Two resistances are connected in the two gaps of a meter bridge. The balancing point is obtained at $20…

Two resistances are connected in the two gaps of a meter bridge. The balancing point is obtained at $20 \mathrm{~cm}$. When a resistance of $15 \Omega$ is connected in series with the smaller resistance of the two, the balancing point shifts to $40 \mathrm{~cm}$. The value of smaller resistance is
  1. $9 \Omega$
  2. $12 \Omega$
  3. $6 \Omega$
  4. $3 \Omega$

Solution

$\begin{aligned} & \mathrm{L}=20 \mathrm{~cm} ; \mathrm{L}^{\prime}=40 \mathrm{~cm} \\ & \text { Using } \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{L}}{100-\mathrm{L}}=\frac{20}{100-20}=\frac{20}{80} \\ & \Rightarrow \mathrm{R}_2=4 \mathrm{R}_1 \\ & \frac{\mathrm{R}_1^{\prime}}{\mathrm{R}_2^{\prime}}=\frac{\mathrm{L}^{\prime}}{100-\mathrm{L}^{\prime}}=\frac{40}{100-40}=\frac{40}{60}=\frac{2}{3} \\ & \text { Here } \mathrm{R}_1^{\prime}=\mathrm{R}_1+15 \text { and } \mathrm{R}_2{ }^{\prime}=\mathrm{R}_2=4 \mathrm{R}_1 \\ & \frac{\mathrm{R}_1+15}{4 \mathrm{R}_1}=\frac{2}{3} \\ & 3 \mathrm{R}_1+45=8 \mathrm{R}_1 \\ & \Rightarrow \mathrm{R}_1=9 \Omega \text { Smaller resistance }\end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 2)

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