Question

For two resistance wires joined in parallel, the resultant resistance is $ \frac{6}{5}\Omega . $ When one of the resistance wires breaks the effective resistance becomes $ 2\,\Omega $ The resistance of the broken wire is

• A. $ \frac{3}{5}\,\Omega $
• B. $ 2\,\Omega $
• C. $ \frac{6}{5}\,\Omega $
• D. $ 3\,\Omega $

Answer 1

The Correct Option is D

Let $R_{1}$ and $R_{2}$ be the resistances of two wires. In parallel combination, $\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{6}{5}\,\,\,... (i)$ If one of the wires breaks, then effective resistance becomes $2 \,\Omega$, that is, $R_{1}=2 \,\Omega$ Substituting in E (i), we have $\frac{2 R_{2}}{2+R_{2}}=\frac{6}{5}$ or $10\, R_{2}=12+6R_{2}$ or $4 \,R_{2}=12 $ $\therefore R_{2}=\frac{12}{4}=3\, \Omega$