Question

Effective resistance of three conductors in parallel is $\frac{60}{47}ohm$ . One of the wire breaks and the effective resistance is $\frac{15}{8}ohm$ . The resistance of the wire that broken is

• A. 2Ω
• B. 4Ω
• C. 16Ω
• D. 8Ω

Answer 1

The correct option is B

4Ω

Let $R_1,R_{2}and{R}_3$ be the resistances of the three conductors.

In the first case, $\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{\frac{47}{60}}....\left(1\right)$

In the second case, $\frac{1}{R_2}+\frac{1}{R_3}=\frac{8}{15}....\left(2\right)$

From eq’s (1) and (2) $\frac{1}{R_1}=\frac{47}{60}-\frac{8}{15}=\frac{47-32}{60}=\frac{15}{60}\therefore R_1=\frac{60}{15}=4\Omega$