Question

A sinusoidal voltage source with instantaneous voltage v(t)=120 √2sin(120πt+ 30∘), is applied to a series RL load and the resulting current in Fourier series form is i(t)=4+6 √2sin(120πt)+2 √2sin(240πt).

Determine
a. The average power absorbed by the load

Answer 1

The average power absorbed by the RL load, when a sinusoidal voltage source is applied, is calculated using the rms values of the voltage and the in-phase component of the current. In this case, the average power absorbed by the load is 720 watts.

Step-by-step explanation:

To determine the average power absorbed by the load when a sinusoidal voltage source with instantaneous voltage given by v(t) = 120 √2sin(120πt+ 30°), is applied to a series RL load, and the current is given by the Fourier series i(t) = 4 + 6 √2sin(120πt) + 2 √2sin(240πt), the only contribution to average power comes from the component of the current that is in phase with the voltage. In this case, it would be the term 6 √2sin(120πt).

Using the formula for average power Pave = Irms Vrms cos(φ) and noting that the phase angle φ between the voltage source and the in-phase current component is 0 degrees (cos(0) = 1), we can calculate the average power.

The root mean square (rms) values of the voltage and current are determined by dividing the amplitude by √2. Therefore, Vrms = 120 V and Irms = 6 A. Substituting these into the average power formula yields Pave = 120 V × 6 A × 1 = 720 W. So the average power absorbed by the load is 720 watts.