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Measuring the e.m.f. And Internal Resistance of a Cell

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Introduction

Measuring the e.m.f. And Internal Resistance of a Cell Planning I wish to find out the electromotive force (e.m.f.) and Internal Resistance of a direct current battery cell. Circuit Diagram 1. Set up apparatus as shown. 2. Use ammeter to adjust variable resistor to vary of currents. 3. Read voltmeter value and record. 4. Disconnect circuit and take e.m.f reading from voltmeter across cell. 5. Repeat for another 9 resistor settings. Use variable resistor and ammeter to adjust the set up to get a range of 0.1 to 1.0A. 6. Repeat entire experiment to get more accurate average. Safety Aspects 1. Do not have circuit connected up for too long, this will run the battery down and may cause it to heat up and maybe explode. 2. Wear safety goggles and lab-coats during experiment. 3. Follow standard laboratory safety procedures. Apparatus 0 - 1.0A analogue ammeter 2 digital voltmeters 1 variable resistor 1 cell 8 connecting wires. Safety goggles and lab-coat Paper and pen Arranging Evidence Results Current (Amps) Potential Difference (V) 1 2 3 Average 0.1 1.46 1.47 1.46 1.46 0.2 1.42 1.42 1.41 1.42 0.3 1.36 1.39 1.36 1.37 0.4 1.33 1.33 1.33 1.33 0.5 1.27 1.29 1.28 1.28 0.6 1.24 1.23 1.23 1.23 0.7 1.18 1.19 1.20 1.19 0.8 1.13 1.15 1.14 1.14 0.9 1.10 1.09 1.11 1.10 1.0 1.04 1.03 1.06 1.04 Using these results I will be able to draw a graph to find the e.m.f. ...read more.

Middle

Therefore, if the equation of the line can be found then the e.m.f. can also be found. By using the results from the gradient calculations I know that the gradient is -0.47 � 0.03. If I use a pair of co-ordinates that lie on the line I can find the constant value, C, using the equation y - y1 = m (x - x1). Using co-ordinates (0.2,1.415) y - 1.415 = -0.47 (x - 0.2) y - 1.415 = -0.47x + 0.094 y = -0.47x + 1.509 This means that C must equal 1.51 (3sf). To find the error in this value I substituted the gradients and a set of points for the minimum line and maximum line in to the equation E = Ir + V. For the maxm line: E = 0.86 x 0.49 + 1.1 E = 1.52 (3sf) For the minm line: E = 1.0 x 0.45 + 10.5 E = 1.50 (3sf) The difference between these two values and the values calculated is 0.01. Therefore the electromotive force of the cell = 1.51�0.01V(3sf) Evaluating Evidence These were the possible sources of error: * As I was using an analogue ammeter the needle that is used to read the values off is slightly higher than the actual scale so to enable a completely accurate reading you have to look directly down at it. ...read more.

Conclusion

Although the ammeter could have been more accurate by having a smaller error and it may even be more precise if two or even three people were to perform the experiment together. One to turn the circuit on and off, one to take the voltmeter reading and one to take the ammeter reading. This way there is no loss of accuracy due to the time delay between taking readings that is caused by the operator. As the technique used was very accurate and precise, this made the conclusions drawn also very accurate. Although there may have been a human error made when drawing the best-fit, maximum and minimum lines on the graph and reading the results off accurately. To confirm my final value of e.m.f. I found the e.m.f. of the cell in the circuit each time I took a reading. These were the results. Current (Amps) E.M.F. taken from Voltmeter 1 2 3 Average 0.1 1.50 1.51 1.51 1.51 0.2 1.51 1.51 1.50 1.51 0.3 1.51 1.52 1.50 1.51 0.4 1.50 1.51 1.50 1.50 0.5 1.52 1.50 1.50 1.51 0.6 1.51 1.50 1.50 1.50 0.7 1.50 1.50 1.50 1.50 0.8 1.49 1.50 1.49 1.49 0.9 1.50 1.49 1.48 1.49 1.0 1.48 1.49 1.48 1.48 These show that the e.m.f value is actually around 1.50V, compared with my final result of 1.51V. This shows that the techniques appear to be very accurate and precise. This table also shows that at times the value overlaps the result I found which adds weight to the conclusion that my answer is accurate. Alice Salt ...read more.

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