Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


Presentation on theme: "Name: ________________ Class: _________________ Index: ________________ D.C. Circuit."— Presentation transcript:

1

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

3 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

4 -- 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

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

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

7 2V 2V 2V 6 V 2V -2V 2V 2 V  Examples of cell arrangement

8 Resistors Arrangement

9 Resistance: In series(connected side by side) In parallel (connected in parallel) Series and Parallel Circuits

10 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, R total = R 1 + R 2 + R 3 Resistors in series

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

12 If two resistors of resistance R 1 and R 2 are connected in parallel, Then Hence Two resistors connected in parallel

13 I Current flow through R 1, R 2 and R 3 with the same current, I. Current through resistors in series

14 I I1I1 I2I2 I3I3 R1 = R2 = R3: I = I 1 + I 2 + I 3 R 1 > R 2 > R 3 : I 3 > I 2 > I 3 Current through resistors in parallel

15 Voltage – Series/parallel circuit E E V2V2 V2V2 V1V1 V1V1 E = V 1 + V 2 E = V 1 = V 2

16 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 ? P Q 3A 5A 7A Solution Resultant current = (7 + 3) - (5) = 5 A (in the direction QP)

17 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. I 66 22 4V Solution (a) combined resistor = 6 + 2 = 8  (b) since V= RI 4 = 8 I, I = 0.5 A (c) V 6  = 6 x 0.5 = 3 V Example

18 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 Example

19 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 88 36  18  Continue on next slide Example

20 Solution Continue... (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

21 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 short circuit. A B Short Circuit

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

23 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 12  = 4.0V * Total potential difference for 4.0  and 8.0  resistor = 4.0V + 8.0V = 12V

24 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

25 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.

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

27 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.

28

29 Solution: (a) T = 14 x 5 ms = 70 ms (b) f = 1/T = 1 / 70 ms = 0.014 ms (c) Peak voltage = 4 x 20 V = 80 V

30 Solution: Peak voltage = 2 x 20 V = 40 V

31 Solution: Assume each small square is 1cm by 1 cm (a)Time to complete 1 cycle = 15 ms f = 1/T = 1 / 15 ms = 0.067 Hz (b) T = 15 ms (c) When frequency is doubled, period is halved and the length between the 2 crests is halved.

32 Solution: Assume each small square is 1cm by 1 cm Time between X and Y = 8 x 50 ms = 400 ms Distance = 3 x 10 8 m/s x 400 ms = 1.2 x 10 8 m

33 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_B oxes_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


Download ppt "Name: ________________ Class: _________________ Index: ________________ D.C. Circuit."

Similar presentations


Ads by Google