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Q. Two unknown resistances are connected in two gaps of a meter-bridge. The null point is obtained at $40\,cm$ from left end. A $30\,\Omega$ resistance is connected in scries with the smaller of the two resistances, the null point shifts by $20\, cm$ to the right end. The value of smaller resistance in $\Omega$ is

MHT CETMHT CET 2017Current Electricity

Solution:

let the reistance at right be $R$ and at left be $r$
$\frac{r}{R}=\frac{1_{x}}{l_{R}}=\frac{40}{60}$,
$r$ is smaller
$\frac{r+30}{R}=\frac{60}{40}$
From above equation we get $r =24 \Omega$