1
\$\begingroup\$

The task looks quite simple it says:

Find the current that ammeter shows. \$ R = X_c = 10 Ω \$ \$ X_L = 20 Ω\$

The picture of circuit:

enter image description here

I tried to solve it using method of superposition:

In first case If I look at just one current source (I plug out another current source from circuit and make short circuit out of voltage source) I will have:

\$ I_A^, = I* \frac{R- jX_c + jX_l}{R - jX_c + jX_l -jX_c} = 1 + i\$.

How I think current goes for \$I_A^,\$ :https://imgur.com/a/ee8ixi8

The same thing for another voltage source I would have

\$ I_A^{,,} = 1 + i\$

https://imgur.com/a/JOMsEK5

And for voltage source (when I plug out current sources ) I would have:

\$I_A^{,,,} = \frac{U}{R - 2jX_c + jX_L} = 1\$

https://imgur.com/a/zj7fWyq

Now by the method of superposition \$ I_A = I_A^, + I _A^{,,} + I_A^{,,,} = 3 + 2 i\$ But in the textbook it says that the solution is 1.

What did I do wrong to get the incorrect solution?

\$\endgroup\$
  • \$\begingroup\$ Please add seperate pictures for each current (\$ I_A^{,}\$, etc) in which you show how the current is running. I think you've simplified the one for \$ I_A^{,,}\$ too much and you can check the signs of the summation again. \$\endgroup\$ – Huisman Aug 4 '19 at 21:01
  • \$\begingroup\$ I just did now. \$\endgroup\$ – Gustav Robert Kirchhoff Aug 4 '19 at 22:23
  • \$\begingroup\$ Check the polarity / direction for all 3 equations. \$\endgroup\$ – Kripacharya Aug 5 '19 at 4:57
  • \$\begingroup\$ You posted 3 the same links/pictures. \$\endgroup\$ – Huisman Aug 5 '19 at 9:44
  • \$\begingroup\$ Yes I did I saw that just now, sry I will change that \$\endgroup\$ – Gustav Robert Kirchhoff Aug 5 '19 at 16:06
1
\$\begingroup\$

What did I do wrong to get the incorrect solution?

Two problems. Problem #1 is a trivial thing -- in your third diagram you show current flowing into the positive terminal of your voltage source, but current should flow out of your source. (Your equation says that all the reactive elements cancel out, so if we're only worrying about the voltage source and the resistor we can see that \$I_A^{,,,}\$ should be flowing in a direction opposite of what you've drawn here.)

The second issue is that in your analysis here you get the same answer for \$I_A^{,}\$ and \$I_A^{,,}\$, but they are in fact different. It would be nice for you to deduce the answer on your own, so I'll use the handy "spoilers" feature so that you can have a chance to try it on your own first, if you'd like to.

Hint 1:

If you redraw your first two diagrams, but show the complete path taken by the current, you will see that these two diagrams have different current divider circuits at work.

Hint 2 (more or less a full spoiler):

First diagram: $$I_A^{,}=I∗\frac{R−jX_c+jX_l}{R−jX_c+jX_l−jX_c}=1+i$$ Second Diagram: $$I_A^{,,}=I∗\frac{R−jX_c}{R−jX_c+jX_l−jX_c}=1-i $$

Combine that with the result of fixing problem #1 above, and you obtain the correct answer.

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.