Don't Touch It!

Let $V$ denote the transformer's voltage and $V_{R}$ be the voltage at the Return junction. By applying Kirchhoff's Current Law at the Return junction, one obtains the following relationship:

$\frac{V_{R} - V}{(2 + 50) \Omega} = \frac{0 - V_{R}}{1000 \Omega} + \frac{0 - V_{R}}{2 \Omega}$ ;

or $(\frac{1}{52} + \frac{1}{1000} + \frac{1}{2})V_{R} = \frac{V}{52} = \frac{240}{52}$ ;

or $V_{R} = 8.872$ volts. The current running through you (if you dare!) computes to (by Ohm's Law): $I = \frac{V_{R}}{1000 \Omega} = 8.872 \times 10^{-3}$ , or $\boxed{8.872 mA}.$