In an experiment to determine the time period (?) of a...

In an experiment to determine the time period (?) of a RC-circuit, where R = 22 K?, the following table were obtained during the charging. t (sec) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 | 110 | 120 ?V | 0 | 3.8 | 5.8 | 7.1 | 8.1 | 8.8 | 9.1 | 9.5 | 9.6 | 9.7 | 9.8 | 9.8 | 9.9 a- Plot potential difference versus time in the graph paper using the experimental data presented in the table above. b- Calculate the RC-circuit period (?), given that: t_max = 5 ? c- Calculate the mean value of the capacitance (C) of the capacitor, given that: ? = RC

Transcript

00:01 Hello students to plot the potential difference versus time using the experimental data.

00:06 So we have the data set given in the question.

00:08 There is a data set.

00:11 So we need to go with the time on the x -axis and delta v on the y -axis, right? so i've just plotted it.

00:22 I'll show you now.

00:23 So this is the potential difference versus time graph for the given data.

00:29 So it's supposed to be starting from 0 and it is increasing hyperbolically and then it comes back after a particular point.

00:42 This is a graph now to calculate the rc circuit rc circuit period.

00:47 We can use the formula t is equal to t max t max by n.

01:01 So this is this is can be used for calculating the the rc circuit period t max is the maximum time value and n is the number of cycles.

01:10 This case t max is given as 5 so 5 and the number of cycle is equal to 1.

01:17 So that is equal to 5 seconds.

01:22 So that is the t value right? that is the t rc circuit period t is equal to 5 seconds.

01:29 It is the b part now coming to the c part to calculate the mean value of capacitance of the capacitor.

01:35 We are given that t is equal to rc.

01:39 We can say c is equal to t by r.

01:42 So t is already given as 5 seconds and we need to find out the value of r here.

01:50 So from the given data, we can set see that t is equal to 0 v is equal to 0.

01:57 Therefore at the beginning of the charging the capacitor is essentially acting short circuit.

02:01 So the current is at maximum value.

02:03 So according to ohm's law, we can find out at let's take the graph at a medium value, right? so at medium value, it is linearly increasing in the middle portion.

02:18 So that means let's take 60 and then find out r...

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