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AS and A Level: Electrical & Thermal Physics
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Doing circuit calculations
To find the total resistance of a circuit follow these steps.
1) Replace any parallel network with a single equivalent resistor, REQ using 1/REQ= 1/R1 + 1/R2.
Tip: REQ will be lower than either of the parallel resistors R1 or R2 so you can check your calculation.
2) Add all of the series resistors together (including REQ) to find the total resistance of the circuit RT.
- 2 Calculate the total circuit current, IT using IT = V/RT. This current flows through all of the series resistors so the p.d. across each series resistor is given by V = IT R. The p.d. across any parallel network will be IT REQ.
- 3 A potential divider circuit consists of two resistors in series. Follow the same steps as above to find the p.d. across each resistor. Alternatively, R1/R2 =V1/V2 or V1 = V *R1/(R1 +R2) [V = supply voltage]
Which bulb is brightest?
1) If two bulbs are in series, they have the same current. The brighter bulb is the one with greatest power, P. Use P = I2R. The bulb with largest R is brightest.
2) If two bulbs are in parallel, they have the same p.d. across them. Use P=V2/R. The bulb with the lowest R has the highest power and is therefore brightest.
- 1 Use the correct units. If diameter is given in mm, convert to metres before calculating area, A. e.g. d = 1mm so r = 0.5mm = 0.5 x 10-3 m. So A = x (0.5 x 10-3)2 = 7.9 x 10-7 m2.
- 2 Typical questions involve proportions such as what happens to R if the diameter of the wire is doubled? Doubling the diameter would double the radius. Doubling the radius would quadruple the area. So the resistance would decrease to ¼ of the original resistance. The same argument explains why a thinner wire has a higher resistance.
Applications of resistivity:
1) A rheostat is a resistor made by winding a wire around a cylindrical tube. A sliding contact changes the length of the wire carrying current and therefore changes the resistance, R.
2) A strain gauge, has a resistance that increases when it is stretched because the wire from which it is made increases in length.
3) The battery tester on the side of some AA batteries works by using a shaped conductor. The thin end has lowest A, therefore highest R. Current is the same at all points, the thin end gets hottest (P = I2R) and a thermochromic ink becomes transparent, revealing a display.
- 1 Many students find internal resistance a difficult concept. However the circuit is similar to a potential divider. Think of the circuit as a cell of emf E, in series with an internal resistance, r and an external resistance R. When current, I flows through the circuit, E = Ir + IR. This is Kirchhoff’s 2nd law.
- 2 Using a voltmeter to measure the terminal p.d. V, we can rewrite the equation E = Ir + IR as E = Ir + V and then rearrange to give V = rI + E which is the equation of a straight line. A graph of V against I gives a straight line of gradient -r and intercept E. This is how to find the emf experimentally.
- 3 When the current through the cell is high, there is a large drop in the terminal p.d. The difference between the cell emf and the terminal p.d. is called the ‘lost volts’ and equals Ir.
- 4 Short circuiting the cell will lead to a large drop in external voltage and large amount of power dissipated in the cell as P = I2r.
- 5 A car battery (lead acid) is designed to supply large currents. When switching on the engine the current is large and there will be a large drop in terminal p.d. and this will cause lights to dim momentarily.
three graphs above it can be seen in all of them that there is a firm relationship between the two factors, one being voltage and second one being current. Therefore I have accomplished my aim, as I have found what is the relationship between the current through a resistor and the voltage across it. The relationship as can be seen from the graph is positive, and thus directly proportional. This can be seen from the graph itself, because firstly, the best-fit line is a straight line going through all the points, in a positive direction.
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The offender could also face which I think is a very harsh five-year prison sentence and all for what could have been an innocent kiss with your partner. A recent national survey Shows the while the current laws aren't perfect they aren't doing a bad job as it showed us that 33% of boys have had sex under the age of 16 and that only 25% of girl have, and this shows that the current laws are keeping at least two-thirds of boys from having illegal sex and three-quarters of girls and means that a larger percentage of teenagers are obeying the current laws and waiting until they have sex.
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This causes the charge circulating in the circuit to dissipate some electrical energy in the power supply itself. The power supply becomes warm when it delivers a current. Diag 3: The circuit above shows a power supply which has e.m.f (E) and internal resistance (r). It delivers a current I when connected to an external resistor of resistance R, also called the load. Vr is the potential difference across the internal resistance. Kirchoffs second law states: E = Vr + VR The potential difference VR across the load is thus given by VR = E - Vr VR is called the terminal potential difference.
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The primary coil will be wound around a laminated soft iron core (the lamination prevents eddy currents from developing) and this will provide the link between the primary and secondary coil. The secondary coil will have a number of turns between 1 and 50 but I will alter the number of turns in order to investigate the change in the voltage, I will record both the number of turns and also the voltage output as read by a digital voltmeter. Fair Test S- I will keep the input supply and the number of turns on the supply the same A- I will alter the number of turns of wire on the secondary coil M- I will measure the output voltage from the second coil.
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Variables The variables are as follows: Independent variables: * The water temp: This will be allowed to change and measured using a Thermometer reading to 1 degree. * The volume of the water: This will be kept constant. Any loss by evaporation will be assumed too small to effect results. * The actual thermistor: Different thermistors may have different responses so the same one will be used for all readings. * The thermometer: Different thermometers may have different responses so the same one will be used for all readings.
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I carried out the plan using the following apparatus Light gate Scalectrix track Scalectrix car Variable resistor Ammeter Crocodile clips Wire Solder and soldering iron Wire clippers Clamps Stand I will set up my apparatus as shown in the diagram below What I am going to measure, count and observe I'm going to measure, the current using an ammeter and the speed of the car at three points using an electronic timer. I also measured the distances of the light sensors from the start.
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Investigate the relationship between electromagnet strength and amount of current flowing through the wire.
Then, I tried to measure the load as accurately as possible. However, this will not be fair as there is a huge error range because I only used 50g and 100g weights. This means that a measurement of weight can only be to an accuracy of 50g. This is highly inaccurate and so made the test unfair. There were many safety aspects while carrying out this experiment. I had to take many precautions to ensure that I did not get hurt, nothing got damaged and the experiment remained a fair test.
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I also predict my graph will look like this: I don't think I will get results which are directly proportional (This is when one of the quantities is doubled, and the other quantity also doubles. E.g. If Resistance is doubled, length would also be doubled). I also don't think I will get results which are inversely proportional (This is when one of the quantities is doubled, the other quantity halves. E.g. If length is doubled, the resistance would half). Instead, I think I would get results which are neither of these but have a linear relationship (i.e.
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This is because the filament lamp gives out so much of light and heat. This one reason that graph of a filament lamp is curved. The other reason that graph is curved is because the steeper the gradient is the more lights the gives out. So at 12 volts I believe that it produce very strong light. I predict that my graph will be curved because this is based on two theories. First the temperature will not be constant. Secondly the resistant of the filament lamp will not be constant. If the voltage was to be increased the graph would be come more curved and the gradient will be positive and it will be steeper.
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As the number of free electrons increase, resistance decreases. Ohmic Conductors Metallic Conductors A metal contains a large number of almost 'free' electrons. This property makes metals extremely good conductors of electricity. But as the temperature of the metal increases the ions in the metal vibrate vigorously, colliding with the free flowing electrons, obstructing the flow of electricity, thus as temperature increases, resistance increases as well. Metallic conductors are Ohmic Conductors. In ohmic conductors, Current is always proportional to the voltage.
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Also, these sources of energy give endless possibilities of how and where electrolysis can be used. Hypothetically, electrolysis can quite possibly be used to fuel industry, homes, and even create fuel for automobiles during long trips. Depending on the size and efficiency for which this technology allows, electrolysis can be used quite extensively. Electrolysis is currently being used to produce extremely pure Hydrogen for use by the national space program (NASA) and by the private sector in the pharmaceutical, electronics, and food industry (DOE 1995). NASA uses electrolysis to produce hydrogen for use on the space shuttles. The hydrogen is used to fuel the rocket boosters as well as power the ships internal systems, fuel cells.
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When the current is 2 amps, the flow of charge is two coulombs per second. Current (I) = charge flowing (Q) time (t) I = Q/t Voltage In an electrical circuit, the battery provides the electrons with electrical potential energy. It turns chemical energy from the materials in the cell, into electrical energy, in the electrons. The electrons move through the circuit from the negative terminal where they have high electrical potential energy towards the positive terminal. When the electrons reach a bulb, they lose some of their electrical potential energy. This lost energy is then turned into heat and light. Finally, the electrons return to the positive terminal of the battery with less energy.
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This meant that now we could see how many degrees movement gives a certain out put. There are a number of things which we can investigate about the potentiometer: Resolution - the smallest change that can be detected; Sensitivity - the ratio of change in input to change in output; Response time - time taken for the sensor to respond to a change in input; Systematic error - whether there is a zero error or not. To calculate the resistance we will use the formula R=V/I this means that we have to be able to take readings for the current and potential difference.
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and longer fusing time. Cartridge fuses Fuse wire is enclosed in an inert environment. Once burnt, the fuse has to be replaced i.e. a repetitive cost when replacing. High Rupture Capacity (HRC) fuses These are used for high current protection. These have a ceramic outer barrel. The fuse wire is covered with sand in order to quench the arc produced when the fuse burns. The ceramic barrel can withstand the shock of the interruption of the high current. Usually semi-enclosed rewireable fuses have about 1.7-2 times the rated current carrying capacity.
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The aim of this experiment is to investigate the relationship between the current, voltage and resistance through the use of a fixed resistor and a filament lamp.
Safety Precautions Bags stools and other possible obstructions were removed from gangway. Worktops were cleaned and dried to reduce risk of electrocution and prevent damage or interference to experimental equipment. Standard lab safety was enforced. In order to keep errors to minimum and to ensure fair test, equipment such as wires, voltmeters, ammeters, resistors, PSU's and filament lamps remained the same throughout the experiment. Method The work surface was ensured to be dry and clean beforehand. Bags, stools and other possible obstacles were removed from the work area.
