A single phase half bridge inverter has load R = 2 Ω and a dc voltage source Vs/2 = 115 V. Find the rms value of the fundamental load current.

To find the rms value of the fundamental load current for a single-phase half-bridge inverter, we first need to determine the amplitude of the output voltage.

Given:

– The DC voltage source is ( V_s = 2 times 115 , V = 230 , V ).

– The load resistance ( R = 2 , Omega ).

The output voltage for a single-phase half-bridge inverter can be approximately considered as ( V_{o,text{peak}} = frac{V_s}{2} ), so:

[

V_{o,text{peak}} = frac{230 , V}{2} = 115 , V

]

For a half-bridge inverter, the rms value of the fundamental output voltage ( V_{o,text{rms}} ) is given by:

[

V_{o,text{rms}} = frac{V_{o,text{peak}}}{sqrt{2}} approx frac{115 , V}{sqrt{2}} approx 81.02 , V

]

Now, using Ohm’s law, the rms current ( I_{text{rms}} ) through the load can be calculated as:

[

I_{text{rms}} = frac{V_{o,text{rms}}}{R} = frac{81.02 , V}{2 , Omega