# 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

1
2020-09-29T14:08:00-0400

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

"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\\\\\n=\\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.

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