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ELEC 273 |LAB EXAM TIPS EXPERIMENT 5 |AC POWER MEASUREMENTS| CONCORDIA UNIVERSITY CA$15.85   Add to cart

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ELEC 273 |LAB EXAM TIPS EXPERIMENT 5 |AC POWER MEASUREMENTS| CONCORDIA UNIVERSITY

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ELEC 273 |LAB EXAM TIPS EXPERIMENT 5 |AC POWER MEASUREMENTS| CONCORDIA UNIVERSITY

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  • May 15, 2024
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  • 2023/2024
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  • elec 273
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ELEC 273 LAB EXAM TIPS EXPERIMENT 5 AC POWER
MEASUREMENTS CONCORDIA UNIVERSITY

, OBJECTIVES:

1- To measure AC Power on Resistive, Inductive and Capacitive

Loads 2- The Power-Factor-Correction of an Inductive Load



INTRODUCTION:



AC power absorbed by a load is expressed by the following equation:

P(t) = V(t) i(t)

The instantaneous power is the rate at which an element (load) absorbs energy. A load is
composed of dissipative and reactive elements, and as a result, it can be expressed by an
impedance Z = R +jX = |Z|∟θ .˚

Working in the time domain, if the voltage and current associated with the load

are: v(t) = Vp cos (t + v ) and i(t) = Ip cos(t + i)

where Vp and Ip are peak values and v and i are phase angles, then:

p(t) = VpIp cos (t + v) cos(t + i) = (VpIp/2) cos(v - i) + (VpIp/2) cos(2t +v + i).

The difference between the voltage phase angle and current phase angle, in the time
domain, is actually the angle of the impedance Z in the phasor domain. By expressing the
voltage and current in their RMS values, p(t) can be re-written as:

p(t) = VI cos θ + VI cos (2t + θ) = P + VI cos (2t + θ)

Vp Ip
where the RMS values of the voltage and current are : V
= √2 and I = √2

and where (v - i) = θ and P = VI cos θ.

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