The condition for a balanced Wheatstone bridge is:
$$\frac{P}{Q} = \frac{R}{S_{\text{eff}}}$$
Given $P = 3\,\Omega, Q = 3\,\Omega, R = 3\,\Omega$:
$$\frac{3}{3} = \frac{3}{S_{\text{eff}}} \implies S_{\text{eff}} = 3\,\Omega$$
The arm has resistance $S = 4\,\Omega$. To reduce its equivalent resistance to $S_{\text{eff}} = 3\,\Omega$, a shunt resistance $S'$ is connected in parallel with $S$:
$$\frac{1}{S_{\text{eff}}} = \frac{1}{S} + \frac{1}{S'}\implies \frac{1}{3} = \frac{1}{4} + \frac{1}{S'}$$
$$\frac{1}{S'} = \frac{1}{3} - \frac{1}{4} = \frac{4 - 3}{12} = \frac{1}{12} \implies S' = 12\,\Omega$$
Therefore, $S$ must be shunted with a $12\,\Omega$ resistor.
Hence, option 2 is the correct answer.
संतुलित व्हीटस्टोन ब्रिज की शर्त:
$$\frac{P}{Q} = \frac{R}{S_{\text{eff}}} \implies \frac{3}{3} = \frac{3}{S_{\text{eff}}} \implies S_{\text{eff}} = 3\,\Omega$$
प्रतिरोध $S = 4\,\Omega$ के साथ समानांतर में शंट प्रतिरोध $S'$ जोड़ने पर:
$$\frac{1}{S_{\text{eff}}} = \frac{1}{S} + \frac{1}{S'} \implies \frac{1}{3} = \frac{1}{4} + \frac{1}{S'}$$
$$\frac{1}{S'} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \implies S' = 12\,\Omega$$
अतः $S$ को $12\,\Omega$ के प्रतिरोधक के साथ शंट किया जाना चाहिए।
अतः विकल्प 2 सही उत्तर है।
