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

Name: ________________ Class: _________________ Index: ________________

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D.C. Circuit. Name: ________________ Class: _________________ Index: ________________. Objectives - PowerPoint PPT Presentation

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Page 1: Name: ________________ Class: _________________ Index: ________________

Name: ________________

Class: _________________

Index: ________________

D.C. Circuit

Page 2: Name: ________________ Class: _________________ Index: ________________

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

Page 3: Name: ________________ Class: _________________ Index: ________________

-- 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

Page 4: Name: ________________ Class: _________________ Index: ________________

Electrical Circuit Symbols

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

LED

Bell

Page 5: Name: ________________ Class: _________________ Index: ________________

• 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)

(b)

A cell = 2V; emf = 2V

Cells in Series and Parallel

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2V 2V 2V

6 V

2V -2V 2V

2 V

2V

2V

2V

2V

Examples of cell arrangement

Page 7: Name: ________________ Class: _________________ Index: ________________

Resistors Arrangement

Page 8: Name: ________________ Class: _________________ Index: ________________

•Resistance:In series(connected side by side)

In parallel (connected in parallel)

Series and Parallel Circuits

Page 9: Name: ________________ Class: _________________ Index: ________________

• 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

Resistors in series

Page 10: Name: ________________ Class: _________________ Index: ________________

• When resistor connected in parallel:

The total resistance (or effective resistance or resultant resistance) is

Resistors in parallel

Page 11: Name: ________________ Class: _________________ Index: ________________

• If two resistors of resistance R1 and R2 are connected in parallel,

Then

Hence

Two resistors connected in parallel

Page 12: Name: ________________ Class: _________________ Index: ________________

I

Current flow through R1, R2

and R3 with the same current,I.

Current through resistors in series

Page 13: Name: ________________ Class: _________________ Index: ________________

I

I1

I2

I3

R1 = R2 = R3: I = I1 + I2 + I3

R1 > R2 > R3: I3 > I2 > I1

Current through resistors in parallel

Page 14: Name: ________________ Class: _________________ Index: ________________

Voltage – Series/parallel circuit

E

E V2

V2

V1

V1

E = V1 + V2

E = V1 = V2

Page 15: Name: ________________ Class: _________________ Index: ________________

Example

The diagram shows the magnitude and directions ofthe electric currents entering and leaving junction P.What will be the magnitude and direction of the current in the wire PQ ?

P

Q3A

5A

7A

Solution

Resultant current = (7 + 3) - (5) = 5 A (in the direction QP)

Page 16: Name: ________________ Class: _________________ Index: ________________

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 24V

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

Example

Page 17: Name: ________________ Class: _________________ Index: ________________

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

Page 18: Name: ________________ Class: _________________ Index: ________________

The battery in the circuit illustrated has an e.m.f. of16 V and negligible internal resistance.Calculate(a) the combined resistance of the system.(b) the current flowing through the 8 resistor.

16V

8 36

18

Continue on next slide

Example

Page 19: Name: ________________ Class: _________________ Index: ________________

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

Page 20: Name: ________________ Class: _________________ Index: ________________

• 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

Page 21: Name: ________________ Class: _________________ Index: ________________

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.

Page 22: Name: ________________ Class: _________________ Index: ________________

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

Page 23: Name: ________________ Class: _________________ Index: ________________

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

Page 24: Name: ________________ Class: _________________ Index: ________________

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.

Page 25: Name: ________________ Class: _________________ Index: ________________

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.

(a) 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

Page 26: Name: ________________ Class: _________________ Index: ________________

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.

Page 27: Name: ________________ Class: _________________ Index: ________________
Page 28: Name: ________________ Class: _________________ Index: ________________

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

Page 29: Name: ________________ Class: _________________ Index: ________________

Solution:

Peak voltage = 2 x 20 V = 40 V

Page 30: Name: ________________ Class: _________________ Index: ________________

Solution:

Assume each small square is 1cm by 1 cm

(a) 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.

Page 31: Name: ________________ Class: _________________ Index: ________________

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

Page 32: Name: ________________ Class: _________________ Index: ________________

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