11 Internal resistance, series and parallel circuits, and the potential divider Exam practice questions

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1 Pages Start by working out the resistance of each of the combinations: W: In series R = R + R 2 + R 3 = 5 Ω + 5 Ω + 5 Ω = 45 Ω X: Start by adding the two series resistors: 5 Ω + 5 Ω = 30 Ω Then combine this with the parallel resistor: = + = + = +2 = 3 R R R 2 R 30 Ω 5 Ω 30 Ω 30 Ω R = 0 Ω Y: Combining the two parallel resistors: = + = + = 2 R R R 2 R 5 Ω 5 Ω 5 Ω R = 7.5 Ω Then adding the 5 Ω in series gives 5 Ω Ω = 22.5 Ω Z: Combining the three parallel resistors: = + + = + + = 3 R = 5 Ω R R R 2 R 3 R 5 Ω 5 Ω 5 Ω 5 Ω We can now see that Z = W/9 the answer is A. [Total Mark] 2 W = 2Y the answer is D. [Total Mark] 3 X = 2 Z the answer is D. [Total Mark] 4 a) Each arm has two 4.7 Ω resistors in series, giving 9.4 Ω in total for each arm. These arms are in parallel. [] So: = + = + = 2 R R R 2 R 9.4 Ω 9.4 Ω 9.4 Ω R = 4.7 Ω [2] b) This network of four resistors has a resistance equal to that of each of the single resistors. It might be preferable to use this network rather than a single resistor as the current will split, with half going through each arm. This means that the current in each resistor is only half what it would have been for a single resistor. As the power depends on the square of the current (P = I 2 R), the power in each resistor will be a quarter of what it would be in a single resistor. This means that the resistors will not heat as much. [2] Note that the total power developed in the network is the same as for a single resistor it is shared equally by the four resistors. [Total 5 Marks]

2 5 a) For the bulb: i) R = V = 6 V I 0.5 A ii) P = IV = 0.5 A 6 V = 3 W [2] b) As the battery consists of four cells in series: Total internal resistance r = Ω = 0.60 Ω [] I = e.m.f. = 6.0 V = 6.0 V Total circuit resistance ( ) Ω (2.60) Ω = A [2] V = ε Ir = 6.0 V (0.476 A 0.60 Ω) = ( ) V = 5.7 V 5.7 V [2] c) Power wasted = power developed in the internal resistances P = I 2 r = (0.476 A) Ω = 0.4 W (40 mw) [2] d) The original 3 W of power dissipated in the filament is only reduced by 0.4 W as a result of the increase in internal resistance. This will not significantly reduce the temperature of the filament so, although the resistance of the filament will decrease slightly, the assumption that it remains constant is not unreasonable. [2] [Total Marks] 6 If the battery is short-circuited, the only resistance will be the internal resistance of the battery. Using I = V R : I = 9 V 0.50 Ω = 8 A This is a large current and will generate a power of P = I 2 R = (8A) Ω = 62 W inside the battery, which will make the battery hot. [2] [2] [Total 4 Marks] 7 a) A 50 MΩ resistor in series with the output will keep the current very small. As the severity of an electric shock depends on the current passing through your body to earth, the resistor acts as a safety device. [2] b) From I = V R the maximum current will be: I = 5 03 V = Ω 0 4 A = 0.0 ma [2] c) The total resistance between the terminal of the supply and the ground will be 50 MΩ + 0 kω. As 0 kω is only 0.0 MΩ, the girl s resistance has very little effect and so the current in the girl would be virtually the same as in part b). [2]

3 d) Using P = I 2 R, the power dissipated in the girl would be: P = ( A) Ω = W = 0.0 mw This is considerably less than the 4 mw from the car battery, and so shows the effectiveness of the 50 MΩ resistor as a safety device. [3] Warning: high voltage supplies, even when safety protected, are still very dangerous and must be treated with the utmost caution. [Total 9 Marks] 8 The table is completed by adding values for the resistance and power, for example V =.44 V and I = 0.20 A gives: R = V I =.44 V 0.20 ma = 7.2 Ω and P = VI =.44V x 0.20A = 0.29W I/A I/A V/V R/Ω P/W Note the the maximum power generated in the load resistor is when its resistance is equal to the internal resistance of the cell (0.60 ohm) [Total 7 Marks]

