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Physics Factsheet www.curriculum-press.co.uk Number 91 Accuracy in Electrical Investigations This Factsheet is not meant to be a survey of practical investigations concerning electricity, or to provide complete methods for these investigations. Real and Ideal Meters An ammeter is placed in series with a resistor. An ideal ammeter has zero resistance, and does not reduce the current flow. It is instead intended to look at the causes of inaccuracy in electrical work, and to suggest some techniques for minimising and measuring inaccuracy. I R In practise, the ammeter will have some resistance. But in modern electronics, current flow is tiny (implying a large circuit resistance), so the resistance of the ammeter is not too important. Resistors and Combinations of Resistors Standard resistors (e.g. carbon film) are manufactured to stated tolerances (maximum percentage errors) often described by a coloured band on the resistor. A gold band means 5% maximum error. However, voltmeters are placed in parallel to the component being measured, leading to real problems at times. R I1 Example 1: What is the maximum possible error, in ohms, for a 22kΩ resistor with a stated tolerance of 5%? Answer: I A A I I 5 × 22000 = 1100Ω. 100 V I2 Often we must combine resistors to achieve required values. It is worth seeing the effect on the percentage error of such combinations. The ideal voltmeter would have infinite resistance, drawing no current away from the main circuit. But a moving coil meter requires a current through it to produce a deflection. If the current flow through R is very small, the current sidetracked to the parallel voltmeter could be significant, and lead to a measurable error in the voltage reading. Example 2: A 1000Ω resistor and a 2200Ω resistor (each of 5% tolerance) are wired in series, then in parallel. Find the maximum percentage error in each combination. Example 3 (a): Look at this circuit: 1000Ω 1000Ω 12V 2200Ω 2200Ω 50kΩ Answer: Series: Maximum value = (1000 × 1.05) + (2200 × 1.05) = 3360Ω . Nominal value = 1000 + 2200 = 3200Ω. 160 Percentage error = × 100 = 5%. 3200 100kΩ Find the p.d. across the 100kΩ resistor. (b) Compare this second circuit. 12V 100kΩ Parallel: Maximum value: 1 1 + = 1.385 × 10-3, R = 722Ω . 1050 2310 1 1 Nominal value: + = 1.455 × 10-3, R = 688Ω . 1000 2200 Percentage error = 5% again. 50kΩ V 100kΩ Now find the p.d. across the 100kΩ resistor. Answer: 12 = 8.0 × 10-5 A. 150 000 V(across 100kΩ) = IR = 8.0 × 10-5 × 1.0 × 105 = 8V. 1 1 2 (b) Parallel: = = 100 000 100 000 100 000 R = 50kΩ (same as other resistor) V(across parallel combination) = 6V. (a) I = Combining resistors with the same tolerance should lead to an effective resistance with the same tolerance. It should be noted that these are maximum percentage errors. In practice, some of the errors might well cancel each other. The effective resistance of the combination would probably be closer to the nominal value. But there is no guarantee of this. 1 Physics Factsheet 91. Accuracy in Electrical Investigations Introducing the moving coil voltmeter across the 100kΩ resistor has changed the p.d. across it from 8V to 6V. This can lead to Joule heating problems and is very wasteful of power. The solution to this voltmeter problem is to use a digital voltmeter in electronic circuits. These meters can have effective resistances up in the megohm region. Calculations with potential dividers are only accurate if the load is much larger than the resistances in the divider itself. In practical situations, a voltmeter must be used to provide an accurate value for the output p.d. Example 4: This circuit uses a digital voltmeter with an input resistance of 20MΩ. 12V R1 E 100kΩ R2 50kΩ R V V 20MΩ Find the p.d. across the 100kΩ resistor. Resistance and Temperature We tend to assume that resistors have fixed resistances. But their resistance values increase with temperature. As an example, for copper the temperature coefficient of resistance can be approximately given as: ∆R where R is the resistance at 0oC α = (R ∆θ) ∆θ is theo temperature rise. o Answer: 1 1 + = 1.005 ×10-5, R = 9.95×104 Ω. 2.0×107 1.0×105 12 I= = 8.03×10-5 A. ( 5.0×104 + 9.95×104 ) V(across parallel combination) = IR = 8.03×10-5 × 9.95×104 = 8.0 V. (the same as without the meter in the circuit) Parallel: Example 5: What is the percentage increase in the resistance of a length of copper wire when its temperature goes up by 25oC? (α = 0.0043oC-1) Exam hint: When doing practical work, always consider the effects of the measuring instruments on the values being obtained. High input impedance devices, like digital voltmeters and cathode ray oscilloscopes, can significantly increase accuracy. ∆R = α∆θ = 0.0043 × 25 = 0.11 Ro The percentage increase is 11%. Answer: Potential Dividers A potential divider allows us to produce