Current Electricity and Electrical InstrumentMedium
Question
Two resistors $2\Omega$ and $3\Omega$ are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire XY . When an unknown resistor is connected in parallel with $3\Omega$ resistor, the null point is shifted by 22.5 cm toward Y . The resistance of unknown resistor is $\Omega$.
Options
A.3
B.2
C.4
D.1
Solution
Initially, $\frac{2}{3} = \frac{x}{100 - x}$
$$\Rightarrow x = 40\text{ }cm $$Now when ' $R$ ' connected in parallel
$${\frac{2}{\frac{3R}{3 + R}} = \frac{40 + 22.5}{60 - 22.5} = \frac{62.5}{37.5} }{\therefore R = 2\Omega}$$
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