There is a multiple choice question: $$V_{G1}=10cos(10t+60°)$$ and


The resistance of the wire is \$R=10Ω\$, which one is the power that \$G_1\$ transmits?

A. 1.25W

B. 1.5W

C. 2.5W

D. 3.25W

The answer for this question is (C)2.5W

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Solution provided from the book


So the \$V_{rms}\$ of \$V_{G1}\$ is \$V^{(1)}_{rms}=\frac{10∠60°}{\sqrt{2}}=5\sqrt{2}∠60°V\$

$$V_{G2}=5cos(10t)=5∠0°$$ So the \$V_{rms}\$ of \$V_{G2}\$ is \$V^{(1)}_{rms}=\frac{5∠0°}{\sqrt{2}}=2.5\sqrt{2}∠0°V\$

Now the \$S_{12}=V^{(1)}_{rms}I^*=V^{(1)}_{rms}(\frac{V^{(1)}_{rms}-V^{(2)}_{rms}}{R∠0°})^*=5∠0°-2.5∠60°\$,so \$P_{12}=5cos0°-2.5cos60°=3.75W\$,

But there is no option that \$P_{12}=3.75W\$, so this question has a little mistake

Now my classmate said no, there is no mistake in this question. The book provided the wrong solution so the author thinks this question has a little mistake. My classmate's solution is as below:

My classmate's solution

\$I=\frac{10-5}{10}=0.5\$,So \$P=I^2R=(0.5^2) \times 10 =2.5W\$,so the answer for this question should be (C)2.5W

The author of this book didn't design this question,and I want to ask which solution is right because I think both of their solutions make sense. What I don't know is which one is correct for this question? If the answer is 2.5W,and what should we modify the question to let the answer become \$P_{12}=3.75W?\$?

my friend teaches me how to calculate this question,so my solution is the same as my classmate,but not the same as the solution from the book.Both solutions can convince me that their solution are right,however,there is only one true answer,that is why i ask here

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    \$\begingroup\$ Why is your classmate the only one who attempted this? You haven’t explained what your solution is. Also saying “what should we modify the question to let the answer become” is a red flag that you really don’t care about the subject at hand... \$\endgroup\$ – user103380 May 20 '20 at 14:51
  • \$\begingroup\$ It's been long since I've done something like that, but neglecting the phase difference between the voltages seems a bit too easy. \$\endgroup\$ – Arsenal May 20 '20 at 14:56
  • \$\begingroup\$ @KingDuken I don't know why are you so angry,i ask here just because i don't know which one is correct,because "i think" both of their solution are right. \$\endgroup\$ – XM551 May 20 '20 at 15:26
  • \$\begingroup\$ Also i do care,so i ask in here,i think the author's solution and my classmate solution's should be both right ,however,there is only one true answer,so i want to ask the different between them \$\endgroup\$ – XM551 May 20 '20 at 15:28
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    \$\begingroup\$ The reason they are annoyed is because this is homework but you have not shown your attempts, only other people's and they are basically wondering why. \$\endgroup\$ – DKNguyen May 20 '20 at 15:30

Unless you know for a fact that you can ignore the phase relationship when the frequencies are the same, do not use a method that assumes that. I could not quite visualize if it was true or not visualizing how the two voltage sine waves overlap in my head, but writing out the phasor math makes it pretty clear that you cannot.

The author's answer is right. The answer list is wrong. I was too lazy to write out my own math but read the author's answer and checked in a simulator:


There are multiple ways to get this answer and most of them do not need to assume anything about the phase. The first is the author's method using complex power calculations.

The second way is to use do a circuit analysis with phasors (i.e. frequency domain circuit analysis where voltage sources appear as phasors \$V_{pk}\angle\theta = V_{pk}e^{j\theta}\$ , inductors appear as \$j\omega\ell\$ and capacitors appear as \$\frac{1}{j\omega C}\$).

The third way is to just do a time-domain circuit analysis which is easy since you just have a resistor and no capacitors or inductor. You would work out the equations for time-dependent equations for current and along with the time dependent equations for both voltage sources:

Instantaneous power: \$ p(t) = v(t) \times i(t)\$

Energy over one period = \$ \int_0^T{p(t)}dt\$

Average Real Power = \$\frac{Energy}{T}\$

  • \$\begingroup\$ wow,what is the name of that simulation tool? \$\endgroup\$ – XM551 May 21 '20 at 3:46
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    \$\begingroup\$ @XM551 Falstead circuit simulator. It's free and online. \$\endgroup\$ – DKNguyen May 21 '20 at 3:56

I did the phasor analysis myself (shown below) and obtained 3.75W. Sometimes books have mistakes in them. All the author has to do is accidentally write a 2 instead of 7 in the ten’s decimal place and have editor not acknowledge the mistake. When I was in school my differential equations book had mistakes in it that the professor has to correct!

Unfortunately I am unable to rotate image. enter image description here


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