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(C) The circuit is a Wheatstone bridge. Let the resistors be $R_1 = 3\, \Omega$,$R_2 = 4\, \Omega$,$R_3 = 6\, \Omega$,and $R_4 = 8\, \Omega$. The cent

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$14/3\, \Omega$

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(C) The circuit is a Wheatstone bridge. Let the resistors be $R_1 = 3\, \Omega$,$R_2 = 4\, \Omega$,$R_3 = 6\, \Omega$,and $R_4 = 8\, \Omega$. The central resistor is $R_5 = 7\, \Omega$.Check the ratio of the arms: $R_1/R_3 = 3/6 = 1/2$ and $R_2/R_4 = 4/8 = 1/2$.Since $R_1/R_3 = R_2/R_4$,the bridge is balanced. Therefore,no current flows through the $7\, \Omega$ resistor.The circuit simplifies to two parallel branches: the upper branch with $3\, \Omega$ and $4\, \Omega$ in series $(3+4 = 7\, \Omega)$ and the lower branch with $6\, \Omega$ and $8\, \Omega$ in series $(6+8 = 14\, \Omega)$.The equivalent resistance $R_{eq}$ is given by $1/R_{eq} = 1/7 + 1/14 = (2+1)/14 = 3/14$.Thus,$R_{eq} = 14/3\, \Omega$.

For the given circuit,the equivalent resistance between $A$ and $B$ is:

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