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### 5.2 2012

1. 1. Topic 5.2 Electric Circuits
2. 2.  Defining potential difference The coulombs entering a lamp have electrical potential energy; those leaving have very little potential energy. There is a potential difference (or p.d.) across the lamp, because the potential energy of each coulomb has been transferred to heat and light within the lamp. p.d. is measured in volts (V) and is often called voltage.
3. 3.  The p.d. between two points is the electrical potential energy transferred to other forms, per coulomb of charge that passes between the two points.
4. 4.  Resistors and bulbs transfer electrical energy to other forms, but which components provide electrical energy? A dry cell, a dynamo and a solar cell are some examples. Any component that supplies electrical energy is a source of electromotive force or e.m.f. It is measured in volts. The e.m.f. of a dry cell is 1.5 V, that of a car battery is 12 V
5. 5.  A battery transfers chemical energy to electrical energy, so that as each coulomb moves through the battery it gains electrical potential energy. The greater the e.m.f. of a source, the more energy is transferred per coulomb. In fact: The e.m.f of a source is the electrical potential energy transferred from other forms, per coulomb of charge that passes through the source. Compare this definition with the definition of p.d. and make sure you know the difference between them.
6. 6.  The cell gives 1.5 joules of electrical energy to each coulomb that passes through it, but the electrical energy transferred in the resistor is less than 1.5 joules per coulomb and can vary. The circuit seems to be losing energy - can you think where?
7. 7.  The cell itself has some resistance, its internal resistance. Each coulomb gains energy as it travels through the cell, but some of this energy is wasted or `lost as the coulombs move against the resistance of the cell itself. So, the energy delivered by each coulomb to the circuit is less than the energy supplied to each coulomb by the cell.
8. 8.  Very often the internal resistance is small and can be ignored. Dry cells, however, have a significant internal resistance. This is why a battery can become hot when supplying electric current. The wasted energy is dissipated as heat.
9. 9.  The diagram shows three resistors connected in series There are 3 facts that you should know for a series circuit:  the current through each resistor in series is the same  the total p.d., V across the resistors is the sum of the p.d.s across the separate resistors, so: V = Vl + V2 + V3  the combined resistance R in the circuit is the sum of the separate resistors
10. 10.  R = Rl + R2 + R3 Suppose we replace the 3 resistors with one resistor R that will take the same current I when the same p.d. V is placed across it
11. 11.  This is shown in the diagram. Lets calculate R. We know that for the resistors in series:  V = Vl + V2 + V3 But for any resistor: p.d. = current x resistance (V = I R). If we apply this to each of our resistors, and remember that the current through each resistor is the same and equal to I, we get: IR = IRl+IR2+IR3 If we now divide each term in the equation by I, we get:  R = R1 + R2 + R3
12. 12.  When one component fails the whole circuit fails. The current is the same at all points and the voltage is divided between the bulbs. The more bulbs added the dimmer each one is.
13. 13.  We now have three resistors connected in parallel: There are 3 facts that you should know for a parallel circuit:  the p.d. across each resistor in parallel is the same  the current in the main circuit is the sum of the currents in each of the parallel branches, so:  I = I1 + I2 + I3  the combined resistance R is calculated from the equation:
14. 14.  Suppose we replace the 3 resistors with one resistor R that takes the same total current I when the same p.d. V is placed across it.
15. 15.  This is shown in the diagram. Now lets calculate R. We know that for the resistors in parallel: I = I1+I2+I3 But for any resistor, current = p.d. = resistance (I = V/R ). If we apply this to each of our resistors, and remember that the p.d. across each resistor is the same and equal to V, we get:V/R=V/R1 + V/R2 + V/R3 Now we divide each term by V, to get: 1/R=1/R1 + 1/R2 + 1/R3
16. 16.  You will find that the total resistance R is always less than the smallest resistance in the parallel combination.
17. 17.  When one bulb fails the rest of the circuit continues to work. The more components, the lower the resistance. The total current drawn increases. Voltage in each branch is the same as the supply voltage therefore bulbs in parallel will each be as bright as a single bulb.
18. 18.  You need to be able to recognize and use the accepted circuit symbols included in the Physics Data Booklet
19. 19.  In order to measure the current, an ammeter is placed in series, in the circuit. What effect might this have on the size of the current? The ideal ammeter has zero resistance, so that placing it in the circuit does not make the current smaller. Real ammeters do have very small resistances - around 0.01 Ω.
20. 20.  A voltmeter is connected in parallel with a component, in order to measure the p.d. across it. Why can this increase the current in the circuit? Since the voltmeter is in parallel with the component, their combined resistance is less than the components resistance. The ideal voltmeter has infinite resistance and takes no current. Digital voltmeters have very high resistances, around 10 MΩ, and so they have little effect on the circuit they are placed in.
21. 21.  A potential divider is a device or a circuit that uses two (or more) resistors or a variable resistor (potentiometer) to provide a fraction of the available voltage (p.d.) from the supply.
22. 22.  The p.d. from the supply is divided across the resistors in direct proportion to their individual resistances.
23. 23.  Take the fixed resistance circuit - this is a series circuit therefore the current in the same at all points. Isupply = I1 = I2 Where I1 = current through R1 I2 = current through R2
24. 24.  Using Ohm’s Law
25. 25.  A thermistor is a device which will usually decrease in resistance with increasing temperature. A light dependent resistor, LDR, will decrease in resistance with increasing light intensity. (Light Decreases its Resistance).
26. 26.  Calculate the readings on the meters shown below when the thermistor has a resistance of a) 1 kW (warm conditions) and b) 16 kW. (cold conditions)