Two resistances are connected in two gaps of a meter bridge. The balance point is 20cm from the zero end. A resistance of 15$\Omega$ is connected is series with the smaller of the two. The null point shifts to 40cm. The value of the smaller resistance in ohm is

Question

Dr. Abubacker · Accepted Answer

(c) Let S be the large and R be the smaller resistance.
From formula for meter bridge
$S = \left( \frac{100 - l}{l} ight)R = \frac{100 - 20}{20}R = 4R$
Again, $S = \left( \frac{100 - l}{100} ight)(R + 15)$
$= \frac{100 - 40}{40}(R + 15) = \frac{3}{2}(R + 15)$
$	herefore\ \ \ \ 4R = \frac{3}{2}(R + 15)$
$\ \ \ \ \ \ \frac{8R}{3} - R = 15\ $
$\Longrightarrow \frac{5R}{3} = 15$
$\ \ \ \ \ \ \ \ R = 9\Omega$

Q. Two resistances are connected in two gaps of a meter bridge. The balance point is 20cm from the zero end. A resistance of 15$\Omega$ is connected is series with the smaller of the two. The null point shifts to 40cm. The value of the smaller resistance in ohm is

Explanation

Reviewed by Dr. Abubacker

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