an adjustable output p.d. which is less than the fixed input p.d. from the supply. Joule heating causes significant changes in resistance in metallic conductors. There are various ways of minimising Joule heating effects: (a) keep the current very small. (b) ensure adequate ventilation or use a cooling fan (c) use physically large components, or combinations to share the heating effect. R1 E R2 V 10Ω 10Ω R The output p.d. is given by V = R + 2R × E. 1 2 10Ω However the current drawn by the load can affect the size of the output voltage. E 10Ω (d) use cooling e.g. immersing a resistance wire in water. R1 R2 10Ω R The load, R, now forms a parallel branch with R2, lowering the output voltage. (This theory is exactly the same as in the previous situation where the voltmeter is in parallel with the 100kΩ resistor.) To minimise this effect, R2 must be much smaller than the load, R. However, if R2 is very small, then the current and power dissipation through R1 and R2 will be large. 2 Physics Factsheet 91. Accuracy in Electrical Investigations 3. Assume the temperature coefficient of resistance of copper is 0.0043oC-1. (a) Find the increase in resistance when a length of copper wire of resistance 0.24Ω at 0oC is heated from 10oC to 45oC. (b) Express this as a percentage change. Metre bridges Wheatstone bridges and metre bridges are very useful in making accurate determinations of resistance and of the emf of cells. Their accuracy relies on the fact that at the balance point, there is no deflection on the meter. X (Unknown) 4. If you decided to reduce Joule heating in a circuit by using very thin resistance wire to reduce the current flow, suggest two other factors you would have to consider (in terms of accuracy). R 5. Here is another look at a metre bridge. A !1 !2 B X (Unknown) Null deflection means no errors due to current flowing through the meter. In our diagram, X l = l1 R 2 However there can still be sources of inaccuracy. At A and B there can be contact resistances. These add to the resistances of wire lengths l1 and l2. These end errors are most significant if l1 or l2 is very small. A !2 B The wire has a total resistance of 6.0Ω. There is a contact resistance of 0.010Ω at A and at B. The wire is 1.00m long. (a) If the balance point is at the centre of the wire, find the resistances of l1 and l2. (b) Find the percentage errors in Rl1 and Rl2 due to the contact resistances. Rl (c) Find the percentage error in Rl1 . 2 (d) If the unknown resistance is changed, and the balance point is just 1.0cm from A, find the resistances of l1 and l2. (e) Again find the percentage errors in Rl1 and Rl2. Rl (f) Now find the percentage error in Rl1 2 When using metre bridges, make R approximately equal to X. This will give a balance point near the middle of the wire, reducing the effect of end errors on l1 and l2. Graphical work Quite often, we can use graphical work to improve accuracy, making use of best fit lines. 5. (a) l1 = 3.0Ω, l2 = 3.0Ω. (b) % error l1 = % error (l2 = (0.010 / 3.0) × 100 = 0.33% (c) % error l1 / l2 = 0.66% (d) l1 = 1/100 × 6.0 = 0.060Ω l2 = 99/100 × 6.0 = 5.94Ω (e) % error l1 = (0.010 / 0.060) × 100 = 16.7% % error l2 = 0.17% (f) % error l1 / l2 = 16.9%. This is, of course, applicable to all fields of physics and other sciences. But it is always best to improve the accuracy of the investigation itself. Don’t rely on graphical and other techniques to sort out experimental weaknesses. Questions 1. Use the method of maximum and minimum values to find the maximum percentage error for this combination. The tolerance of each resistor is 5%. 4. Examples might be (a) increased percentage errors from electrical measuring instruments when dealing with very small currents, and (b) increased percentage errors in measuring the diameter of the wire. 100Ω !1 R 100Ω 3. (a) DR = αRoθ = 0.036Ω (b) Percentage change = (0.036 / 0.24) × 100 = 15%. 2. (a) R = V/I = 120 000Ω. (b) 1/R = 1 / 100 000 + 1 / 120 000. R = 54 500Ω. Percentage change = 46%. (c) 1/R = 1 / 100 + 1 / 120 000. R = 99.9Ω Percentage change = 0.1%. 100Ω 100Ω 2. (a) A voltmeter draws a current of 0.1mA when a p.d. of 12V is applied across it. Find its resistance. (b) Assume the resistance it exerts is a constant. Find the change in effective resistance when it is connected across a 100kΩ resistor. Express this as a percentage change. 1. Nominal value: Series 200Ω in each branch, then Parallel gives R = 100Ω. Maximum value: Increase each resistor to 105Ω. Find that effective R = 105Ω. Maximum percentage error is 5Ω for the combination, as well as for each resistor. 100kΩ Answers: V (c) Find the percentage change in effective resistance when the voltmeter is connected across a 100Ω resistor. 3 Acknowledgements: This Physics Factsheet was researched and written by Paul Freeman The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. ISSN 1351-5136