Answer to Question #135244 in Electrical Engineering for bella

Question #135244

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

Expert's answer

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),\\\\\nv(t)=V\\text{ sin}(\\omega t+\\theta_v).\\\\"

The power for the pure resistive load will be

"p(t)=v(t)i(t)=\\\\\n=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\u00b0", 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


The phase shift then is


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

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