4 9 a) I/mA V/V R/kΩ P/mW [3] b) i) For small currents the graph, as shown above, is linear, indicating a constant internal resistance. For larger currents the graph clearly curves downwards, showing that the internal resistance increases as the current gets larger. ii) The e.m.f. of the cell will be the voltage when the current is zero, i.e. the intercept on the voltage axis. From the graph this is 6.5 V. The internal resistance for low current values is given by the numerical value of the gradient of the linear part of the graph. Extending the linear part to give a large triangle: r = ( ) V ( ) ma = 3.7 kω [9]

5 c) i) ii) From the graph, the maximum power is approximately 2. mw. [4] [Total 6 Marks] 0 a) i) Combining the 22 Ω and 33 Ω parallel resistors: R = + R R 2 R = 22 Ω + 33 Ω = Ω = 5 66 Ω R = 66 Ω 5 = 3.2 Ω The total circuit resistance is therefore 47 Ω Ω = 60.2 Ω. The circuit current will be given by: I = V R = 6.02 V 60.2 Ω = 0.0A The current through the 47 Ω resistor is therefore 00 ma. When this current comes to the parallel network, it will split in the inverse ratio of the resistances, i.e. 22/55 (= 40 ma) through the 33 Ω resistor and 33/55 (= 60 ma) through the 22 Ω resistor. Alternatively, the potential difference across the parallel network is: V = IR = 0.0 A 3.2 Ω =.32 V giving the current in the 22 Ω resistor as: I = V R =.32 V = A = 60 ma 22 Ω and the current in the 33Ω resistor as I = V R =.32 V = A = 40 ma 33 Ω b) The power generated is given by P = I 2 R in each case: 47 Ω: (0.0 A) 2 47 Ω = 0.47 W

6 22 Ω: (0.06 A) 2 22 Ω = W 33 Ω: (0.04 A) 2 33 Ω = W [8] a) Before the voltmeter is connected, the total circuit resistance is [Total 0 Marks] R = 22 kω + 33 kω = 55 kω [] This gives a circuit current of: I = V = 7.5 V = 0.36 ma [] R 55k Ω The potential difference across the 33 kω resistor is therefore: V = IR = 0.36mA 33kΩ = 4.5 V [] Alternatively, if you are good at ratios you might spot that the 7.5 V will split up in the ratio of the two resistors: V across 33 kω = 33/ V = 4.5 V b) i) When the switch is closed, the voltmeter is connected in parallel with the 33 kω resistor. The resistance of this parallel combination will be less than 33 kω, so the voltage dropped across the combination will be less than the previous 4.5 V. [2] ii) If the voltmeter reads 4.0 V, the potential difference across the 22 kω resistor must be ( ) V = 3.5 V. [] The current in the 22 kω resistor, which is also the circuit current, will be: I = V = 3.5V = 0.59mA [] R 22kΩ The combined resistance of the parallel arrangement of the voltmeter and the 33 kω resistor will therefore be: R = V I = 4.0 V 0.59 ma For this parallel arrangement: = 25. kω [] R = + = R 33 R V R V R = R kω 33 kω R V = kΩ kΩ = kΩ R V = 06 kω If the voltmeter is rated as 0 V/00 μa, its resistance should be: R = V = 0V =.0 I A 05 Ω = 00 kω [2] This differs by 6% from the experimental value. As each resistor has a tolerance of 5%, the experimental value is within the overall tolerance and so is compatible with the stated rating of the voltmeter. [2] [2] [Total 4 Marks]

