C) Capacitación y asistencia técnica
11. PAPEL INSTITUCIONAL Y GESTIÓN DE LAS INNOVACIONES EN LA EMPRESA
Example
From the diagram given in the figure (below), calculate
I1 I2 I
6 Ω
3 Ω
12 VI
4 Ω
I
(a) the total resistance of the circuit (b) the current flowing in the main circuit (c) the current in each resistor
Solution
(a) For the parallel circuit we have, 1 Reff --- 1 3 --- 1 6 --- + 3 6 ---⇒Reff 2 = = = So that , Reff = 2Ω
Thus the total resistance (of the parallel and series arms), RT, is given by
RT = 2 Ω + 4 Ω
= 6 Ω
That is, the effective resistance is 6 Ω.
(b) To determine the current, once again we use, V = IR, so that, I V RT --- 12 6 --- 2 A = = =
The current flowing in the main circuit is 2 A.
(c) The current in the 4 Ω resistor is 2 A. The potential difference across the 4 Ω resistor = IR = (2 A) . (4 Ω) = 8 V
The potential difference across the parallel network = 12 V – 8 V
= 4 V
The current in the network resistors is given by,
I I1 +I2 V R1 --- V R2 --- + = = I 4 3 --- A 4 6 --- A + 1.33 A +0.67 A = =
The current in the 4 Ω resistor is 2 A. The currents in the 3 Ω and the 6 Ω resistors are 1.33 A and 0.67 A respectively.
5.2.4 ELECTRICCIRCUIT
SYMBOLS
The circuit symbols shown in Table 522 are used for drawing circuit diagrams in this course. This standard set of symbols is also provided in the data booklet that is used during examination sessions. You should use these symbols in place of any other that you may have learnt.
5.2.5 AMMETERSANDVOLTMETERS
Galvanometers are used to detect electric currents. They use a property of electromagnetism – a coil with a current flowing in it experiences a force when placed in a magnetic field. Most non-digital ammeters and voltmeters consist of a moving coil galvanometer connected to resistors. Digital meters are becoming more common, and the digital multimeter can act as an ammeter, a voltmeter or an ohmmeter.
A voltmeter
1. is always connected across a device (in parallel).
2. has a very high resistance so that it takes very little current from the device whose potential difference is being measured.
3. has a high resistor (a multiplier) connected in
series with a galvanometer.
4. an ideal voltmeter would have infinite resistance with no current passing through it and no energy would be dissipated in it.
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Example
A galvanometer has a resistance of 1.0 × 103 Ω (mainly
due to the resistace of the coil), and gives a full-scale deflection (fsd) for 1.0 mA of current. Calculate the size of a multiplier resistor that would be needed to convert this to a voltmeter with a fsd of 10.0 volts.
Solution
The Figure below demonstrates a possible set up.
G
RG
10 V I
RS
joined wires wires crossing (not joined) cell
battery lamp a.c. supply
switch ammeter voltmeter
galvanometer resistor variable resistor
potentiometer heating element fuse
transformer oscilloscope
A V
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Because the resistors Rg and Rs are in series then the potential difference across the resistors will be 10 volts. Therefore, I (Rg + Rs) = 10
0.001 × (1000 + Rs) = 10 0.001Rs = 9
Rs = 9000 Ω
The multiplier resistor = 9.0 × 103 Ω
An ammeter
1. is always connected in series with a circuit.
2. has a very low resistance compared with the resistance of the circuit so that it will not
significantly alter the current flowing in the circuit.
3. has a low resistor (a shunt) connected in parallel with a galvanometer.
4. would ideally have no resistance with no potential difference across it and no energy would be dissipated in it.
Example
A galvanometer has a resistance of 1.0 × 103 Ω (mainly due
to the resistace of the coil), and gives a full-scale deflection (fsd) for 1.0 mA of current. Calculate the size of a shunt resistor that would be needed to convert this to a meter with a fsd of 5.0 A.
Solution
The Figure below demonstrates a possible set up.
G RG I i I - i RS I
Because the resistors Rg and Rs are in parallel then the potential difference across the resistors is the same. Therefore, iRg = (I – i) Rs.
Rs = (i / I – i) Rg = [10-3 A / (5 – 0.001) A] × 1000 Ω = 0.20 Ω
The shunt resistor = 0.20 Ω
5.2.6 A POTENTIALDIVIDER
In electronic systems, it is often necessary to obtain smaller voltages from larger voltages for the various electronic circuits. A potential divider is a device that produces the required voltage for a component from a larger voltage. It consists of a series of resistors or a rheostat (variable resistor) connected in series in a circuit. A simple voltage divider is shown in Figure 525.
V2 V1 V R1 R2 I
Figure 525 A simple voltage divider Using Ohm’s Law, V1 = IR1 and V = I(R1 + R2).
