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Simple Circuit Analysis |
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Basic ideas to grasp before you look at circuit diagrams: Voltage (V ) is an electric potential difference. It is measured in volts (V) with a voltmeter connected in parallel across the points in the circuit you wish to compare. If there is a potential difference across two points in the circuit current will flow between them. Charges ‘fall’ from high electric potential to low electric potential. A power supply therefore provides a ‘slope’ or potential gradient down which the charged objects (electrons or ions) will flow.
Resistance (R) is a measure of the reluctance of the conductor to allow the charges to move through it. It is measured in ohms ( Good conductors have loosely held outer shell electrons that they are 'happy' to allow to move away from the parent atom - they have low resistivity making their resistance lower than that of an insulator of the same dimensions. Insulators hold on tightly to their electrons and do not let them wander therefore there are no free electrons to carry the charge and the resistivity of the material is high. The structure of the atom the material is made of therefore has a big effect on the resistance of the material. A ‘wider’ wire (a wire of bigger diameter) has more electrons moving down the potential gradient provided by the power supply, therefore a wire of bigger diameter will allow a bigger current to flow through it if a given voltage is put across it. This makes its resistance smaller. The average drift velocity of the electrons does not change (the ‘slope’ is still the same steepness) – it is the increase in number moving that alters the current. A longer wire has less volts per metre as the volts are shared out across more wire – the potential gradient is therefore not as steep and the average drift velocity of the electrons will be less making the current smaller. This means that a longer wire has a bigger resistance. The number moving in a given length of the wire is the same – it is the change in their drift velocity that changes the current.
A useful shortcut - if you have N identical resistors of value R Ammeters have very low resistances. (They are made by connecting a low resistance shunt in parallel with a galvanometer to give them a very low resistance). They can therefore be connected in series with a component in a circuit without changing the resistance on that strand of the circuit by very much at all. Voltmeters have a very high resistance, (They are made by connecting a high resistance shunt in series with a galvanometer to give them that very high resistance). They can therefore be connected in parallel with a component in a circuit without changing the resistance of the circuit by very much at all. Step 1: Split it into strands A ‘strand’ is a route to/from the power supply that does not branch off midway. Therefore if you have parallel components in series with other components you have to simplify that arrangement before you can do this step. In the above example: Strand 1 has two resistors in series they will therefore share the voltage drop from the supply. Strand 2 only has one resistor it will therefore get the entire voltage drop from the supply. Strand 3 poses a problem. The equivalent resistance of the two in parallel has to be found first. Once this has been done the strand will be like strand 1 – two in series.
Each strand has access to the full potential difference provided by the power supply. Whatever a voltmeter around the power supply reads is what a voltmeter around the components on that strand will read. We say that the potential or voltage drop across the components is the same as the potential drop across the power supply. That means that the whole of strand 1 (the one with R 1 in it) will have the same voltage across it as the battery does – the reading on the voltmeter. The ammeter and resistor together share the potential drop from the battery. The same is true for each of the three strands in our circuit (from above). Rather than draw in lots of voltmeters get into the habit of putting arrows across each strand to show how the voltage is distributed. It is a good idea to have a marker pen (not green or red!) with you in the exam to do this when you have circuit diagrams to deal with. Strand 1 is easy – two equal resistors so they get half each. Strand 2 is easy too, as the entire potential drop is across one resistor. Strand 3 is a bit trickier. V is shared across 1.5R, therefore each R gets V / 1.5 = 2 / 3V . That makes R get 2 / 3V and R / 2 get 1 / 3V Why ammeter voltage drops can be ignored The resistance of an ammeter is negligible. Therefore the voltage share that it gets when it is in a strand is negligible also and generally we can ignore the voltage drop that an ammeter would get and assume that for all intents and purposes it is zero. This is not really the case. It does have a resistance and therefore does take some of the voltage drop but if your voltmeter can only be read to three or four significant figures (which is usually the case) it will not record that difference and read zero when placed in a circuit. Let’s do a calculation to work out what that voltage value would be. Suppose an ammeter has a resistance of 6.3 m
