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Charging and Discharging a Capacitor at Constant Rate

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Introduction

Charging and Discharging a Capacitor at Constant Rate Name: Chow Ching Yue Class: 6S Class Number: 7 A. Objectives * Learn how to calculate the charge stored in a capacitor or discharging it at a constant rate. * Illustrate how to keep the charging and discharging currents constant. * Show that the voltage across a capacitor is proportional to the change stored in it. * Determine the capacitance of a capacitor directly from its definition. B. Preview Questions 1. If a capacitor is charged at a constant rate, what do you say about the current through the capaciotor? Ans: The current is constant. 2. If the capacitor discharges at a constant rate, what do you say about the current through the capacitor? Ans: The current is constant. 3. What is the charge stored in a capacitor if a steady current of 90 �A flows through it for 110 s? Ans: Applying Q = It Q = (110) (90 x 10-6) = 9900 �C 4. If the terminals 1 and 2 of the variable resistor shown in Fig. 1 are used, how can you tell experimentally which direction the knob should be turned to give the highest resistance? Ans: Connect p he circuit as shown. Set the variable resistor at the middle position initially. Slightly turn the variable resistor and observe how the meter deflects. The turning direction that gives a smaller current represents increasing resistance. ...read more.

Middle

Fig. 1 Measuring the e.m.f. of the battery Fig. 2 Discharging the capacitor Fig.3 Adjusting the resistance of variable resistor to keep current constant F. Precautions * If you are not sure which direction the knob of the variable resistor that you should turn to give the highest resistance, you may either use a multimeter to test or connect up the circuit (using an ammeter) as shown. Note that you should set the variable resistor at the middle position initially. Slightly turn the variable resistor and observe how the meter deflects. The turning direction that gives a smaller current represents increasing resistance. * Check the correctness of the polarities of the capacitor and the ammeter. Avoid causing the ammeter to deflect in the opposite direction. G. Results, Calculations and Graphs Experiment 1 Experimental set-up: Step 2: e.m.f. of the battery, Vo = 2.96 � 0.01V Step 4: Initial charging current, Io = 68 � 2�A Steps 7 - 9: Trial 1 Trial 2 Trial 3 Average charging period Time recorded (s) 104 102 102 103 � 0.01 Step 10: Total charge stored, Q = It 68 x 103 = 7004 �C capacitance C, C = Q / V = 7004 / 2.96 = 2366�F Error: By Q = CV and Q = It C= It / V ?C = ?I + ?t + ?V C I t V ?C / C = (2/68) ...read more.

Conclusion

Resistance of digital voltmeter is not infinity Precautions 1. Take repeated readings and averaging to reduce systematic errors 2. Do not put the capacitor near any conductors to reduce stray capacitance 3. Select only one student for recording data Suggestions for improvements 1. Use data logger to collect data 2. Use electronic device to keep the current constant Comparison of result with that expected From the experimental result, four values of capacitance C obtained are 2366 � 78�F, 2267 � 78�F, 2281 � 75�F and 2267 � 100�F respectively, while the actual value of the capacitance C is 2200�F. The obtained values of 4 experiments are all higher than the expected value. This is because the there are errors existed during experiments, such as stray capacitance, reading errors, ammeter and voltmeter were not ideal. I. Conclusion By collecting data from 4 experiments, the average capacitance C is: C = [(2366 � 78�F) + (2267 � 78�F) + (2281 � 75�F) + (2267 � 100�F)] / 4 = 2295 �A Error = (2295 - 2200) / 2200 x 100% = 4.33% From the experiment, we can find that the current of charging and discharging can be kept constant by adjusting the resistance of the variable resistance. Also, form experiment 2 & 4, straight lines passed through origins are obtained in the graphs of Q against V. therefore, the voltage across a capacitor is proportional to the charged stored in it. Compare with the actual value of capacitance C, the experimental result C is higher than the actual one because of the presence of errors in the experiment. ?? ?? ?? ?? 1 ...read more.

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