As before this equation can be expressed in terms of

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. As before, this equation can be expressed in terms of the voltage V c across the capacitor, V c = Q/C = V o e -t/RC ……. 5) This equation shows that the charge and voltage on the capacitor decay exponentially with time. The time constant of the circuit (t 1 = RC) is the time necessary for the voltage (or charge) to decay to 1/e (= 0.368) of its original value V o . A related quantity is the half-life, t 1/2 , which is the time required for the voltage (charge) to decrease to just one-half the original value. This is given by: V c /V o = ½ = e -t1/2/RC or t 1/2 = RC ln 2 = 0.693 RC …….. 6)
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Apparatus The equipment used in this experiment are: 1) Two capacitors (big and small) 2) Red and black banana jacks 3) Switch 4) Power supply of 10V 5) 10 Ω Resistor 6) Voltmeter 7) Extra Wires to create the curcuit Procedure The circuit was set up while we used the biggest of the two capacitors (C 1 ) for C. The switch was closed and the output voltage of the power supply was adjusted to V o = 10V, as measured by the meter. The switch was opened, and we started the timer at the same instant. The time taken for the capacitor voltage to drop to 3.68v was measured. This was equal to the time constant of the circuit we connected. And if we use t’ to represent this, therefore t = R m C 1 ………… 7) This procedure was repeated for three seperate measurements of t , and the average value was gotten. An external resistor of R x = 10 M     Ω was connected in parallel with the capacitor and meter. We also repeated this to measure a time constant of t’’
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that relates to the resistance of the new circuit. The value of R m can be found by the equation below; R m = t – t ’’ R x t ’’ Capacitance Measurement While making use of the calculated value of R m for the meter resistance and the measured time constant t ,the capacitance of the Capacitor (C1) was gotten. We used the same settings as of the figure 1 and later on we changed the larger capacitor to the smaller one. The switch was then closed and charged up to 10V. The switch was opened and the time for the capacitor to drop to 3.68v was measured. This was repeated thrice and the average value was used to compute the capacitance C 2 . Both capacitors C 1 and C 2 were connected in series with the voltmeter across both of them. The circuit capacitance C s was then measured. The procedure was done over and over again to determine the combined capacitance C p with both capacitors connected in parallel. After the values of C1 and C2 were determined, the predicted combined capacitances when connected in series and parallel was solved for.
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  • Spring '12
  • Hardy

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