part 1. Charging the capacitor. Half-life. The time it takes to charge the capacitor to half full is called the half-life and is related to the time constant t in the following way: t1/, = t (In2) In this activity the charge on the capacitor will be measured indirectly by measuring the voltage across the resistor since the charge is proportional to voltage q = CV and the voltages are connected with each other (see eq.1). The half-life time is to be determined. Procedure: It is recommended to operate within the optimal range which is 500[k(2] to 2[MS] and 5[MF] to 20[MF ]. Choose more resistance for less capacitance and choose more capacitance for less resistance. If you are given multiple capacitors of smaller capacitances then you might want to connect them in parallel to get the maximum equivalent capacitance. If you are given multiple resistors of smaller resistances then you might want to conne them in series to get the maximum equivalent resistance. Switch the multimeter to the capacitance mode and measure the actual capacitance o given capacitor (or the system of smaller capacitances). This is your accepted truth to compare the experiment to. Actual capacitance is: 10.01 M [MF ] Switch the multimeter to the resistance mode and measure the actual resistance given resistor (or the system of smaller resistances). 1. 996 M [AS] Actual resistance is: KIX // /setUP a circuit shown below. Make sure there is no charge stored on the capacitor before the experiments (connect the capacitor's terminals with a wire for a short time). Ask your instructor to double check your connections before turning the power \"ON\" on the power supply. Make sure the switch connects the RC circuit to the power supply (position 1). Reset your time measuring device to zero. Set the battery voltage to 9[V]. Simultaneously, turn the power on and start measuring the time and voltage across the resistor. Your goal is to measure the half life time. You may repeat the procedure so that you get the best results. Experimental Data (charging): Half-life voltage Vi, = 4.5 [v 1 12,630 g (e ot 1. Uk 8 (Rnd o 528 s Time to half-max ty,=_'2:52% | ] o 1% G Analysis of Data: Since half life is defined as: t1/2 = T In2 = (RC) In2 then Cexperimental t1/2 = R In2 Calculate the percent difference between the actual value of the capacitance (from the digital multimeter measurement) and the experimental value: Given the = 12 . 639 2 12 . 528 9 12.416 R = 1. 996 MR Capacitance ( from digital multimeter measurement ) = 10. 01 MF Cexperimental = + 1 /2 12.528 s - = 9.06 X 10 F = 9.06 M RIn ( 2 ) ( 1 . 9 9 6 x 10 6 52 ) x In ( 2 ) %% error = I actual value - experimental valuel x 10 0 actual value = 1 ( 10 . 01 x 10- 6 ) - ( 9. 06 * 10 - )1 x 100 ( 10. 01 x 10 - 6 ) = 9.49%. Conclusion:Part 2. Determination of the unknown capacitance. What is the capacitance of an unknown capacitor? Procedure: Use the same circuit but with a different capacitance and resistance values. The optimal range is 500[kQ] to 1 [MQ] and 5[JF] to 20[MF]. Choose more resistance for less capacitance and choose more capacitance for less resistance. Actual resistance is: 2. 095 [ M 52 ] The capacitance value is considered unknown at this time. R C switch Set the battery voltage to 9[V] and let the battery charge the capacitor to 9 [ (as close as possible). Ask your instructor to double check your connections. Reset your time measuring device to zero. Simultaneously, flip the switch to position 2 (no more connection to the power supp and start measuring the time and voltage across the resistor. Every 2-3 [s] is best. Write down the measured parameters below. Repeat the pro so that you get the best results.4.1 MF Experimental Data (discharging): Time Voltage across the In (AVR) resistor, AVR 6. 81 V 1.92 V 7 . V 4 . 9 9 V 1.60 V 5 . 721 8. 69 V 1. 31 V . 12s 4 . 9 7 V 2 . 7 7 V 1.02 V S 15's 2. 18 V 0 . 78 1 185 L . 26 V 1. 52 V 0 . 42 12 215 2 . 7 7 V 1. 2 2 0. 20 245 2. 40 V 0 . 9 1 V - 0 . 1 30s 0. 52 V - 0. 65 0. 38V - 1.10 9 3 5s 40s 1 3 1 V O. 20V The capacitor discharges through the resistor with time constant t = RC. The resist in parallel with the capacitor and both have the same potential difference at all time The capacitor charge and the resistor (and capacitor) voltage decays exponentially AVR = Voe It To analyze exponential decays, we take the natural logarithm of both sides: 1 In(AVR) = In(AVo) + In (e /t) = In(AV.) --t TInterpret In(AVR) = In(AVo) - =t as y(t) = b+ mt and get the result: graph of In(AVR) versus t (called a semi-log graph) should be linear with y-intercept b = In(AV.) and slope m = - - We can determine t and hence C from an experimental measurement of the slope. Draw a graph of In (AVR) (log of measured voltage across the resistor, vertical axis) vs time (horizontal axis). From the y-intercept of the best-fit line, find AVo = 9 [V ] , as expected. Using the slope, find T = - 1 slope Using this calculate the unknown capacitance: Cexperimental = R Now, measure the actual value of the capacitance with the digital multimeter and compare it to the value obtained from the graph: Actual capacitance is: 6. 19 [MF ] Conclusion