Question

Effective resistance of parallel combination is 6/5 $\Omega$. If one of the resistances is broken, then the resultant resistance becomes 2 $\Omega$. Then other resistance is

• A. 4 $\Omega$
• B. 3 $\Omega$
• C. 6 $\Omega$
• D. 5 $\Omega$

Answer 1

Answer: (B)

3 $\Omega$

Answer 2

Let the two resistances be $R_1$ and $R_2$ with

$$\frac{R_1 R_2}{R_1 + R_2} = \frac{6}{5} \quad (1)$$

When one resistance fails, the effective resistance becomes $2\,\Omega$. This implies that the remaining resistor is $R_1 = 2\,\Omega$ (assuming $R_1$ is the one still in the circuit). Substituting into (1):

$$\frac{2\,R_2}{2 + R_2} = \frac{6}{5}$$

Cross-multiply:

$$10\,R_2 = 6(2 + R_2) = 12 + 6R_2$$

Subtract $6R_2$ from both sides:

$$4\,R_2 = 12 \quad \Rightarrow \quad R_2 = 3\,\Omega$$