V1 / V = IR1 / I(R1 + R2). Therefore, V1 = R1 / (R1 + R2) × V
This is known as the potential divider equation.
Example
In the potential divider shown below, calculate:
(a) the total current in the circuit
(b) the potential difference across each resistor (c) the voltmeter reading if it was connected
between terminals 2 and 6.
1 2 3 4 5 6 7
12 V
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Solution
(a) The total resistance = 12 Ω. From Ohm’s law V = IR and I = V / R
I = 12 V / 12 Ω = 1 A
(b) The 12 V is equally shared by each 2 Ω resistor. Therefore, the potential difference across each resistor = 2 V. Alternatively, V = IR = 1 A × 2 Ω = 2 V
(c) Between terminals 2 and 6 there are 4 resistors each of 2 Ω. Therefore, the potential difference the terminals
= 4 × 2 V = 8 V.
Because resistance is directly proportional to the length of a resistor, a variable resistor also known as a potentiometer or colloquially as a “pot” can also be used to determine the potential difference across an output transducer (device for converting energy from one form to another) such as a filament lamp in Figure 527.
If the pointer was at A then the potential difference would be zero as there is no power dissipated per unit current in the potentiometer (the load), and there would be no output voltage. However, if the pointer is moved up to two-thirds the length of the potentiometer as in the figure, then the output voltage across the filament lamp would be ⅔ × 6V = 4V.
6V
potentiometer
V A
Figure 527 Laboratory potentiometer
Pots have a rotating wheel mounted in plastic and they are commonly used as volume and tone controls in sound systems. They can be made from wire, metal oxides or carbon compounds.
5.2.7 SENSORSIN
POTENTIALDIVIDER
CIRCUITS
A number of sensors (input transducers) that rely on a change in resistance can be used in conjunction with potential dividers to allow for the transfer of energy from one form to another. Three such sensors are:
• the light dependent resistor (LDR)
• the negative temperature coefficient thermistor (NTC)
• strain gauges.
A light dependent resistor (LDR) is a photo-conductive cell whose resistance changes with the intensity of the incident light. Typically, it contains a grid of interlocking electrodes made of gold deposited on glass over which is deposited a layer of the semiconductor, cadmium sulfide. Its range of resistance is from over 10 MΩ in the dark to about 100 Ω in sunlight. A simple LDR and its circuit symbol are shown in Figure 528.
Figure 528 A light dependent resistor
There are only two ways to construct a voltage divider with an LDR sensor – with the LDR at the top, or with the LDR at the bottom as shown in Figure 529.
I Vin Vin I 0V 0V Vout Vout
Figure 529 (a) and (b) two ways to construct an L.D.R.
Light dependant resistors have many uses in electronic circuits including smoke detectors, burglar alarms, camera
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light meters, camera aperture controls in automatic cameras and controls for switching street lights off and on.
Resistors that change resistance with temperature are called thermistors (derived from thermal resistors). They are made from ceramic materials containing a semiconductor the main types being bead and rod thermistors. The NTC (negative temperature coefficient) thermistor contains a mixture of iron, nickel and cobalt oxides with small amounts of other substances. They may have a positive (PTC) or negative (NTC) temperature coefficient according to the equation:
Rf = R0(1+αt)
where R0 equals the resistance at some reference temperature say 0 °C, Rf is the resistance at some temperature, t °C, above the reference temperature,and α is the temperature coefficient for the material being used.
The circuit symbol for a thermistor is shown in Figure 530.
Figure 530 The circuit symbol for a thermistor
For a NTC thermistor, the resistance decreases when the temperature rises and therefore they can pass more current. This current could be used to to operate a galvonometer with a scale calibrated in degrees as used in electronic thermometers or car coolant system gauges. Thermistors are also used in data-logging temperature probes but the analogue signal has to be converted to digital signal using an analogue to digital converter (ADC).
With normal resistors the resistance becomes higher when the temperature increases and they therefore have a small positive temperature coefficient. Figures 531 and 532 demonstrate how the resistance changes with temperature for both types of thermistors.
20 40 60 80 100 Temperature / ˚C R e sistance Ω 50 10 Figure 531 A PTC thermistor 20 40 60 80 100 200 Temperature / ˚C R e sistance Ω Figure 532 An NTC thermistor
An electronic thermometer can be made using an NTC thermistor as shown in Figure 533.
