Solveeit Logo

Question

Question: The values of current \( I \) flowing in a given resistor for corresponding values of potential diff...

The values of current II flowing in a given resistor for corresponding values of potential difference VV across resistor are given below:

II (Amperes)0.50.511223344
VV (volts)1.61.63.43.46.76.710.210.213.213.2

Plot the graph between VV and II . Calculate the resistance of that resistor.

Explanation

Solution

Hint : Here, the values of II and VV are given and we have to plot these values in the graph and let us find the resistance by using the formula of R=1slopeR = \dfrac{1}{{slope}} , resistance is the inverse of the slope of the graph between II and VV .
Slope in the coordinate system is given by: slope=y2y1x2x1slope = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}

Complete Step By Step Answer:
Here, let us first put the above given data in the graph such that take current II on xaxisx - axis and potential difference VV on yaxisy - axis . As shown below:

In the above figure we see a straight line between the current and potential difference. We know that if we have to find the slope of the straight line we have to consider any two points such that (1,3.4)(1,3.4) and (3,10.2)(3,10.2) .
Resistance is actually the inverse of the slope between the potential difference and current as:
R=1slopeR = \dfrac{1}{{slope}} …. (1)(1)
Let us first calculate the slope of the straight line as:
slope=y2y1x2x1slope = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}
slope=10.23.431\Rightarrow slope = \dfrac{{10.2 - 3.4}}{{3 - 1}} …. (We considered the points above)
slope=6.82\Rightarrow slope = \dfrac{{6.8}}{2}
slope=3.4\therefore slope = 3.4
So let us now put this value in equation (1)(1) and calculate resistance as:
R=13.4\therefore R = \dfrac{1}{{3.4}}
R=0.294Ω\Rightarrow R = 0.294\Omega
Thus, we calculated the resistance of the above given data as 0.294Ω0.294\Omega
Hence, we have obtained the answer.

Note :
We have plotted the graph as shown above and we know that the resistance is given by the inverse of the slope of the straight line between the potential difference and current. Calculate the terms carefully and write the decimal nos. precisely.