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(A) To obtain an equivalent resistance of $6 \ \Omega$,we analyze the circuit in image $A$:In the upper branch,$R_1$ and $R_2$ are in series,so $R_{up

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(A) To obtain an equivalent resistance of $6 \ \Omega$,we analyze the circuit in image $A$:In the upper branch,$R_1$ and $R_2$ are in series,so $R_{up} = R_1 + R_2 = 5 \ \Omega + 5 \ \Omega = 10 \ \Omega$.In the lower branch,$R_3$ and $R_4$ are in series,so $R_{low} = R_3 + R_4 = 5 \ \Omega + 10 \ \Omega = 15 \ \Omega$.These two branches are in parallel,so the equivalent resistance $R_P$ is given by:$\frac{1}{R_P} = \frac{1}{R_{up}} + \frac{1}{R_{low}} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} \ \Omega^{-1}$.Therefore,$R_P = 6 \ \Omega$.Thus,the circuit in image $A$ provides the required equivalent resistance.

From the combination of resistors with resistance values $R_1 = R_2 = R_3 = 5 \ \Omega$ and $R_4 = 10 \ \Omega$,which of the following combinations is the best circuit to get an equivalent resistance of $6 \ \Omega$?

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