Deleted

 

No, inductors are like resistors when you combine them - paralleling reduces the value.

If Rs = 0 then that capacitor to ground doesn't factor into the transfer function.

It's just:

\(
\frac{Vo}{Vin} = \frac{R}{0.5sL || \frac{1}{sC} + R}
\)

 

Yes, a 0 ohm resistor is a short circuit.
Since the voltage source is right across that capacitor to ground you know the voltage across it - it's exactly Vin at all times.
It has no effect on the other components.

 

The transfer function is the relationship between input and output.
You're looking at voltages and the only way that capacitor could affect the output voltage is if its current caused a voltage drop somewhere between the input and output.

In this case the capacitor would increase the current coming from the voltage source but since there's no resistance it doesn't do anything else. The current flows only through the capacitor and the voltage source because it's a 0 resistance path, it doesn't go through the output resistor at all.

If Rs wasn't 0 the capacitor current would add a voltage drop between the input and output, or equivalently the current would split between the voltage source and the output resistor's branch.

 

You're right, that transfer function is from the capacitor to the output but with Rs = 0 that's identical to Vin/Vout.