00:01
Hello students, in this question we have a data point for the voltage.
00:06
So current and voltage is varying.
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So it starts from 0 .1 and 0 .56 then 0 .2, 1 .1 and 0 .3, 1 .6, 0 .4, 2 .1 and 0 .5, 2 .6.
00:27
So that is the plot.
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Now we need to find out the slope.
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We need to plot this graph.
00:35
First of all, that is the first thing we need to do.
00:38
We have to plot the vi graph or vi graph.
00:43
Right.
00:44
The first point is let's take from 0 .1.
00:48
So that is 0 .56.
00:50
So let's start here.
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0 .56 and then we get 0 .2.
00:57
So that goes to 1 .1.
01:00
So double of that.
01:02
So that is 1 .1.
01:04
Right.
01:05
So that will be here.
01:09
Now 0 .3, 0 .4 and 0 .5 coming in.
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So 0 .3 goes for 1 .6.
01:20
So that's around here and 2 .1 again in here and 2 .6 again here.
01:28
So it's a straight line.
01:30
Ok.
01:31
So this is the vi graph that is tempting to be a straight line.
01:36
So now what we need to find out to determine the data collected.
01:41
If the data collected in the experiment supports the ohm's law.
01:45
So we just look at the graph and we can see that straight away that this is originating.
01:52
This is going as a straight line.
01:53
So v is v by i is a constant that is called r.
02:01
So constantly you can see individually v by i that is 0 .56 by 0 .1 is equal to.
02:12
So let's take individual values here.
02:16
So this plot looks something within the limit.
02:19
The ratio of v by i within this plot is within the range of 5 .3 to 5 .6.
02:25
So it has an accuracy of 0 .3 plus or minus or 0 .03 .1 .5.
02:33
That is to be accurate...