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The question

The circuit

I've been trying this question for a while. I'm not sure where to start. I tried redrawing the circuit differently but that did not help much.

Please ignore the fact that my drawing is not neat, it was just a rough one, I was checking if it would help before actually doing it neatly.

enter image description here

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    \$\begingroup\$ Use KCL and KVL then solve. \$\endgroup\$ Dec 28 '21 at 6:01
  • \$\begingroup\$ Thanks, but to find I1, I would first need to find the total impedance right? I don't know how to simplify this further :/ \$\endgroup\$
    – Crab00189
    Dec 28 '21 at 6:19
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    \$\begingroup\$ There are 2 loops and 3 nodes in this schematic. So 5 equations. Just write these. with 5 currents unknown. \$\endgroup\$
    – Antonio51
    Dec 28 '21 at 8:09
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    \$\begingroup\$ 2 loops for KVL. 3 nodes for KCL. Confirmed. See answer. I did not simplify anything. Let it as it is. \$\endgroup\$
    – Antonio51
    Dec 28 '21 at 10:11
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    \$\begingroup\$ Sorry @TonyStewartEE75 . You are right. I have a bad habit of not "including" current sources in loops. So, I counted (for necessary equations), 2 loops and 3 nodes. \$\endgroup\$
    – Antonio51
    Dec 29 '21 at 17:11
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I don't know what techniques you are using. But what I see quickly transforms into this:

schematic

simulate this circuit – Schematic created using CircuitLab

I converted the Thevenin source plus resistance on the left side into a Norton equivalent and labeled the nodes. Both of the current sources are in phase with each other, so that simplifies the process below.

The KCL is easy:

$$\begin{align*} \frac{V_x}{4}+\frac{V_x}{-j\,2} &= 5 + 5 + \frac{V_y}{-j\,2} \\\\ \frac{V_y}{-j\,2}+\frac{V_y}{-j\,2}+\frac{V_y}{j\,10}&=\frac{V_x}{-j\,2}+\frac{V_z}{j\,10} \\\\ \frac{V_z}{8}+\frac{V_z}{j\,10}+5&=\frac{V_y}{j\,10} \end{align*}$$

From this, it solves out as: \$V_x=\frac1{17}\left(360-j\,240\right)\$, \$V_y=\frac1{17}\left(240-j\,80\right)\$, and \$V_z=\frac1{17}\left(-360-j\,480\right)\$.

That's also the same as: \$V_x\approx 25.451 \;\angle {-}33.69^\circ\$, \$V_y\approx 14.881 \;\angle {-}18.435^\circ\$, and \$V_z\approx 35.294 \;\angle {-}126.87^\circ\$. Note that these are RMS values!!

You can work out the rest of the details, I suspect. Including the time-domain equations if you need them. (Don't forget to convert from RMS to peak values, if you attempt that.)

A short run from LTspice (using your voltage source and not my Norton equivalent) shows the following results:

enter image description here

Which is close enough to satisfy me that I didn't badly mess up the Thevenin to Norton or the following KCL.

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    \$\begingroup\$ Saying "nice, +1" would be a bit redundant at this point, but I'll just add that the measurements would more precise with an .ac list {f}. \$\endgroup\$ Dec 28 '21 at 9:59
  • \$\begingroup\$ Thanks jonk! Appreciate it! \$\endgroup\$
    – Crab00189
    Dec 28 '21 at 10:55
  • \$\begingroup\$ @aconcernedcitizen I honestly don't know how to apply that in this case. I'd like to learn more about what you mean, too. \$\endgroup\$
    – jonk
    Dec 28 '21 at 17:32
  • \$\begingroup\$ @jonk You performed a .TRAN analysis and then you used Ctrl+LClick on the waveform's label, which gave you the three shown measurements, but those will be an RMS performed on a waveform that's made of discrete points; it may suffer from timestep, compression, etc. With an .AC analysis at one frequency, you get a precise measurement of magnitude and phase, an intrinsic result, not calculated based on a measurement (I hope I manage to get some message across, I can't find all the good words now). Nothing new, just .AC insteaad of .TRAN (so nothing you didn't know, already :-) ). \$\endgroup\$ Dec 28 '21 at 19:41
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    \$\begingroup\$ @aconcernedcitizen That is one step beyond my knowledge. I never tried two AC sources before!!! Thanks! \$\endgroup\$
    – jonk
    Dec 29 '21 at 9:36
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Good job, @jonk.

Applying directly KVL and KCL.

enter image description here

Calculated directly currents. Voltages are obvious. (unless error).

Adapted for RMS values. Sources (phase=0).

enter image description here

enter image description here

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  • \$\begingroup\$ Thanks for this!! \$\endgroup\$
    – Crab00189
    Dec 29 '21 at 4:16
  • \$\begingroup\$ is V1-4= voltage across 4 ohm resistor? \$\endgroup\$
    – Crab00189
    Dec 29 '21 at 5:19
  • \$\begingroup\$ V1-4? Where? I did not use any voltage except the voltage source v(t)=V1. For the voltages, calculate only i*Z. So Voltage across 4 Ohm is = i1*4. The voltage across the inductor is = i5*j10. Be careful about the direction of polarity of the voltage vs current. \$\endgroup\$
    – Antonio51
    Dec 29 '21 at 9:46
  • \$\begingroup\$ Ohh thanks! What's Iinj tho? \$\endgroup\$
    – Crab00189
    Dec 29 '21 at 11:54
  • \$\begingroup\$ Iinj is i(t), source current "injected" in the circuit. \$\endgroup\$
    – Antonio51
    Dec 29 '21 at 12:06

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