Question

Pankaj records the measurement of resistance and current to find the potential across A and B. Estimate the resistance for a current of 7.5 amps.

R (Resistance)	5	10	15	20	25	30
I (current)	10	4.9	3.2	2.4	1.9	1.7

• A. 6.5 ohms
• B. 4.5 ohms
• C. 1.5 ohms
• D. 2.75 ohms

Answer 1

The correct option is A 6.5 ohms

$x\left(R\right)$	$y\left(I\right)$	$x^2$	$y^2$	$x^{2}y$
5 10 15 20 25 30	10 4.9 3.2 2.4 1.9 1.7	25 100 225 400 625 900	100 24.01 10.24 05.76 03.61 02.89	50 49 48 48 47.50 51

$\sum x=105,\sum y=24.1,\sum x^2=2275,\sum y^2=146.51$

$\sum xy=293.50$

$cov(x,y)=\frac{\sum xy}{N}-\frac{\sum x}{N}.\frac{\sum y}{N}=\frac{293.5}{6}-(\frac{105}{6}.\frac{24.1}{6})$

$cov(x,y)=-21.36$

so, $(y-\overline{y})=r\frac{\sigma y}{\sigma x}(x-\overline{x})$

$(y-\overline{y})=\frac{cov(x,y)}{{\sigma }^{2}x}(x-\overline{x})$

$so,(y-\overline{y})=r\frac{\sigma y}{\sigma x}(x-\overline{x})$

$(y-\overline{y})=\frac{cov(x,y)}{{\sigma }_x^2}(x-17.5)$

$(y-4.016)=\frac{-21.36}{{\sigma }_x^2}(x-17.5)$

Now,

${\sigma }^{2}x=\frac{\sum x^2}{N}-\left({\frac{\sum x}{N}}^2\right)=72.91$

so, $(y-4.016)=\frac{-21.36}{72.91}(x-17.50)$

$0.292x+y-9.1424=0$

Now at $y=I=7.5$ Amps.

$x=5.624$ Ohm