Um circuito é alimentado por uma fonte de tensão alternada com forma de onda dada por v(t) = 100.cos (377.t).
Suponha que foi observada uma corrente na carga com forma de onda dada por i(t) = 1.cos (377.t + 60o). Assinale a
alternativa que representa a potência ativa consumida pela carga.
Dados: sen(30o) = 0,5; cos(60o) = 0,5; sen(60o) = 0,866; cos(30o) = 0,86; √2 = 1,41.
a) 100 W.
b) 70,71 W.
c) 50 W.
d) 43,3 W.
*e) 25 W.
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We have an AC circuit problem where a voltage source with waveform v of t equals 100 cosine 377 t feeds a load. The current through the load has waveform i of t equals 1 cosine 377 t plus 60 degrees. Our goal is to calculate the active power consumed by this load.
To calculate AC power, we start with instantaneous power P of t equals v of t times i of t. For our circuit, this becomes 100 cosine 377 t times 1 cosine 377 t plus 60 degrees. The average power formula uses RMS values: P average equals V RMS times I RMS times cosine phi, where phi is the phase difference between voltage and current. RMS values relate to peak values by dividing by square root of 2.
Now we calculate the RMS values from the given peak values. For voltage, V RMS equals V peak divided by square root of 2, which is 100 divided by 1.41, giving us 70.92 volts. For current, I RMS equals I peak divided by square root of 2, which is 1 divided by 1.41, giving us 0.708 amperes. The waveform shows the relationship between peak and RMS values geometrically.
Now we analyze the phase difference between voltage and current. The voltage has zero phase angle, while the current has a positive 60 degree phase angle, meaning the current leads the voltage by 60 degrees. This is shown in the phasor diagram where the current phasor is rotated 60 degrees ahead of the voltage phasor. The power factor is cosine of 60 degrees, which equals 0.5.
Now we calculate the final active power using our formula P average equals V RMS times I RMS times cosine phi. Substituting our values: P average equals 70.92 times 0.708 times cosine 60 degrees, which equals 70.92 times 0.708 times 0.5. This gives us 50.21 times 0.5, which equals 25.1 watts, approximately 25 watts. Comparing with the multiple choice options, our answer matches option e, which is 25 watts. This confirms our calculation is correct.