7 2 a) As the temperature of the thermistor falls, its resistance increases as there will be less charge carriers per unit volume. If the resistance of the thermistor increases, the proportion of the supply voltage dropped across it will also increase and so the proportion of the supply voltage across the resistor R will decrease and V out will fall. [3] b) i) Reading from the graph, at 0 C the thermistor has a resistance of 8.0kΩ. [] ii) When V out is 5.0 V, the voltage across the thermistor and the potentiometer will be ( ) V = 4.0 V. Assuming that the potentiometer is set at zero, the voltage across the thermistor will be 4.0 V. [] The current in the thermistor (and therefore the circuit current as it is a series circuit) will be given by: iii) I = 4.0V = 0.22mA [] 8.0kΩ The value of R is then given by: R = V out I = 5.0 V = 22.5 kω 0.22 ma The best value to use would therefore be the 22 kω resistor. [2] The purpose of the potentiometer is to fine tune the circuit. It is adjusted so that the lamp is switched on at exactly 0 C. [2] [Total 0 Marks] 3 a) The electromotive force of an electrical source is defined as the energy per unit charge converted into electrical energy by the source, in this case the cell, whilst the potential difference between two points in a circuit is the electrical energy per unit charge converted into other forms of energy, in this case in the resistor. In the equation V = ε Ir: ε is the chemical energy per coulomb converted into electrical energy by the cell Ir is the energy per coulomb used up by the charge doing work against the internal resistance of the cell V (= ε Ir) is the energy per coulomb that will be dissipated in the resistor R. [4] b) i) I = e.m.f. = ε Total circuit resistance R + r εr + r) εr =ε(r R + r R + r ii) V = ε Ir = ε = εr R + r [3]

8 c) From I = ε R + r ε = I(R + r) so: ε = 66 ma (2.2 Ω + r) ε = 308 ma (4.7 Ω + r) Dividing we get: = Ω + r 4.7 Ω + r (4.7 Ω + r) = 2 (2.2 Ω + r) 4.7 Ω + r = 4.4 Ω + 2r r = 0.3 Ω Substituting into ε = I (R + r): ε = 66 ma ( ) Ω =.54 V [4] d) From (b) we have: I = ε R + r P = IV = and V = εr R + r ε εr = R + r R + r ε 2 R (R + r) 2 [2] e) To find a maximum we need to differentiate with respect to R: dp = d ε2 R dr dr (R + r) 2 = ε 2 2ε 2 R (R + r) 2ε 2 R = ε2 (r R) (R + r) 2 (R + r) 3=ε2 (R + r) 3 (R + r) 3 This will be zero (i.e. the power will be a maximum) when R = r [3] [Total 6 Marks] 5 a) Conservation of charge means that we cannot gain or lose charge. Consider the 3-way junction shown in the diagram. As current is the rate of flow of charge, by conservation of charge I 2 + I 3 = I. If this were not so, charge would be gained or lost at the junction which, by the conservation of charge, is not possible. [3]

9 b) This question is really about symmetry, something we come across a lot in physics. The first thing to do is draw a diagram showing the current in each resistor. If a current I enters the network at point X, by symmetry it will split equally in each of the three resistors connected to point X. We can now put ⅓I in each of these resistors. [] By the same argument, the current in each of the three resistors connected to point Y will also be ⅓I so that a current I leaves at point Y. [] But what about the remaining six resistors? Again, by symmetry, the ⅓I splits equally so that the current in the remaining six resistors is ⅙I. [] Your distribution of current should now look like the figure below.

10 What we now have is three resistors in parallel at X (combined resistance R/3) connected in series with the middle six resistors in parallel (combined resistance R/6), which in turn are connected in series the three parallel resistors at Y (combined resistance R/3). This is shown in the diagram below. [3] The network resistance is therefore R 3 + R 6 + R 3 = 5R 6 [] 6 a) From the graph of R against L we can take two convenient values of R and L and substitute into the equation logr = nlogl + loga The obvious points to take are: L =.0, R = 00 and L = 0 000, R = 0.. We then have: log(00) = n log(.0) + log a 2 = 0 + log a log a = 2 log(0.) = n log(0 000) + log a = 4n + log a Substituting that ln a = 2 into the second equation gives us = 4n + 2 4n = 3 n = ¾ If log a = 2 a = 00 From R = al n R = 00 L 3 4 R = 00 4 [4] L 3

11 b) When L = 50 lux: R = 00 4 = 00 L 3 4 = 5.3 kω [2] 50 3 (You should check from the graph that this looks about right.) c) If the LDR is 0.5 m from the lamp (i.e. its distance from the lamp is halved), then four times as much light will fall on the LDR, giving L = 200 lux. R = 00 4 = L =.9 kω [2] d) We have to assume that the lamp acts as a point source and that the illumination is inversely proportional to the square of the distance from the lamp. We also have to assume that the only light falling on the LDR is the light from the lamp, i.e. there is no background illumination. [2] [Total 0 Marks]

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