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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.
- Marked by Teachers essays 5
- Peer Reviewed essays 9
Where Q= energy transfer C= specific heat capacity We can rearrange this to give: C= Q/(m(T2-T1)) And as power = energy/time Therefore E= Pt = Q And P = IV therefore Q = IVt Hence C= IVt/(m(T2-T1)) Which is rearranged to the form y=px + c to give: T2= (IVt / (m.C)) + T1. Where p is the gradient, and equals 1/C, therefore x = IVt/m = Q/m, and y = T2 the y intercept is equal to T1 Therefore I have calculated this table: Energy transfer Errors (J)
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The conductivity is usually small, but can be increased if the body or clothing is wet. The risk of injury also increases according to the size of the voltage or current, or the duration of contact. There is a risk of electrocution (death by electric shock) if current passes across the heart. For example, if one foot is touching wet ground, the risk is greater if the arm on the opposite side touches a high-voltage source than it would be if the arm on the same side did so. Current passing into the body generates heat, which burns the tissue.
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However ,if the cell is connected across an external resistor as shown amount of energy per unit charge is wasted in getting through the cell. V = ? - Ir Therefore ,V is less than ?,then not all the energy per unit charge supplied by the cell is transformed in the external circuit into other forms of energy . It implies that a certain amount of energy per unit charge is wasted in getting through the cell. The internal resistor of a cell depends on several factors and is seldom constant .
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If it is a liquid the viscosity will increase, if you decrease the temperature. However in a gas, the viscosity will increase if you increase the temperature. In the production of chocolate the temperature must be average, so that it flows quickly while producing, but must be viscous enough to be moulded to the desired shape. Also the cross sectional area affects the stress required to break it, this will in turn affect the texture.
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A certain portion of the photons from the light that strikes the cell is absorbed and then transferred into the semiconductor material. The electrons then flow freely (DC current), having the energy being shifted in to them. The current produced is directly dependent on the amount of light that reaches the cell. Therefore; the more light energy, the more current and potential difference, and therefore more power. P = I x V The energy given to excite the electrons away has to be greater than the energy given to the electrons if they excite and then relax to return to
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Investigate how the charge on a capacitor is related to the potential difference applied across it by charging the capacitor at a constant rate.
Then, . If a capacitor is charged up at a constant rate, i.e.=I, where I is a constant. Then= is also a constant. Hence, the potential difference across the capacitor increases linearly with time. Data Analysis: The result of this experiment is shown as below: p.d. across capacitor V 1V 2V 3V 4V 5V 1st attempt Time t 5.92s 11.73s 17.67s 23.67s 29.42s 2nd attempt Time t 5.90s 11.30s 16.87s 22.71s 28.59s 3rd attempt Time t 5.19s 10.75s 16.94s 22.85s 28.72s Discussion: When the capacitor is being charged up, the microammeter reading decrease with time, and the CRO trace reading increase constantly.
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The charge on the capacitor is : When the switch is in contact with C, the capacitor discharges through the micro ammeter. This process of charging and discharging occurs times per second. In each second, pulses of charge () flow through the micro ammeter. Hence, the current () through the micro ammeter is given by : Procedures 1. The circuit was connected as shown in the setup of the diagram. The resistance of the variable resistor was set at its maximum value.
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The horizontal component of the magnetic field will be measured. This is not the maximum magnetic field that we want. 8 Why the sensitivity of the search coil can be increased by increasing the frequency? The more frequent the voltage across the wire, the more frequent is the change of the magnetic field. The magnetic field across the search coil is proportional to the sensitivity. 9 Tabulate the length l of vertical trace on CRO and current I measured in the table below: Length of vertical trace on CRO l/cm 1.2 2 3.4 4.6 6.2 7.6 8 Current I/A 0.1 0.2 0.3 0.4 0.5 0.6 0.7 10 Plot a graph of l against I.
