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Transcript
Some Useful Electrical Circuit Information An Analogy .. the flow of water (e.g. in a central-heating system) Voltage corresponds to Pressure. Electric Charge corresponds to the volume of water. When it starts to flow, it becomes .. Electric Current which corresponds to the Volume Flow Rate of the water. Circuit Principles A current will only flow if it has a complete circuit. Ohm's Law: V = I R, where V = voltage, I = current, and R = resistance. The units are volt (V), ampere (A) and ohm (often spelled out but also often abbreviated to the Greek ) respectively. Note that it follows by Sums that I = V/R and R = V/I. Kirchhoff's Laws: 1. The total current flowing into any junction in a circuit is equal to the total current flowing out of the junction. 2. Round any complete loop in a circuit, the sum of the voltages generated by voltage generators (e.g. batteries or the mains) is equal to the sum of the products of current times resistance for all the resistances in the loop. This circuit should illustrate the idea. I1 3 ohm + 3V - 4 ohm I2 + 5 ohm 4V - I3 The currents I1 and I2 are flowing into the junction at the top, whilst I3 is flowing out of it .. so I1 + I2 = I3. Going round the left-hand loop clockwise, we encounter one voltage (3 V) and two resistances of 3 and 5 ohm respectively .. so 3 = 3I1 + 5I3. Likewise anticlockwise round the right-hand loop .. 4 = 4I2 + 5I3. We can solve the three equations "simultaneously" to obtain I1, I2 and I3 (they tell me it is easy to do so if we use a TI85 or something). We may revisit such situations later on when we are more in practice but we will keep things simpler for the moment. But we can apply Kirchhoff with a minimum of sums to see what will happen in the following cases: 3 ohm I (a ) 4 ohm + - 5 ohm 3V - ohm 5 3V + (b) - + 1.5 V I In Circuit (a), we just have one loop, so we first add up the voltages; an easy task as there is just one of them … 3 V. Then we look at the deadly “IR Drops” and see to our horror that there are three of them, but at least the current I is the same in each (why ?). So we have to add up 3 x I for the 3-ohm, 4 x I for the 4-ohm and 5 x I for the 5-ohm, giving (3 + 4 + 5) x I or 12 I. Mr. (Herr Doktor ?) Kirchhoff says that these two answers are equal, so 12 x I = 3 and I = 3/12 = 0.25 A. We could have got the same answer by just adding up the resistances and using Ohm’s Law (see below under Resistances in Series) In Circuit (b), we have two voltage sources and only one resistance. In this case, we add up the voltages, noting that both are driving the current the same way round the circuit; 1.5 + 3 = 4.5 V. There is only one resistance, so the “sum of IR drops” is just the IR drop, or 5 x I. So 5 x I = 4.5 and I = 4.5/5 = 0.9 A. So it looks as if we can add up the voltages of generators in series in such a situation. Do we know any practical situations in which we might have two voltage sources (e.g. batteries) in series, either the same way round or opposing each other ? Resistances in Series and Parallel Series: R1 R2 R3 Total Resistance = R1 + R2 + R3 Parallel : R1 R2 Total R is calculated from R3 1 1 1 1 R R1 R2 R3 . Power in Electrical Circuits Power = Voltage x Current (applies generally) Power = V2/R = I2R (only applies to resistances) Internal Resistance We find that all practical sources of voltage behave as if they have some resistance in series with their terminals. The practical effect of this (at least partly imaginary) resistance is that the terminal voltage of the source falls off as the current it is supplying increases. A frustrating manifestation is that of a failing car battery which may well be giving 12 V when no current is drawn but, when we operate the starter, we hear an ominous "rr .. rr .. rr .." instead of "rrrrrrrVROOM". We can represent such a practical voltage source by an ideal one in series with the resistance (an Equivalent Circuit). Electromotive force (e.m.f.) E Internal resistance R First Electrical Examples Sheet 1. State and explain Ohm's Law and Kirchhoff's laws. 2. Complete the following table. V, volt 10 240 ? 15 110 ? I, ampere 2.5 ? 3 3 ? 17 R, ohm ? 48 20 ? 5.5 60 What is the power in each of the above cases ? (25, 1200, 180, 45, 2200, 17340 w) 3. What is the total resistance of three 30-ohm light-bulbs if they are connected (a) in series, (b) in parallel ? (90 ohm, 10 ohm) 4. If the above lamps are intended to work at 24 volt, what will be their power rating? If we connected some of them in series to a 240-volt power supply, how many would we have to connect in series if their correct current was to flow ? (19.2 watt, 10 lamps) 5. If we connect two 60-watt lamps and a 1-kw electric fire in parallel across the 240volt mains, what total current will be drawn ? (4.67 A) 6. 2 ohm + 12 ohm 20 volt - 24 ohm Determine the current in each resistor. (Hint: work out the parallel resistance of the 12- and 24-ohm combination and use it to work out the total series resistance and therefore the current in the 2-ohm resistance. Then continue to use Ohm's Law again to calculate the voltage across the parallel combination .. and therefore the currents) (2 A, 1.333 A, 0.667A) 7. This circuit represents a system in which a light at a remote road junction can be supplied either by a 24-volt battery of internal resistance 4 ohm, or by a wind generator of internal resistance 2 ohm which generates a voltage of 2u volt on opencircuit where u is the wind speed in km/hr. The generator also serves to keep the battery charged. (a) If the wind generator were disconnected, what voltage would be supplied to the lamp ? (20 V) (b) At what wind speed would the generator begin to supply any current to the lamp ? (10 km/hr) (c) At what wind speed would the wind generator begin to charge the battery ? (13.2 km/hr) (d) As it stands, the circuit will suffer from a serious problem when the wind is light. What is the problem and what can be done about it ? Wind Generator 2 ohm + - 2u volt 4 ohm Light .. resistance 20 ohm Battery + 24 - volt 8. A 12-V car battery supplies two 6-w rear light bulbs and two 30-w headlamp bulbs. Sketch the circuit and calculate the total current drawn from the battery. What would the current and battery terminal voltage be if the battery had an internal resistance of 0.5 ohm ? (6 A, 4.8 A, 9.6 V)