The Norton equivalent resistance ($$Omega$$) at Terminals A-B is most nearly: 

A.   50
B.   60
C.   100
D.   120

Norton's theorem, like Thévenin's theorem, solves for the equivalent resistance in the circuit by shorting voltage sources and placing open circuits where ideal current sources are located. Once the $15V$ source is shorted, we see that the Norton equivalent resistance, $R_N$, is calculated as follows:

Note: Resistors in parallel are represented with two verticle lines ($\left|\right|$) and the equivalent resistance of N resistors in parallel is calculated as follows:

$R_1\left|\right|R_2\left|\right|...\left|\right|R_{N-1}\left|\right|R_N = {\left(\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_{N-1}}+\frac{1}{R_N}\right)}^{-1}$

$$\begin{gathered} R_N = \left(\right(80\Omega + 120\Omega ) || 50\Omega )+60\Omega \\ = (200 ||50) +60 = {\left(\frac{1}{200}+\frac{1}{50}\right)}^{-1}+ 60 \\ = {\left(\frac{1}{200}+\frac{4}{200}\right)}^{-1}+ 60 \\ = \frac{200}{5} + 60 \\ = 40 + 60 \\ = 100\Omega \end{gathered}$$

Therefore we say that the answer is C. 100.