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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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ELEC 273 LAB EXAM TIPS EXPERIMENT 5 AC POWERMEASUREMENTS 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(2t +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 (2t + θ) = P + VI cos (2t + 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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