A modified Wheatstone bridge is used to measure the value of load resistance $R_{L}$ shown in figure. The resistance $R_{X}$ is varied such that balance is obtained, then value of branch parameters are as follows

$R_{a} = 140 Ω;$ $R_{b} = 0.75 Ω;$

$R_{1} = 70 Ω;$ $R_{2} = 180 Ω;$

$R_{3} = 0.375 Ω$

Determine the value of $R_{X}$

0.1246 Ω

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0.2438 Ω

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0.4625 Ω

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0.9642 Ω

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Solution

The correct option is D $0.9642 Ω$
The modified bridge can be drawn after delta to star transformation

$R_{10} = \frac{R_{a} R_{b}}{R_{a} + R_{b} + R_{L}} = \frac{140 \times 0.75}{140 + 0.75 + R_{L}} = \frac{105}{140.75 + R_{L}}$

$R_{20} = \frac{R_{b} R_{L}}{R_{a} + R_{b} + R_{L}} = \frac{0.75 \times R_{L}}{140 + 0.75 + R_{L}} = \frac{0.75 R_{L}}{140.75 + R_{L}}$

$R_{30} = \frac{R_{a} R_{L}}{R_{a} + R_{b} + R_{L}} = \frac{140 \times R_{L}}{140 + 0.75 + R_{L}} = \frac{140 R_{L}}{140.75 + R_{L}}$

Applying balance condition for wheatstone bridge,

$(R_{2} + R_{30}) R_{3} = (R_{20} + R_{x}) R_{1}$

$[180 + \frac{140 R_{L}}{140.75 + R_{L}}] 0.375 = [180 + \frac{0.75 R_{L}}{140.75 + R_{L}} + R_{x}] 70$

$67.5 + \frac{52.5 R_{L}}{140.75 + R_{L}} = \frac{52.5 R_{L}}{140.75 + R_{L}} + 70 R_{x}$

$67.5 = 70 R_{x}$

$R_{x} = 0.9642 Ω$

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