Final answer:
The real power P of the source is calculated to be -234.317073 W, indicating that the source absorbs the power, while the reactive power Q is 1329.27439 var, indicating that the source delivers reactive power.
Step-by-step explanation:
To calculate the real and reactive power of a single-phase voltage source, we use the voltage V and the current I as well as the phase difference between them. The voltage is given as V=45∠120° volts and the current is I=30∠20° amps.
The real power P (in watts) is calculated using the formula:
P = Vrms * Irms * cos(φ), where Vrms and Irms are the rms values of the voltage and current, and φ is the phase difference between the voltage and the current.
For a sine wave, the rms value is equal to the peak value divided by √2. However, since we are given the peak values in rms form, we do not need to convert them.
To find the phase difference φ, we subtract the angle of the current from the angle of the voltage: φ = 120° - 20° = 100°.
Now we calculate the cos(φ): cos(100°) = -0.1736481777.
The real power P is:
P = 45 * 30 * (-0.1736481777) = -234.317073 W
The negative sign indicates that the source is actually absorbing power rather than delivering it.
The reactive power Q (in volt-amperes reactive, var) is given by:
Q = Vrms * Irms * sin(φ).
sin(100°) = 0.984807753.
So, the reactive power Q is:
Q = 45 * 30 * 0.984807753 = 1329.27439 var
The source delivers reactive power since the sine of the phase angle is positive.