Average Current for switched circuit
Your maths looks fine, but in order to know whether it is correct or not we'd need to know the shape of the wave. Your equation is true for a simple uni-polar pulse waveform, so I'll start there.

Suppose we had the following circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

And SW1 was closed with duty 0.0324 = 3.24%. We'd get a waveform like this:



Then \$I_{peak}\$ is clearly 10A. \$I_{rms}\$ is always given by this equation:

\$I_{rms} = \sqrt{\frac{1}{T}\int_{0}^{T} i(t)^2 dt}\$

The integral for most of the 1 second period is zero. For the brief on time (0.0324 seconds) \$i(t)\$ is 10, so the integral reduces to:

\$I_{rms} = \sqrt{\frac{1}{1}\int_{0}^{0.0324} 10^2 dt} = \sqrt{\frac{1}{1}100\times0.0324} = \sqrt{3.24} = 1.8\$

which checks out pretty nicely.

\$I_{average}\$, on the other hand is always given by this equation:

\$I_{average} = \frac{1}{T}\int_{0}^{T} i(t) dt\$

This integral reduces to:

\$I_{average} = \frac{1}{1}\int_{0}^{0.0324} 10 dt = \frac{1}{1}10\times0.0324 = 0.324\$

So, for a uni-polar pulse waveform I'd say yep, your maths looks right. For any other waveform, all bets are off.