200 Ω milliammeter
with fsd 0 – 10 mA
Figure 533 An electronic thermometer
When a metal conducting wire is put under vertical strain, it will become longer and thinner and as a result its resistance will increase. An electrical strain gauge is a device that employs this principle. It can be used to obtain information about the size and distribution of strains in structures such as metal bridges and aircraft to name but two. A simple gauge consists of very fine parallel threads of a continous metal alloy wire cemented to a thin piece of paper that are hooked up to a resistance measuring device with thick connecting wires as shown in Figure 534. When it is securely attached on the metal to be tested, it
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will experience the same strain as the test metal and as this happens the strain gauge wire become longer and thinner and as such the resistance increases.
thin threads of wire
thin paper
connecting lead structure under strain
Figure 534 A resistance measuring device
5.2.8 SOLVEPROBLEMSINVOLVING
ELECTRICCIRCUITS
Many problems have been done with their worked examples. Now try to solve the following exercise.
Exercise 5.2
1. Three identical resistors of 3 Ω are connected in parallel in a circuit. The effective resistance would be
A. 1 Ω B. 3 Ω C. 6 Ω D. 9 Ω
2. In a television tube the picture is produced in a fluorescent material at the front of the picture tube by
A. an electrical discharge B. a beam of positive ions
C. an electrolytic deposition of metal atoms D. a stream of electrons
3. A 100 W light globe gives out a brighter light than a 60 W globe mainly because
A. a larger potential difference is used to run it B. its resistance wire filament is longer
C. more electric current flows through it D. it has a higher amount of inert gas in it
4. In the circuit below, a heater with resistanceIn the circuit below, a heater with resistance R is connected in series with a 48 V supply and a resistor S.
48 V
S
R
If the potential difference across the heater is to be maintained at 12 V, the resistance of the resistor S should be
A. R / 2
B. R / 4
C. R / 3
D. 3R
5. The diagrams below show circuits X, Y and Z of three resistors, each resistor having the same resistance.
circuit X circuit Y circuit Z
Which one of the following shows the resistances of the circuits in increasing order of magnitude?
lowest → highest A. Y Z X B. ZZZ X YXX YY C. XXX Y ZYY ZZ D. ZZZ Y XYY XX
6. The graph below shows the current/voltage characteristics of a filament lamp.
24 16 8 0 0 2 4 6 8 10 Voltage /V Curren t A
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The resistance of the filament at 3.0 V is
A. 0.25 Ω B. 250 Ω C. 4000 Ω D. 8000 Ω
7. Metal X has half the resistivity of metal Y and length three times that of Y. If both X and Y have the same surface area, the ratio of their resistance RX / RY is:
A. 3 : 4 B. 2 : 3 C. 1.5 : 1 D. 2 : 9
8. Three identical resistors of 3 Ω are connected in parallel in a circuit. The effective resistance would be
A. 1 Ω B. 3 Ω C. 6 Ω D. 9 Ω
9. If 18 J of work must be done to move 2.0 C of charge from point A to point B in an electric field, the potential difference between points A and B is:
A. 0.1 V B. 9.0 V C. 12 V D. 20 V
10. The fundamental SI unit – the ampere – is defined in terms of:
A. potential difference and resistance
B. the time rate of change of charge in a circuit C. the force acting between two current
carrying wires
D. the product of charge and time
11. A 100 W light globe gives out a brighter light than a 60 W globe mainly because
A. a larger potential difference is used to run it B. its resistance wire filament is longer
C. more electric current flows through it D. it has a higher amount of inert gas in it
12. Three identical lamps L1, L2 and L3 are connected as shown in the following diagram.
L
1L
3L
2S
When switch S is closed
A. L1 and L3 brighten and L2 goes out B. all three lamps glow with the same
brightness
C. L2 brightens and L1 and L3 remain unchanged
D. L1 and L3 go dimmer and L2 goes out 13. Two resistors are connected in parallel and have
the currents I1 and I2 as shown in the diagram
If the effective resistance of the circuit is R then
A. R = (R1 + R2) / R1R2 B. I1/R1 = I2 / R2
C. R = R1R2 / R1 + R2 D. R = I (R1 + R2) / R1R2
14. The following circuit was set up to determine the internal resistance r of a dry cell. The load resistor was varied from 100Ω to 150Ω, and the current in the circuit was measured using an ammeter. The two respective values of the current are given in the circuits. I 1 I2 R 1 R2
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The internal resistance r of the dry cell is
A. 3.0 Ω B. 7.1 Ω C. 58.8Ω D. 250 Ω
15. This question concerns the electric circuit below
I1
2Ω
4Ω
12V
The cell voltage is 12 V. The value of current I1 is A. 2 A
B. 4 A C. 6 A D. 9 A
16. Consider the circuit below that contains a 15 V battery with zero internal resistance.
4Ω 15 V 4Ω 4Ω X Y 3Ω
A voltmeter connected between the points X and Y should read:
A. 0 V B. 3 V C. 6 V D. 9 V
17. Determine the equivalent resistance when 12 Ω, 6 Ω and 4 Ω are placed in
(a) series (b) parallel
18. Calculate the work done in moving a 12.0 μC through a p.d of 240 V.
19. The diagram shows resistances joined in a compound circuit. I1 I2 I 4 Ω 6 Ω 12.2 V I 2 Ω 20 Ω
(a) Determine the total resistance of the circuit. (b) Calculate current flows through the 2.0 Ω
resistor.
