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
Expert's answer
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.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS