A single phase half bridge inverter has load R = 2 Ω and a dc voltage source Vs/2 = 115 V. Find the power delivered to the load due to the fundamental component.

To find the power delivered to the load due to the fundamental component for a single-phase half-bridge inverter, we can use the following steps:

1. Determine the peak phase voltage (Vp) delivered to the load:

[

V_p = frac{V_s}{2} = 115 , V

]

2. The RMS value of the fundamental component for a half-bridge inverter is:

[

V_{rms} = frac{V_p}{sqrt{2}} = frac{115 , V}{sqrt{2}} approx 81.02 , V

]

3. The power ( P ) delivered to the load can be calculated using the formula for power in terms of resistance and voltage:

[

P = frac{V_{rms}^2}{R}

]

4. Substituting the values we have:

[

P = frac{(81.02 , V)^2}{2 , Omega} approx frac{6564.32}{2} approx 3282.16 , W

]

Thus, the power delivered to the load due to the fundamental component is approximately 3282.16 W.