To find the fundamental load power of a single-phase full bridge inverter with an RLC load, we first need to determine the total impedance (Z) of the load.

1. Calculate the reactance (X):

[

X = X_L – X_C = 11 , Omega – 20.54 , Omega = -9.54 , Omega

]

2. Calculate the total impedance (Z):

[

Z = sqrt{R^2 + X^2} = sqrt{4^2 + (-9.54)^2} = sqrt{16 + 91.52} = sqrt{107.52} approx 10.37 , Omega

]

3. Calculate the current (I) using the RMS value of the voltage:

[

I = frac{V_{dc}}{Z} = frac{230 , V}{10.37 , Omega} approx 22.17 , A

]

4. Calculate the power factor (pf):

[

pf = cos(theta) = frac{R}{Z} = frac{4 , Omega}{10.37 , Omega} approx 0.385

]

5. Calculate the fundamental load power (P):

[

To find the rms value of the fundamental load current for a single-phase full bridge inverter with an RLC load, we can follow these steps:

1. Calculate the output voltage of the inverter (Vm): For a full bridge inverter, the peak output voltage is equal to the DC input voltage. Thus,

[

V_m = V_{dc} = 230 , V

]

2. Calculate the angular frequency (ω): The angular frequency for the output frequency (f) is calculated as:

[

omega = 2pi f = 2pi times 50 approx 314.16 , text{rad/s}

]

3. Calculate the reactance of the inductor (XL): The reactance (X_L) is calculated as:

[

X_L = omega L = 314.16 times 35 times 10^{-3} approx 10.99 , Omega

]

4. Calculate the reactance of the capacitor (XC): The reactance (X_C) is calculated as:

[

X_C = frac{1}{omega C} = frac{1}{314.16 times 155 times 10^{-6}} approx 20.43 , Omega

]

5. **Calculate the total impedance (Z

To find the power delivered to the load in watts due to the fundamental component of the load current in a single-phase full-bridge inverter circuit, we can use the formula for power:

[

P = I_{rms}^2 times R

]

However, we first need to determine the root mean square (rms) value of the fundamental component of the current. For a full-bridge inverter with a resistive load, the rms current can be derived from the dc source voltage (( V_s )) and the load resistance (( R )).

For a full-bridge inverter, the output fundamental voltage ( V_{1} ) is given by:

[

V_{1} = frac{V_s}{pi}

]

Substituting ( V_s = 230 , V ):

[

V_{1} = frac{230}{pi} approx 73.24 , V

]

Now, we can find the fundamental component of the current:

[

I_{1} = frac{V_{1}}{R} = frac{73.24}{2} approx 36.62 , A

]

The rms current ( I_{rms} ) is currently the same as the effective current flowing through the resistive load since it primarily consists of the fundamental component in a purely resistive load. To find power:

[

P = I_{r

To find the rms value of the fundamental load current in a single-phase full bridge inverter, we can use the formula for the maximum output current, which is given by:

[ I_{text{max}} = frac{V_s}{R} ]

where:

– ( V_s ) is the DC voltage source,

– ( R ) is the load resistance.

Given:

– ( V_s = 230 , V )

– ( R = 2 , Omega )

First, we calculate ( I_{text{max}} ):

[

I_{text{max}} = frac{230 , V}{2 , Omega} = 115 , A

]

The rms value of the fundamental load current for a full bridge inverter is given by:

[

I_{rms} = frac{I_{text{max}}}{sqrt{2}} approx frac{115 , A}{sqrt{2}} approx 81.02 , A

]

Thus, the rms value of the fundamental load current is approximately:

81.02 A