Here's how I might approach this problem:
Use superposition to write
$$V_x = 5\angle30\cdot j||(2 - j) + 2V_x \cdot \frac{2}{j - j + 2} \cdot j $$
Gather terms
$$V_x (1 - j2) = 5\angle30\cdot j||(2 - j) $$
Isolate the desired variable
$$V_x = 5\angle30\frac{j||(2 - j)}{(1 - 2j)}$$
Since the complex power delivered is
$$S = V_x \cdot 5\angle{-30}$$
See that
$$S = 25\frac{j||(2 - j)}{(1 - 2j)} = (-7.5 + j10)VA$$
The current through the load resistor can be obtained by finding the THEVENIN EQUIVALENT across it.If you are unaware about do check out thevenin intro or any other online material on it.It is very useful in solving networks for currents and voltages.
The thevenin equivalent across 120 ohm is :
From the above circuit the current through load can be easily calculated as:
I(Load) = [ 0.385Vs /(110.76 + 120) ] = 0.0016684 Vs ( where Vs is the supply voltage at input)
Now, to get the current drawn from the supply voltage Vs at the input,
Step 1: Find the equivalent resistance across the input source which involves star to delta conversion as you have mentioned.
The R(equivalent) = [ 37.9 + ( (56.8 + 80) || (25.3 + 180) ) ] = 120 ohms.
Step 2:
The current drawn from the input is : I(supply) = Vs/ Req = 0.00833 Vs.
Now that we have both currents in terms of the input supply voltage,
The ratio of load current to the current drawn from the supply is:
I(Load) / I(supply) = 0.001664 /0.00833 = 0.2
If the answer is 5 then it should be the inverse of the above as current through load is always less than supply current.
Edit: Since you wanted to solve it by mesh analysis here is an easier way:
Best Answer
You don't need meshes. Just calculate everything one source at a time, while keeping the other sources zeroed out (i.e. a voltage source is replaced with a wire, and a current source is replaced with an open (cut) circuit). Then sum the results from each source.
So in this case, the voltage source would see nothing but four resistors in series, and both current sources would see (different sets of) 2 series resistors connected in parallel with 2 other series resistors. Hopefully this helps you see how to "zero" the sources.
I'm not familiar with MIT's Circuit Sandbox, but it shouldn't matter where you put the ground since you just need to check the current flowing in different places - so just pick a spot and call it ground!