(c) Deduce the potential difference across the 20.0 Ω resistor.
(d) Determine is the potential difference across the 6.0 Ω resistor.
(e) Calculate is the current through the 4.0 Ω resistor.
20. A photo-electric cell draws a current of 0.12 A when driving a small load of resistance 2 Ω. If the emf of the cell is 0.8 V, determine the internal resistance of it under these conditions.
21. In terms of emf and internal resistance, explain why is it possible to re-charge a nickel-cadmium cell while normal dry cells have to be discarded once they are flat.
22. When a dry cell is connected to a circuit with a load resistor of 4.0 Ω, there is a terminal voltage of 1.3 V. When the load resistor is changed to 12 Ω, the terminal voltage is found to be 1.45 V. Calculate
(a) the emf of the cell.
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23. Calculate the current flowing through a hair drier if it takes 2.40 × 103 C of charge to dry a person’s
hair in 3.0 minutes.
24. An iron draws 6.0 A of current when operating in a country with a mains-supply of 240 V. Calculate the resistance of the iron.
25. An electrical appliance is rated as 2.5 kW, 240 V
(a) Calculate the current it needs to draw in order to operate.
(b) Determine how much energy would be consumed in 2 hours.
26. A 2.5 kW blow heater is used for eight hours. Calculate the cost of running the blow heater if the electricity is sold at 15 cents per kilowatt-hour.
27. The circuit below refers to the following questions:
(i) Determine the current flowing through R, and the value of resistor R.
(ii) Deduce the reading on the voltmeter V.
28. (i) Describe the meaning of the “12 V” on a 12 volt car battery.
(ii) A 14 V car battery drops to 12 V when supplying a current of 5.0 A. Determine the internal resistance of the battery.
29. A circuit was set up to investigate the relationship between the current I through a resistor and the magnitude of the resistance R while a constant electromotive force was supplied by a dry cell. The results of the investigation are given in the following table: R ± 0.5Ω I ± 0.1 A 2.0 5.0 6.0 1.7 12 0.83 16 0.63 18 0.56
(a) Complete the last column for the inverse of the current giving the correct unit.
(b) Plot a graph of R against 1 / I (c) Describe the relationship that exists
between the resistance and the current. (d) Determine the electromotive force of the
dry cell
30. Starting from the laws of conservation of energy and conservation of charge, derive a formula for calculating the effective resistance of two resistors in parallel.
31. The diagram shows a typical circuit.
1.0 Ω 2.0 Ω 0.5 Ω 1.0 Ω 1.0 Ω A B C D 1.5 V
(a) Determine the effective resistance of the whole circuit.
(b) Determine the currents flowing in each network resistor.
(c) Determine the potential differences VAB and VAD.
(d) Determine the potential difference between B and D.
32.. Determine the resistance of the LDR in theDetermine the resistance of the LDR in the diagram below if a current of 4.5 mA is flowing in the circuit.
9 V
1 kΩ
I 35 A 24.0 Ω 3.0 Ω R V 4.0 Ω 1.0 Ω 100.0 VC
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6
6.1.1 State Newton’s universal law of gravitation.
6.1.2 Define gravitational field strength.
6.1.3 Determine the gravitational field due to one or more point masses.
6.1.4 Derive an expression for gravitational field strength at the surface of a planet, assuming that all its mass is concentrated at its centre.
6.1.5 Solve problems involving gravitational forces and fields.
© IBO 2007
6.1.1 NEWTON’S
UNIVERSALLAWOF
GRAVITATION
I
n the Principia Newton stated his Universal Law ofGravitation as follows:
‘Every material particle in the Universe attracts every other material particle with a force that is directly proportional to the product of the masses of the particles and that is inversely proportional to the square of the distance between them’.
r
F12 F21 m1
m2
Figure 601 Newton’s law of gravitation
We can write this law mathematically as:
F12 Gm1m2
r2
---aˆ12 –F21
= =
F12 is the force that particle 1 exerts on particle 2 and F21 is the force that particle 2 exerts on particle 1.
â12 is a unit vector directed along the line joining the
particles.
m1 and m2 are the masses of the two particles respectively and r is their separation.
G is a constant known as the Universal Gravitational Constant and its accepted present day value is
6.67 ×10–11 N m-2 kg-2.
There are several things to note about this equation. The forces between the particles obey Newton’s third law as discussed in Section 2.7. That is, the forces are equal and opposite. The mass of the particles is in fact their gravitational mass as discussed in Section 2.3.3.