In a meter bridge circuit as shown, when one more resistance of 100 Ω is connected is parallel with unknown resistance x, then ratio $\frac{l_{1}}{l_{2}}$ become 2. If l₁ is balance length of the uniform wire A B, then the value of x must be:

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100 Ω

200 Ω

400 Ω

50 Ω

answer is B.

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Detailed Solution

$∵$ wheat stone bridge is in balanced condition
So $\frac{100}{ℓ_{1}} = \frac{\frac{100 x}{100 + x}}{ℓ_{2}}$
$∵ \frac{ℓ_{1}}{ℓ_{2}} = 2 \Rightarrow x = 100 Ω$

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