Consider an ideal full-bridge single-phase DC-AC inverter with a DC bus voltage magnitude of 1000 V. The inverter output voltage \( v(t) \) shown below, is obtained when diagonal switches of the inverter are switched with 50% duty cycle. The inverter feeds a load with a sinusoidal current given by, \( i(t) = 10 \sin(\omega t - \frac{\pi}{3}) \) A, where \( \omega = \frac{2\pi}{T} \). The active power, in watts, delivered to the load is ________. (round off to nearest integer)
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To calculate the active power in a system with a square wave voltage and sinusoidal current, use the formula \( P = V_{\text{dc}} I_{\text{peak}} \cdot \cos(\theta) \), where \( \theta \) is the phase angle between the voltage and current.
The formula for active power \( P \) in an AC circuit is given by:
\[
P = \frac{1}{T} \int_0^T v(t) i(t) \, dt
\]
Given that the inverter voltage is a square wave with a 50% duty cycle, the instantaneous voltage \( v(t) \) is:
\[
v(t) = \pm 1000 \, \text{V} \quad \text{(for half cycles of the square wave)}.
\]
The current is sinusoidal:
\[
i(t) = 10 \sin\left(\omega t - \frac{\pi}{3}\right) \, \text{A}.
\]
Since the square wave voltage and the sinusoidal current are in phase for half the cycle, the average active power delivered is:
\[
P = V_{\text{dc}} I_{\text{peak}} \cdot \cos(\theta)
\]
Where:
- \( V_{\text{dc}} = 1000 \, \text{V} \) is the DC voltage,
- \( I_{\text{peak}} = 10 \, \text{A} \) is the peak current,
- \( \theta = 30^\circ \) is the phase difference between voltage and current.
Substituting the values:
\[
P = 1000 \times 10 \times \cos(30^\circ) = 1000 \times 10 \times 0.866 = 8660 \, \text{W}.
\]
Thus, the active power delivered to the load is approximately \( \boxed{8660} \) W.
