Question 557: Current Electricity

Question

Two resistance $$\mathrm{X}$$ and $$\mathrm{Y}$$ are connected in the two gaps of a meterbridge and the null points is obtained at $$20 \mathrm{~cm}$$ from zero end. When the resistance of $$20 \Omega$$ is connected in series with the smaller of the two resistance $$\mathrm{X}$$ and $$\mathrm{Y}$$, the null point shifts to $$40 \mathrm{~cm}$$ from left end. The value of smaller resistance in ohm is

Accepted Answer

For a meterbridge, $$\frac{\mathrm{X}}{\mathrm{Y}}=\frac{l}{100-l}$$
In the first case, $$l=20 \mathrm{~cm}$$
$$\begin{aligned}
\frac{X}{Y} & =\frac{20}{100-20}=\frac{20}{80}=\frac{1}{4} \
	herefore \quad 4 X & =Y
\end{aligned}$$
In the second case, $$l^{\prime}=40 \mathrm{~cm}$$
$$\begin{array}{ll}
	herefore \quad & \frac{X^{\prime}}{Y^{\prime}}=\frac{40}{100-40}=\frac{40}{60}=\frac{2}{3} \
& 	ext { But, } X^{\prime}=X+20 	ext { and } Y^{\prime}=Y \
	herefore & \frac{X+20}{Y}=\frac{2}{3} \
	herefore \quad & \frac{X+20}{4 X}=\frac{2}{3} \
	herefore \quad & 8 X=3 X+60 \
	herefore \quad & X=12 \Omega
\end{array}$$

Answer

6

Answer

9

Answer

12

Answer

15

Current Electricity

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