Question 735: Current Electricity

Question

An unknown resistance X is to be 
determined using resistances R$$_1$$, R$$_2$$ or R$$_3$$. Their corresponding null points are A, B 
and C. Find which of the above will give the 
most accurate reading and why ?

Accepted Answer

At null points, the Wheatstone bridge will be 
balanced.
Therefore,
$$\frac{\mathrm{X}}{r_{1}}=\frac{\mathrm{R}}{r_{2}} 	ext { (or) } \mathrm{X}=\mathrm{R} \frac{r_{1}}{r_{2}}$$
Where $$\mathrm{R}$$ is a constant and $$r_{1}$$ and $$r_{2}$$ are variable. The maximum fraction error is
$${{\Delta x} \over x} = {{\Delta {r_1}} \over {{r_1}}} + {{\Delta {r_2}} \over {{r_2}}}$$
$$\begin{array}{rr} & \mathrm{R}=\mathrm{R}_{1}, \mathrm{R}=\mathrm{R}_{2}, \mathrm{R}=\mathrm{R}_{3} \ 	ext { Here, } & \Delta r_{1}=\Delta r_{2}=y 	ext { (say), }\end{array}$$
$$y$$ can be considered as least count of the scale,
Then, $$ \frac{\Delta x}{x}=y\left[\frac{r_{2}+r_{1}}{r_{1} r_{2}}ight]$$
For $$\frac{\Delta x}{x}$$ to be minimum, $$r_{1} 	imes r_{2}$$ should be maximum [Since $$r_{1}+r_{2}=\mathrm{c}$$ (constant $$)$$ ]
Let,
$$\begin{aligned}
\mathrm{E} & =r_{1} 	imes r_{2}=r_{1} 	imes\left(c-r_{1}ight) \
& =r_{1} c-r_{1}^{2} \
	herefore \frac{d \mathrm{E}}{d r_{1}} & =c-2 r_{1}=0 	ext { or } \frac{c}{2}=r_{1}\end{aligned}$$
(or) $$r_{2}=c / 2 	ext { (or) } r_{1}=r_{2}$$

Answer

R = R$$_1$$

Answer

R = R$$_2$$

Answer

R = R$$_3$$

Answer

All

Current Electricity

An unknown resistance X is to be determined using resistances R$_1, R_2 or R_3$. Their corresponding null points are A, B and C. Find which of the above will give the most accurate reading and why ?

Choose the correct answer: