Question

Figure shows a meter bridge (which is nothing but a practical Wheatstone bridge), consisting of two resistors X and Y together in parallel with a meter long constantan wire of uniform cross section. Which the help of a movable contact D, one can change the ratio of resistance of the two segments of the wire until a sensitive galvanometer G connected across B and D shows no deflection. The null point is found to be a distance of 33.7 cm. The resistor Y is shunted by a resistance of 12 $\omega$, and the null point is found to shift by a distance of 18.2 cm. Determine the approximate resistance of X and Y in ohm.

Answer 1

As the wire is of uniform cross section, the resistances of the two segments AD and DC of the wire are in the ratio of the lengths of AD and DC. According to the condition of balance of Wheatstone bridge,

$\frac{X}{Y}=\frac{l_1}{l_2}$

Here $l_1=33.7cm$ and $l_2=100-33.7=66.3cm$

$\frac{X}{Y}=\frac{33.7}{66.3}$ (i)

As resistance Y' is due to a parallel combination of resistance Y and a resistance of $12\omega$,

$\frac{1}{Y\text{'}}=\frac{1}{Y}+\frac{1}{12}=\frac{1}{2}+\frac{Y}{12Y}$ or$Y\text{'}=\frac{12}{Y12}+Y$

Since $Y^{\prime }$ is less than $Y,$ the ratio $X/Y^{\prime }$ will be greater than $X/Y$ and the null point should shift toward end C.

$\frac{X}{Y\text{'}}=33.7+$$\frac{18.2}{66.3-18.2}$=$\frac{51.9}{48.1}$

or $\frac{X(12+Y)}{12Y}=\frac{51.9}{48.1}$

or $\frac{12}{Y}=\frac{51.9}{48.1\times 12\times YX}=\frac{\mathrm{51.948.1}\times 12\times 66.3}{33.7}$=$25.47\omega$

or $Y=25.47-12=13.47\omega$

Putting this value in Equation (i), we get

$X=\frac{33.7}{66.3\times Y}=\frac{33.7}{66.3\times 13.47}=6.85\omega$