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To take the voltage of an object, the shunting resistor must be placed in series, as the resistance of the moving-coil galvanometer is not very high. Some formulae which may be useful during this experiment is: Method The experiment firstly started by looking at calculating the resistance of the voltmeter. This was done by setting the circuit up as below: E = d.c. power supply M = Voltmeter being tested Rm = Voltmeter resistance I = Current The avometer was then placed upon successive ranges of 3v, 10v, 30v, followed by the digital voltmeter on ranges 1v and 10v.
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In a superconductor, the resistance drops exactly to zero when the material is cooled below its critical temperature (the temperature at which electrical resistance is zero). So an electrical current flowing in a superconducting wire can persevere with NO power source. In superconductivity materials, its characteristics appear after it's cooled below the critical temperature (varies between materials).This is mostly between 20Kelvin to 1K, e.g. solid mercury has critical temperature of 4.2K. Metals undergo metallic bonding where they have delocalised electrons allowing them to conduct in the form of heat and electricity.
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to reach particular values is recorded. * The experiment is repeated two more times so as to acquire three sets of time (T1, T2, T3) for the current (I) in question and the average time (t) was also recorded. * Using the average time a graph of lnI against t is plotted using the following equation: Where y= lnI, x= t, gradient = and lastly y- intercept = lnImax * Resistance(R) of uncertainty ?2% is measured using the multi-meter and the Capacitance (C) is found by using the gradient = ? Diagrams: Showing setup of circuits in question. Results: I(?A)
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To investigate the relationship between the power consumed by a torch bulb and the resistance, by measuring the potential difference across the bulb and its current.
The voltmeter is set to read volts and ammeter to read micro-amps. * The variable resistor (Rheostat) is adjusted until the current (I) observed on the ammeter is at its minimum value. * Readings are taken from each multi-meter. The ammeter gives the current (I) whilst the voltmeter gives the p.d across the bulb in question. * The variable resistor is adjusted so that the current (I) increases while the resistance (R) decreases. The readings of twelve (12) other points of equally spaced intervals are then taken so as to have thirteen (13)
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1 / r2 Substituting F for d, d ? 1 / r2 To turn that into an equation and not proportionality, we must insert some sort of constant, d = k / r2 This makes sense. If the objects have a greater charge, and thus have a great excess or deficit of electrons, they will have a stronger force of attraction or repulsion with the other object. Similarly, the closer the objects are together, the more unstable the electrons become and thus tend to increase the stronger force of attraction or repulsion between the objects.
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Two widely spaced dots on the tape were marked- this gave the zero position of the pendulum bob. (4) On the tape every 3rd dot from the zero position was marked off. The displacement of these points from the zero position was measure and the corresponding time was worked out. Time interval between successive dots = 3(1/50) =0.06s (5) The data for time and displacement were entered in columns A and B respectively in a spreadsheet program. The following formulae were entered in the cells flagged: C2= (A2 + A3)/2 the mid-point between times A2 and A3 was found; D2= (B2 +B3)/0.06 the average velocity between displacement B2 and B3 was found; E3= (C2 + C3)/2 the mid-point between times C2 and C3 was found; F3= (A2 + A3)/0.06 the average acceleration between velocities D2 and D3 was found.
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(2) The time base was set to any high value so that a steady horizontal trace is displaced. The trace was set to the bottom of the screen. (3) The capacitor was shorted out by connecting a lead across it and the 100k? potentiometer was adjusted for a suitable current, say, 80�A. (4) The shorting lead was removed and the capacitor would up charged up. Note what happened to the microammeter reading and the CRO trace. (5) The procedure was repeated but this time the stop-watch was started and the potentiometer was adjusted continuously to keep the current constant as the capacitor charged up.
