V 36 8 18 Continue on next slide

  19. Continue ... Solution (a) combined resistance = (36 x 18) / (36 + 18) + 8 = 20  (b) since V = RI therefore 16 = 20 I I = 0.8 A hence, current through 8 resistor is 0.8A

  20. A B Short Circuit • In the fig shown, AB is a copper wire which connects two point A and B on the circuit. • Since copper wire has very little resistance, therefore a large amount of current will flow through it. • The lamp then go off. (Why ?) • Therefore we say this circuit is now a shortcircuit.

  21. Potential Divider Circuit A voltage divider (also known as a potential divider) is a simple linear circuit that produces an output voltage (Vout) that is a fraction of its input voltage (Vin). Voltage division refers to the partitioning of a voltage among the components of the divider.

  22. Example Q) Calculate the output voltage from the 4 resistor. Solution: Total resistance = 4.0 + 8.0 = 12 Current = 12V / 12 = 1.0A Output voltage (4.0 ) = 1.0A x 4.0 = 4.0V * Total potential difference for 4.0 and 8.0 resistor = 4.0V + 8.0V = 12V

  23. Input Transducer Input Transducers convert a quantity to an electrical signal (voltage) or to resistance (which can be converted to voltage). Input transducers are also called sensors. Variable resistor converts position (angle) to resistance Thermistor converts temperature to resistance LDR converts brightness (of light) to resistance

  24. Using an input transducer (sensor) in a voltage divider Most input transducers (sensors) vary their resistance and usually a voltage divider is used to convert this to a varying voltage which is more useful. The voltage signal can be fed to other parts of the circuit, such as the input to an IC or a transistor switch. The sensor is one of the resistances in the voltage divider. It can be at the top (R1) or at the bottom (R2), the choice is determined by when you want a large value for the output voltage Vo: Put the sensor at the top (R1) if you want a large Vo when the sensor has a small resistance. Put the sensor at the bottom (R2) if you want a large Vo when the sensor has a large resistance.

  25. Example Suppose the LDR has a resistance of 0.50 k , in bright light, and 200 k in the shade, calculate the magnitude of Vout: (a) under the sun, (b) in the shade. • Total resistance = (10 + 0.5) k = 10.5 k • I = 9.00 V / 10.5 k = 0.86 mA • Vout = 0.86 mA x 0.5 k = 0.43 V • (b) Total resistance = (10 + 200) k = 210 k • I = 9.00 V / 210 k = 0.043 mA • Vout = 0.043 mA x 200 k = 8.6 V

  26. Cathode Ray Oscilloscope An oscilloscope is a test instrument which allows you to look at the 'shape' of electrical signals by displaying a graph of voltage against time on its screen. The graph, usually called the trace, is drawn by a beam of electrons striking the phosphor coating of the screen making it emit light, usually green or blue. Oscilloscopes contain a vacuum tube with a cathode (negative electrode) at one end to emit electrons and an anode (positive electrode) to accelerate them so they move rapidly down the tube to the screen. The tube also contains electrodes to deflect the electron beam up/down and left/right.

  27. Solution: (a) T = 14 x 5 ms = 70 ms (b) f = 1/T = 1 / 70 ms = 14.3 Hz (c) Peak voltage = 4 x 20 V = 80 V

  28. Solution: Peak voltage = 2 x 20 V = 40 V

  29. Solution: • Assume each small square is 1cm by 1 cm • Time to complete 1 cycle = 15 ms • f = 1/T = 1 / 15 ms = 67 Hz • (b) T = 15 ms • (c) When frequency is doubled, period is halved and the length between the 2 crests is halved.

  30. Solution: Assume each small square is 1cm by 1 cm Time between X and Y = 8 x 50 ms = 400 ms Distance = 3 x 108 m/s x 400/2 ms = 0.6 x 108 m

  31. Reference http://www.antonine-education.co.uk/physics_gcse/Unit_2/Topic_6/topic_6_files/image002.gif http://bleex.me.berkeley.edu/ME102/proj_archive/S05/18-Inverted_Pendulum/images/Hardware/Potentiometer.bmp http://physics.kenyon.edu/EarlyApparatus/Electrical_Measurements/Resistance_Boxes_and_Rheostats/Greenslade170a.JPG http://www.jestineyong.com/wp-content/uploads/2008/05/ntc-thermistor.jpg http://www.mstracey.btinternet.co.uk/technical/Theory/ldr.jpg http://www.hobbyprojects.com/dc_theory/potential_dividers.html http://www.antonine-education.co.uk/New_items/DIG/Potential_Divider.gif http://www.kpsec.freeuk.com/vdivider.htm http://www.kpsec.freeuk.com/cro.htm

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