Question

Drive a formula of instantaneous power for single-phase AC power system in the following different loads. i. Pure resistive load ii. Pure inductive load iii. Pure capacitive load iv. RLC load

Answer 1

The instantaneous power is the power at the moment. Assume that we have a current and voltage of

$$i(t)=I\text{ sin}(\omega t+\theta_i),\\ v(t)=V\text{ sin}(\omega t+\theta_v).\\$$

The power for the pure resistive load will be

$$p(t)=v(t)i(t)=\\ =VI[\text{ cos}(\theta_v-\theta_i)-\text{cos}(2\omega t+\theta_v-\theta_i)].$$

Pure inductive load: since current lags voltage for $\theta_v-\theta_i=90°$, the instantaneous active power is 0.

Pure capacitive load: since voltage lags current for 90°, the instantaneous power is 0.

For an RLC circuit, the impedance is

$$Z=\sqrt{R^2+(X_L-X_C)^2}.$$

The phase shift then is

$$\theta=\theta_v-\theta_i=\text{arccos}\frac{R}{Z}=\\\space\\ =\text{arccos}\frac{R}{\sqrt{R^2+(X_L-X_C)^2}}.$$

Substitute this value in the common equation for power above to find the numerical value of instantaneous power.