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

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$10/3 \, \Omega$

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(A) The circuit is a bridge circuit. Let the resistances be $R_1 = 2 \, \Omega$,$R_2 = 3 \, \Omega$,$R_3 = 4 \, \Omega$,$R_4 = 6 \, \Omega$,and the central resistance $R_5 = 7 \, \Omega$.Check the ratio of resistances: $R_1/R_3 = 2/4 = 1/2$ and $R_2/R_4 = 3/6 = 1/2$.Since $R_1/R_3 = R_2/R_4$,the bridge is balanced.In a balanced Wheatstone bridge,no current flows through the central resistance $R_5$. Thus,it can be removed.The circuit simplifies to two parallel branches: the upper branch with $R_1$ and $R_2$ in series,and the lower branch with $R_3$ and $R_4$ in series.Resistance of upper branch $R_{up} = R_1 + R_2 = 2 + 3 = 5 \, \Omega$.Resistance of lower branch $R_{low} = R_3 + R_4 = 4 + 6 = 10 \, \Omega$.Since these two branches are in parallel,the equivalent resistance $R_{AB}$ is given by:$1/R_{AB} = 1/R_{up} + 1/R_{low} = 1/5 + 1/10 = (2+1)/10 = 3/10$.Therefore,$R_{AB} = 10/3 \, \Omega$.

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

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