In the circuit shown in Figure 4 below, how much resistance should be connected to a resistor so…

For points M and N to be at the same potential ($V_M = V_N$), the Wheatstone bridge must be balanced: $\frac{P}{Q} = \frac{R}{S_{\text{eff}}}$ Given $P = 3\,\Omega$, $Q = 6\,\Omega$, and $R = 4\,\Omega$: $\frac{3}{6} = \frac{4}{S_{\text{eff}}} \implies \frac{1}{2} = \frac{4}{S_{\text{eff}}} \implies S_{\text{eff}} = 8\,\Omega$ Since the required effective resistance ($8\,\Omega$) is less than $10\,\Omega$, a resistance $X$ must be connected in parallel with the $10\,\Omega$ resistor: $\frac{1}{S_{\text{eff}}} = \frac{1}{10\,\Omega} + \frac{1}{X}$ $\frac{1}{8\,\Omega} = \frac{1}{10\,\Omega} + \frac{1}{X}$ $\frac{1}{X} = \frac{1}{8} - \frac{1}{10} = \frac{5 - 4}{40} = \frac{1}{40\,\Omega}$ $X = 40\,\Omega$ in parallel. Final answer: 40 \Omega