ExperimTime[s]3na 6ait12151821242935Ischarging):Voltage across theresistor, AVR [v]5.94.3byly 3,12,31.61.20.80.60.40,2In (AVR) InterpretIn (AVR) = In (AV) -t as y(t)=b+mtTand get the result: graph of In (AVR) versus / (called a semi-log graph) should be linearwith y-intercept b = ln (AV) and slope m =1=-²/TWe 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 =Using this calculate the unknown capacitance:TCexperimental R--1slope=Now, measure the actual value of the capacitance with the digital multimeter andcompare it to the value obtained from the graph:Actual capacitance is: 12,4FI

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Answered: Experime Time [s] 3 no 6 а 12 15 18 21…

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Experime Time [s] 3 no 6 а 12 15 18 21 24 29 Data (discharging): 35 Voltage across the resistor, AVR [v] 5.9 4.3 3.1 2,3 ما .ا 1.2 0.8 0.6 0.4 0,2 In (AVR)