Question

The effective resistance of three resistors connected in parallel is 60/47 $\Omega$. When one wire breaks, the effective resistance becomes 15/8 ohms. Find the resistance of the wire that is broken.

Answer 1

At $Q,R_P=\frac{60}{47}⟹\frac{1}{R_P}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{47}{60}...\left(i\right)$

Lets say $R_3$ is breaks, then the net resistance will become

ie., $⟹\frac{1}{R_{P_1}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{8}{15}........\left(ii\right)$

Substituting $\left(ii\right)$ in eqn$\left(i\right)$ we get,

$⟹\frac{8}{15}+\frac{1}{R_3}=\frac{47}{60}⟹\frac{1}{R_3}=\frac{47}{60}-\frac{8}{15}⟹\frac{1}{R_3}=\frac{47-32}{60}⟹\frac{1}{R_3}=\frac{15}{60}=\frac{1}{4}$

Hence the resistance of the that is broken wire = $4\Omega$