A) 50 W
B) 86.6 W
C) 100 W
D) 200 W
Correct Answer: B
Solution :
Average power dissipated in an AC circuit \[{{P}_{av}}={{V}_{rms}}={{I}_{rms}}\cos \phi \] ... (i) where the term \[\cos \phi \] is known as power factor. Given, \[{{V}_{rms}}=100\] volt, \[R=100\,\Omega ,\,\,\phi ={{30}^{o}}\] \[\therefore \] \[{{I}_{rms}}=\frac{{{V}_{rms}}}{R}=\frac{100}{100}=1\,\,A\] Putting the values in Eq. (i), we get \[{{P}_{av}}=100\times 1\times \cos {{30}^{o}}\] \[=100\frac{\sqrt{3}}{2}\] \[=50\sqrt{3}\] = 86.6W Note: The product of \[{{V}_{rms}}\] and \[{{l}_{rms}}\] the apparent power, while the true power is obtained by multiplying the apparent power by the power factor \[\cos \phi \].
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