Total resistance of the strand is 420m Therefore the total resistance of the strand = 0.420 The voltage is shared out across the strand according to the resistance values… each ohm gets the same voltage drop. So, 0.420 006 3 therefore each ohm gets 9 / 0.420 006 3 V and the ammeter gets 0.000 006 3 x 9 / 0.420 006 3 V = 0.000 14 V This value is so low it would not show up on the multimeters we use in class. The voltmeter would read zero and we can therefore ignore the ammeters in the circuit. Step 3: Calculate the current or 'heat dissipation' (power output as heat) in a component Now that we have the value of each resistance and the p.d. across each one it is easy to work out the current passing through a component. We just use V =I R. To calculate the heat dissipated in a component Whenever a current passes through a component electrical energy is changed into heat energy. The dissipation of energy as heat is calculated using P = IV where P is power (energy in unit time – W - watts (J/s - joules per second). Using V = I R we can substitute into this equation:
P = IV but V = I R so, P = I(IR) = I 2 R Also I = V/R so, P = (V/R) V = V 2/R So, P = IV = I2 R = V 2/R When to ignore the current through a voltmeter The total current that flows through a parallel arrangement depends on the voltage drop across it. Voltmeters have very high resistance. They make very little difference to the total resistance of the arrangement. Let us demonstrate this with a calculation: Three resistors in parallel (2.0
1 /R TOTAL = 1/2 + 1/5 + 1/20 = 0.75 So, R TOTAL = 1/0.75 = 1.3 ( or 4/ 3) (2 s.f. because the resistances were given to that standard) (Note that this is smaller than the smallest one in the arrangement! This is always the case – useful tip!!) If we now add a voltmeter of resistance 3 000
1 /R TOTAL = 1/2 + 1/5 + 1/20 + 1/3000 = 0.75033 (working to so many figures purely to illustrate the point!) So, R TOTAL = 1/0.75033 = 1.3 The voltmeter does not interfere in a measurable manner with the resistance of the arrangement. You would need very sensitive meters to notice the difference. The higher the resistance of the voltmeter the less it interferes with the circuit it is measuring potential differences in.
The current that passes through the arrangement will be greatest in the resistor of lowest value. Let us suppose that a voltage of 6.0 V was applied across our arrangement. We can work out the current through the whole arrangement using V = IR:
V = IR I = V / R = 6.0 / 1.33 (here a calculated value is put in at one more sig fig than we must quote to – necessary for accuracy – always work to one more figure!)_ = 4.5 A This current is split into the four branches of the circuit. Using the values of each of the resistors in turn with the p.d. across them we can work out how much current goes into each strand.
V = IR I= V / R = 6.0 / 2.0 = 3.0 A
V = I R I = V / R = 6.0 / 5.0 = 1.2 A
V = IR I= V / R = 6.0 / 20 = 0.30 A
V =IR I = V / R = 6.0 / 3000 = 0.002 A Sum of currents = 0.002 A + 0.30A + 1.2 A + 3.0 A = 4.502 A = 4.5 A (2 sig figs) – we can ignore the current draw by the voltmeter as it is so small The only time a voltmeter will interfere with a circuit is if it is used with resistors of large values similar to that of its own resistance. You should then find out how much current it does draw from the power source as it will be similar to that of the resistors and it will interfere with the resistance of the resistor within the circuit. Always read the question carefully. If it says a ‘high resistance voltmeter…’ it means ignore the current drawn by it and assume it does not affect the resistance of the circuit. BUT if it gives you the resistance you need to do calculations with that resistance to find out what it reads.
Because the resistance of the voltmeter is ‘high’ (compared to the resistances in the circuit) we can ignore it and confidently say that the reading on it will be 3.0V. The two parallel 30 BUT if the value of R was 3000 As a rule of thumb: if the resistance of the meter is more than 100 times that of the resistors then ignore its affect on the circuit. If it is less than that do a calculation.
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When components are connected in series their resistances are added to give a sum total. 
A useful fact - when components are connected in parallel the resistance of the whole parallel arrangement is always smaller than the resistance of the lowest value strand of the arrangement. 
is equivalent to a single resistor of R / 2 (see the ‘tip’ above) we can therefore re-sketch the circuit or (to save time in an exam!) mark on it in such a way as to indicate this. 
Pure parallel components can be split into separate routes back to the power supply; therefore they each become strands of the circuit. You need to simplify resistor arrangements until you have a simple parallel circuit. 
In this circuit let us suppose the value of V is 9 volts, the value of R is 30