In the resistor circuit shown below, the equivalent resistance $$"R"_(eq)(Omega)$$ at Terminals a-b is most nearly:

A.   22
B.   20
C.   4
D.   2

The two main principles we will use to solve this problem include:

1. For resistors in series, add the individual resistances together: $R_{eq}=R_1+R_2+R_3+...$
2. For resistors in parallel, add the reciprocal values of the individual resistances together to find the reciprocal of the total combined resistance: $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...$

We can start the problem by combining the two 3$\Omega$ resistors in series at the right-most part of the circuit.

$R_{eq1}=3\Omega + 3\Omega = 6\Omega$

This resistance is in parallel to two other 6$\Omega$ resistors. Therefore we use:

$$\begin{gathered} \frac{1}{R_{eq2}}=\frac{1}{6\Omega }+\frac{1}{6\Omega }+\frac{1}{R_{eq1}} \\ \frac{1}{R_{eq2}}=\frac{1}{6\Omega }+\frac{1}{6\Omega }+\frac{1}{6\Omega } \\ \frac{1}{R_{eq2}}=\frac{3}{6\Omega } \\ \frac{R_{eq2}}{1}=\frac{6\Omega }{3} \\ {R}_{eq2}=2\Omega \end{gathered}$$

Finally, we combine $R_{eq2}$ with the 2$\Omega$ resistor it is in series with to find the equivalent resistance across terminals a-b:

$$\begin{gathered} R_{eq}=2\Omega +R_{eq2} \\ {R}_{eq}=2\Omega +2\Omega \\ {R}_{eq}=4\Omega \end{gathered}$$

Our answer is choice C, 4.