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To see the effect of volume current on the effectiveness of an immersion heater heating 75 cm3 of water.
Two readings were taken for each set of results. Predictions I predict that the current in the circuit will be proportional to the temperature change i.e. as the current is increased the temperature change will increase as well. This is because if the current increases there will therefore be more heat energy lost from the immersion heater. Results Reading Vol. Of H20 Current (A) Voltage (V) Temp. change 1 Temp. change 2 1 75cm3 1.42 3.77 2oC 3oC 2 75cm3 1.76 4.66 10oC 4oC 3 75cm3 2.16 5.67 6oC 8oC 4 75cm3 2.51 6.62 7oC 8oC 5 75cm3 3.17 8.32 17oC 12oC 6 75cm3 3.84 9.76 23oC 22oC Analysis.
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We obviously need a bit of both, but the two extremes do show us where we need to concentrate. * We need to produce a move generator that only produces good moves; suicidal moves and plain silly moves will get us nowhere. * We need to develop a technique for searching through the resulting possible moves that picks out the most likely helpful moves first and tests these before going onto less beneficial moves - it needs to be able to recognize a good move when it sees one! If we can develop a good move generator then that will help us generate a better evaluation function.
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110.00 173.33 70 13.8 9.4 62.86 134.29 197.14 80 18.8 14.0 60.00 175.00 235.00 90 24.6 19.0 62.22 211.11 273.33 100 31.0 25.6 54.00 256.00 310.00 110 40.0 30.8 83.64 280.00 363.64 120 49.0 40.4 71.67 336.67 408.33 130 60.0 50.3 74.62 386.92 461.54 140 70.0 60.2 70.00 430.00 500.00 150 80.0 70.1 66.00 467.33 533.33 160 90.0 80.0 62.50 500.00 562.50 Load line for the series circuit : VS = VL + I � RS -> VL = -IRS + VS The voltage drop across the lamp can be computed graphically by drawing the load line for the circuit and the V-I curve for the lamp in the same graph and taking the intersection point voltage.
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Also if a different box was used it may be a different colour or shade so would absorb or reflect a different amount of light from the light source, meaning the photodiode would detect more or less light. To limit this interference of light absorption or reflection of light, the same box would have to be used throughout. After discussing my equipment list with the technicians, they do not have a completely opaque, light stopping box, but do have a spare dark room, or an opaque, telescopic black tube in which a light source can be placed.
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I will vary the current over a range of 5 Amps (0 - 5A). I will test the strength of the magnetic field by placing an iron nail at the bottom of the iron core, it will become magnetised and depending on the strength of the field will stick there. I will then add weights to the nail and record how much mass it will hold before it drops off the magnet. Apparatus * Stand and clamp * Power supply * Iron core * Iron nail * Weights (5g, 10g, 100g)
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The Battery The major component in the electrical system of the car is the battery. The battery is used to store power for starting, and for running accessories such as clocks, radios and alarms when the engine is off. The battery allows every component of its electrical circuits able to utilise 12 volts or electricity. A car's battery is designed to provide a very large amount of current for a short period of time. This large amount of current is needed to turn the engine during starting.
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For RTh t1 and 2 I put in 5 and 105, as an RS data sheet showed these to be the values at 100 and 0 degrees. This graph suggests to me that using a fixed resistance of about 25ohms would produce the highest value of DVo, and therefore would produce the most sensitive readings. However, I suspected that the resistance of the fixed resistor should be roughly equal to the thermistors resistance (0.47Kohms), so I used the internet, and found that the ????
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The gauge factor of a strain gauge (G) = (?R/R)/(?l/l) where R = resistance and l = length. Since ?l/l is the strain (e) in the object which the gauge is attached to this can be written as ?R/R = eG, which means that the fractional change in resistance of the gauge is proportional to the strain in the object. To measure the change in resistance I will set up the strain gauge in a wheatstone bridge, the simplest version of which is shown below: Where R1 is the resistance of the unstrained gauge. The voltmeter gives the voltage difference between the gauge and the resistor between B and D (Vo).
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If the dopant is phopherus for example, there will be 5 valence electrons, and a negative (n-type) of semiconductor is created. However, where aluminium or indium is used, there are only 3 valence electrons, and the semiconductor will be positive (p-type). This second type has 'holes' where the missing electrons should be. These holes behave similarly to electrons, but have an opposite - positive - charge. This theory remains theory, as practically we cannot create a positive charge out of nothing, but it fits the explanation of the PV cell. When light strikes a Photovoltaic Cell, atoms are bombarded with photons and they give up electrons.
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By increasing the depth of the electrolyte it increases the surface area or the electrodes. By increasing the surface area the rate of reaction has increased, as there is a greater chance of a collision between the electrons and the electrodes. If I double the depth of the electrolyte, which is doubling the surface area, the current will double. Key Factors: Factors that will affect my experiment are depth, distance between electrolytes, voltage, and size of electrodes, concentration and temperature. My only variable will be the electrodes. I will keep the same all the others.
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