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(B) The circuit is a Wheatstone bridge configuration. The condition for no current to flow through the central resistor $(5 \Omega)$ is that the bridg

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$17$

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(B) The circuit is a Wheatstone bridge configuration. The condition for no current to flow through the central resistor $(5 \Omega)$ is that the bridge must be balanced. For a balanced Wheatstone bridge,the ratio of resistances in the arms must be equal: $\frac{R_{AB}}{R_{AD}} = \frac{R_{BC}}{R_{DC}}$. Here,$R_{AB} = 10 \Omega$,$R_{AD} = 20 \Omega$,$R_{BC} = 15 \Omega$,and $R_{DC} = 30 \Omega$. Checking the ratios: $\frac{10}{20} = 0.5$ and $\frac{15}{30} = 0.5$. Since the ratios are equal,the bridge is balanced,and no current flows through the $5 \Omega$ resistor. In this state,the $10 \Omega$ and $15 \Omega$ resistors are in series,and the $20 \Omega$ and $30 \Omega$ resistors are in series. Resistance of the upper branch: $R_1 = 10 \Omega + 15 \Omega = 25 \Omega$. Resistance of the lower branch: $R_2 = 20 \Omega + 30 \Omega = 50 \Omega$. These two branches are in parallel. The equivalent resistance $R_{eq}$ is given by: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{25} + \frac{1}{50} = \frac{2+1}{50} = \frac{3}{50}$. $R_{eq} = \frac{50}{3} \Omega \approx 16.67 \Omega$. Rounding to the nearest integer,we get $17 \Omega$.

The equivalent resistance of the following circuit when no current flows in the resistance of $5 \Omega$ is nearly (in $Omega$)

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