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(A) The given circuit is a Wheatstone bridge circuit. Let the resistances be $R_1 = 2 \, \Omega$,$R_2 = 4 \, \Omega$,$R_3 = 3 \, \Omega$,$R_4 = 6 \, \

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$\frac{10}{3} \, \Omega$

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(A) The given circuit is a Wheatstone bridge circuit. Let the resistances be $R_1 = 2 \, \Omega$,$R_2 = 4 \, \Omega$,$R_3 = 3 \, \Omega$,$R_4 = 6 \, \Omega$,and the central resistance $R_5 = 7 \, \Omega$. Check the condition for a balanced Wheatstone bridge: $\frac{R_1}{R_2} = \frac{2}{4} = 0.5$ and $\frac{R_3}{R_4} = \frac{3}{6} = 0.5$. Since $\frac{R_1}{R_2} = \frac{R_3}{R_4}$,the bridge is balanced. Therefore,no current flows through the central $7 \, \Omega$ resistor,and it can be removed from the circuit. Now,the circuit consists of two branches in parallel: Upper branch: $R_{up} = R_1 + R_3 = 2 + 3 = 5 \, \Omega$. Lower branch: $R_{low} = R_2 + R_4 = 4 + 6 = 10 \, \Omega$. The effective resistance $R_{eq}$ between $A$ and $B$ is given by: $\frac{1}{R_{eq}} = \frac{1}{R_{up}} + \frac{1}{R_{low}} = \frac{1}{5} + \frac{1}{10} = \frac{2+1}{10} = \frac{3}{10}$. Thus,$R_{eq} = \frac{10}{3} \, \Omega$.

Five resistances are connected as shown in the figure. The effective resistance between points $A$ and $B$ is

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