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In a meter-bridge if the left and right gaps are connected with \(2 \Omega\) and \(3 \Omega\) resistances, respectively then the bridge is balanced. The resistance to be connected with \(3 \Omega\) resistance to get the balancing point at midpoint of the bridge wire is
Options:
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Verified Answer
The correct answer is:
\(6 \Omega\) in parallel
Key Idea In a meter-bridge, if the balancing point is at the midpoint of the bridge wire, then it shows the resistance gaps have equal resistance.
Given,
\(\begin{aligned}
& R_1=2 \Omega \\
& R_2=3 \Omega
\end{aligned}\)
and
\(R_2=3 \Omega\)
For meter-bridge,
\(\frac{R_1}{R_2}=\frac{l_1}{L-l_1}\)
Where, \(l_1\) is balancing length,
\(\begin{aligned}
\frac{R_1}{R_2} & =\frac{l_1}{l_1} \quad\left(\because l_1=L-l_1\right) \\
\Rightarrow \quad R_1 & =R_2 \\
\Rightarrow \quad 2 \Omega & =3 \Omega \| x \Omega
\end{aligned}\)
Here, \(x\) is the parallel balancing resistance to \(3 \Omega\) resistor.
\(\begin{aligned}
\text {Hence, } \frac{3 x}{3+x} & =2 \\
\Rightarrow \quad x & =6 \Omega
\end{aligned}\)
Hence, the correct option is (d).
Given,
\(\begin{aligned}
& R_1=2 \Omega \\
& R_2=3 \Omega
\end{aligned}\)
and
\(R_2=3 \Omega\)
For meter-bridge,
\(\frac{R_1}{R_2}=\frac{l_1}{L-l_1}\)
Where, \(l_1\) is balancing length,
\(\begin{aligned}
\frac{R_1}{R_2} & =\frac{l_1}{l_1} \quad\left(\because l_1=L-l_1\right) \\
\Rightarrow \quad R_1 & =R_2 \\
\Rightarrow \quad 2 \Omega & =3 \Omega \| x \Omega
\end{aligned}\)
Here, \(x\) is the parallel balancing resistance to \(3 \Omega\) resistor.
\(\begin{aligned}
\text {Hence, } \frac{3 x}{3+x} & =2 \\
\Rightarrow \quad x & =6 \Omega
\end{aligned}\)
Hence, the correct option is (d).
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