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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.
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through that circuit defines quantitatively the amount of electrical resistance R. Precisely, R = V/I. Thus, if a 12-volt battery steadily drives a 2-ampere current through a length of wire, the wire has a resistance of 6 volts per ampere, or 6 ohms. Ohm is the common unit of electrical resistance, equivalent to one volt per ampere and represented by the capital Greek letter omega, ?. The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Resistance also depends on the material of the conductor.
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Alternatively a different sensing circuit could be used. A circuit with a thermistor (a component sensing change in temperature) could be used. In normal situations when the window would be fully closed the temperature in the greenhouse would be high so the output voltage from this circuit would be low. However if the windows were to slip open fully then the temperature inside the greenhouse would decrease so the output voltage would increase. The circuit could be linked to an alarm of some type so as when the voltage increases past a certain point when the window is fully open, the alarm rings alerting the gardener to shut the windows again.
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multiplying it by gravity and then the density of steel to calculate the weight of the ball bearing and calculate the downward force on the object. This came about as m=?V and W=mg. Substituting in W= ?Vg. The volume of a sphere = 4/3?r3. Substituting in again, W=4/3?r3?steelg. r= radius of sphere (m) ?steel=density of steel (7.8 g/cm3) g=gravity (9.81m/s2) However, there is more than this acting on the ball bearing. Viscous drag is a force opposing the weight of the object and this is calculated by using Stokes' law.
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Uncertainty of moles Uncertainty of mass = Uncertainty of initial mass + Uncertainty of final mass Uncertainty of mass = 0.005 g + 0.005 g Uncertainty of mass = 0.01 g Percent uncertainty of mass = Percent uncertainty of mass = Percent uncertainty of mass � 1.89% The percent uncertainty for the number of moles is approximately 1.89%. Thus, the number of moles is approximately 0.00834 �1.89%. Current: Average charge = Average current x time After 10 minutes Average current = Average current = 0.9 A Percent uncertainty of average current = Percent uncertainty of average current � 11.1% The percent uncertainty for the average current is approximately 11.1%.
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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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material (measured in ohms, ?); > L is the length of the piece of material (measured in metres, m); > A is the cross-sectional area of the specimen (measured in square metres, m�). From this equation, I can see that the resistivity will be the resistance over length multiplied by the cross-sectional area. I rearranged the equation and substituted into the straight line equation as is shown below. y = mx + c R = ?L /A R?L The value of resistance of the material is depends on the value of the length.
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At any time, Or, , where I0 is the initial current through the resistor. Therefore theoretically, it is known that the decay of charge through constant resistance follows an exponential decay pattern. That is, the discharge rate is always proportional to the charge remained. The time constant reflects the time for the capacitor to discharge. The time required for the capacitor to discharge increases as the time constant increases. In an experiment, the time constant can be estimated by the equations above. Theories about Combination of Capacitors Consider the case when two capacitors are connected in series.
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Measure the diameter in three places, then compare these results. If they are not equal (to within experimental error) measure the wire's diameter more times, to ensure a reliable value for the diameter is found. Then the wire should be attached using sellotape to the metre rule so that it is taught and without kinks, to make measurements as accurate as possible. Set up the experiment as shown in the diagram. Attach one of the crocodile clips at the point marked 5cm by the rule (where its right-hand edge is at the 5cm line) and the other at the point marked 95cm (where its left-hand edge is at the 95cm line)
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Aim of investigation The aim of this work is to investigate the relationship between the resistance and the diameter of the wire. Variables Variable Independent / Controlled / Dependent Resistance D Resistivity C Length of wire C Diameter I Prediction Since the theory suggests that So So the resistance should be inversely proportional to the square of the diameter of the wire. All of these values will be measurable or known, except for the resistivity, . Method Preliminary experiments Determining the size of the wire Use a micrometer to measure the thickness (diameter)
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Another common method is to use combustion turbines to burn oil. In a fossil plant, oil, gas or coal is fired in the boiler, which means that the chemical energy of the fuel is converted into heat. Name of Fuel Calorific Value (J) Cost Coal 15-27 �3.10 per kWh Oil 15 �2.66 per litre Gas (natural) 13.9 �2.78 per litre Nuclear 38-90 �3.40 per litre Wind 47.3 �1.99 per litre Solar 141.7 �12.00 per kWh Geothermal 28-30 �3.05 per kWh Biomass 13 �1.68 per kWh Tidal 15-17 �3.60 per kWh Energy transfer diagram: Sun-->Light Energy-->Photosynthesis-->Plants/Animals-->Fossil Fuels The sun is the starting point for this energy chain, the source of the suns energy if nuclear fusion.
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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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a solar cell. By studying the relationship between external load on current and voltage, the internal resistance of a power supply can be determined. Knowledge gained can be used in designing a circuit that meets or exceed requirements that is required for a space exploration. Figure 1 Internal Resistance In a power supply there are wires and chemical electrolytes and electrodes that make up a cell which all have electrical resistance. Some energy produced by the power supply is transferred to the resistance.
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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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Assuming the capacitor is fully charged and discharged every time, the total charge Q total passing through the milliammeter per second is equal CVf, which is the theoretical current I. And the capacitance of the capacitor can be estimated by the formula C = I/ Vf. Capacitors in parallel If capacitors C1, C2, ..., CN are connected in parallel, the charges stored in each capacitor are shown as below: Q1 = C1V, Q2 =C2V, ..., QN =CNV, where V is the common p.d. across the capacitors. The total charge Q stored in the combination = Q1 + Q2 + ...
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