- #1
MacLaddy
Gold Member
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Homework Statement
Hello folks. I had a recent midterm where I got a problem wrong, but I'm not sure I understand why. I am hoping someone here can shed some light. Please see the attached circuit diagram.
Homework Equations
Phasor notation stuff... I am correct on all except one sign. Please see below.
The Attempt at a Solution
Using superposition, and considering the sources as Case 1, and Case 2, as noted in the image.
Case 1:
[itex]12cos(2000t)=12\angle{0^\circ}[/itex]
[itex]cap= -40j[/itex]
[itex]Ind= 10j[/itex]
Using KVL
[itex]-12\angle{0^\circ}+(-40j)I_1+30I_1+10jI_1=0[/itex]
[itex]I_1=0.28\angle{45^\circ}[/itex]
[itex]I_1=0.28cos(2000t+45^\circ)[/itex]
Case 2:
[itex]18cos(4000t)=18\angle{0^\circ}[/itex]
[itex]cap= -20j[/itex]
[itex]Ind= 20j[/itex]
Using KVL
[itex]-18\angle{0^\circ}+(-20j)I_2+20jI_2+30I_2=0[/itex]
[itex]I_1=0.6\angle{0^\circ}[/itex]
[itex]I_1=0.6cos(4000t)[/itex]
Final answer
[itex]i(t)=0.28cos(2000t+45^\circ)+0.6cos(4000t)[/itex]
However, my instructor says that the answer above should have a negative sign instead of a positive. Like this...
[itex]i(t)=0.28cos(2000t+45^\circ)-0.6cos(4000t)[/itex]
He explained that he is following KVL by using the sign directions that are drawn on the diagram. This seems wrong to me, as that would make every element in the second equation positive.
[itex]18\angle{0^\circ}+(-20j)I_2+20jI_2+30I_2=0[/itex]
But shouldn't the power source always have a different sign than the other elements? If I apply KVL to the above it would indicate that either everything is applying voltage to the element, or everything is removing it.
Any hints on where my thinking is incorrect?
Thanks,
MacLaddy