Question
A meter bridge with two resistances $R_{1}$ and $R_{2}$ as shown in figure was balanced (null point) at 40 cm from the point P . The null point changed to 50 cm from the point P , when $16\Omega$ resistance is connected in parallel to $R_{2}$. The values of resistances $R_{1}$ and $R_{2}$ are $\_\_\_\_$ .
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Solution
$$\begin{array}{r} \frac{R_{1}}{R_{2}} = \frac{40}{60} = \frac{2}{3}\#(1) \end{array}$$
$$\begin{array}{r} \frac{R_{1}}{\left( \frac{R_{2} \times 16}{R_{2} + 16} \right)} = \frac{50}{50} \Rightarrow R_{1} = \frac{16R_{2}}{16 + R_{2}}\#(2) \end{array}$$
$${\frac{2}{3}R_{2} = \frac{16R_{2}}{16 + R_{2}} }{\frac{32}{3} + \frac{2R_{2}}{3} = 16 }{\frac{2R_{2}}{3} = 16 - \frac{32}{3} = \frac{16}{3} }{R_{2} = 8\Omega }$$By equation (1)
$$R_{1} = \frac{2}{3}R_{2} = \frac{16}{3}\Omega$$
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