Question

A voltage e=250sinωt+50sin(3ωt+pi/3)+20 sin(5ωt+(5pi/6)) is applied to a series Circuit of resistance 20Ω and an inductance of 0.05H. Take ω=314rad/s

3.1 Derive an expression for the current.

Calculate the following:

3.2 The rms value of the current and voltage.

3.3 The total power supplied.

3.4 The power factor.

Answer 1

3.1

$e=250\sinωt+50\sin(3ωt+\frac{\pi}{3})+20 \sin(5ωt+(\frac{5\pi}{6})) \\ I_1= \frac{250 \sin ωt}{20+j 15.7}=\frac{250}{\sqrt{20^2+15.7^2}}(\sin ωt-38.13^0)= 9.83 \sin(ωt -38.13)\\ I_2= \frac{50 \sin 3ωt+60^0}{20+j 47.1}=0.977 \sin(3ωt -6.99)\\ I_3= \frac{80 \sin 5ωt+150^0}{20+j78.5}=0.246 \sin(5ωt -74.3)\\ I_{Total}= I_1+I_2+I_3\\ I_{Total}=9.83 \sin(ωt -38.13)+0.977 \sin(3ωt -6.99)+0.246 \sin(5ωt -74.3)$

3.2

$I_{rms}=\sqrt{(\frac{9.83}{\sqrt2})^2+(\frac{0.977}{\sqrt2})^2+(\frac{0.246}{\sqrt2})^2}=6.98 A\\ V_{rms}=\sqrt{(\frac{250}{\sqrt2})^2+(\frac{50}{\sqrt2})^2+(\frac{20}{\sqrt2})^2}=180.83V\\$

3.3

$P_{TOTAL}=V_{rms}*I_{rms}=180.83*6.98=1253.15 VA$

3.4

$250\sinωt\\ \phi= \tan^{-1}(\frac{ωL}{R})=38.13\\ \implies \cos \phi = 0.7866\space lagging\\ 50\sin(3ωt\frac{\pi}{3})\\ \phi= \tan^{-1}(\frac{3ωL}{R})=66.99\\ \implies \cos \phi = 0.3908\space lagging\\ 20 \sin(5ωt+(\frac{5\pi}{6})) \\ \phi= \tan^{-1}(\frac{5ωL}{R})=75.7\\ \implies \cos \phi = 0.2469 \space lagging\\$