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Tabulate the results. Voltage (V) / V 5 10 15 20 25 30 35 Charge (Q) / ? 10?8 C 0.19 0.36 0.50 0.70 0.93 1.05 1.24 10. Plot a graph of the charge stored (Q) in the parallel plate capacitor against the potential difference (V) across it. B. Factors affecting the capacitance 12. Repeat steps 6 to 8 with different numbers of spacers (n) between the metal plates. Tabulate the results. Find the reciprocal of the number of spacers (1/n). No. of spacers (n) 1 2 3 4 5 6 7 1/n 1.00 0.50 0.33 0.25 0.20 0.17 0.14 Charge (Q)
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Repeat steps 3 and 4 with the other frequencies of the signal generator from 400 Hz to 10 Hz. Tabulate the results. Frequency (f) / Hz 100 150 200 250 300 350 Current (I) / ?A 1200 2000 2500 3200 3600 4300 Note that low switching frequency is advised since for high one, the reed switch may not be sensitive enough to respond, and the time may be too short for complete discharge. 6. Plot a graph of frequency (f) against current (I). Results and Discussion 1.
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Objective To investigate the relationship between the charge on a capacitor and the p.d. across the capacitor by charging at a constant rate.
> Clip component holder > Hand-held stop watch > CRO > Connecting leads Procedure 1. Connect the following circuit. Set the CRO to d.c. and the sensitivity to 1 Vcm-1 2. Set the time base of the CRO to a high sweep rate so that a steady horizontal trace is displayed. Shift the trace to the bottom of the screen. 3. Short out the capacitor by connecting a connecting wire across it (XY). Adjust the 100 k? potentiometer to a suitable value for a steady current to flow (e.g. 80�A). 4.
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Two points B. Point and line C. Two parallel lines Disscusion: 1. The role of the conducting plate The conducting plate has nearly zero resistance. When the plate is connected to the power supply by using electrodes, the plate then conducts electricity and has electric field. It provides a surface for us to test the potential difference within the plate by using flying probe. So the equipotential line could be found easily. 2. Comparing the equipotential line patterns with the electric field patterns in Part A of Experiment C3 Experiment C3 is sprinkling semolina powder onto the non-conducting oil in an electric field.
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The majority of superconductors similar to MgB2 have a critical temperature of -269�C, or about 4 Kelvin. Now, a difference of 35 degrees might not seem to be of much consequence, especially when dealing with such extreme temperatures, but because you're dealing with such low temperatures, those 35 degrees make the difference between using liquid helium (a very difficult and volatile substance to work with) and electrical closed-cycle refrigeration (a much less expensive system3). Diamagnetism is a form of magnetism that is only present when a material is put under an external magnetic field. When a magnetic field is applied to a substance, it changes the orbital motion of the electrons; this change is signified by the
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As the temperature of the filament changes the resistance also changes. Problem Identified I have decided to investigate into the resistance of a filament lamp. I will measure the current and potential difference of the filament lamp at various points, after measuring current and potential difference I will draw a graph to show the relationship between current and potential difference. By finding 1 over gradient of the I/V characteristic graph we can work out the resistance of the filament lamp.
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Experiment to investigate how the resistance of a strain gauge attached to a piece of wood varies with the temperature of the wood.
To measure the temperature of the wooden block, I shall use 2 thermocouple devices , one of which I shall attach to the wooden block by using a substance called araldite 2024, which has a good thermal conductivity and a high melting point. ref1. I will use general purpose thermocouples as they can be used to measure high temperatures and are responsive enough to a temperature of 400�C. The ranges of temperature that I will be using are 0�C to 40 �, as these are the extremes the temperatures are likely to be in an average household.
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Data Collection Conditions: 5 volts, 220,000 ohms, starting at 1800 nC Number of Tapping Charge (nC) � 1 (1st try) Charge (nC) � 1 (2nd try) 1 1644 1721 2 1144 1234 3 837 850 4 597 597 5 406 427 6 326 307 7 236 205 8 168 122 9 119 95 10 84 67 11 56 50 12 39 37 13 24 25 14 16 17 15 10 10 16 6 6 17 5 4 18 3 3 29 2 2 20 0 0 Data Processing Using Logger Pro, we determined the rate of discharge, by drawing a curve of best fit.
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