I have this problem:

enter image description here

And I would like some help with how to proceed.

My own thinking:

I think Io = 2cos (wt) mA should be 2 mA. For VR (t) I think it should be VR (t) = 2 mA * 1000 = 2 V and for the capacitor I think I should use 1 / jwC = -j 1 / wC = -j 1/10 ^ 5 * 10 * 10 ^ -9 = -j10 ^ 3 ohm = -j kohm and to get Vc (t) it becomes Vc (t) = 2 mA * -j kohm = -2j V.

Am I thinking right or am I on the wrong path?

  • \$\begingroup\$ Yes, you are thinking right \$\endgroup\$ May 18, 2020 at 16:16

2 Answers 2


You are right but just keep in mind for solving these kind of problems the solution used is called "Sinusoidal Steady-State Analysis" and you can google it if you're uncertain about how it works. But overall what you did is right except that don't forget the voltage across the resistor is not just 2V (it's not a DC voltage). 2V is in phasor, so you have to convert it back to Time-domain form which is 2cos(wt). Just like this convert the voltage for capacitor and inductor too.

General format for time domain is:

$$ A=M\cos(\omega t\;+\;\theta) $$ which is converted to the phasor form: $$ A=M\sphericalangle\theta $$

In order to convert, this is how it's done for the voltage across the resistor:

$$ I_{o}=2\cos(\omega t)\quad mA $$ $$ I_{o}=2\sphericalangle0 \quad mA$$ $$ V_{R}=2\sphericalangle0\;mA\times1k\Omega=2\sphericalangle0\quad (V) $$ $$ \Longrightarrow V_{R}=2\cos(\omega t)\quad (V) $$

Try to do the capacitor and inductor yourself and note that the phase is not zero anymore for those devices.

  • \$\begingroup\$ Hello, how did you convert from phasor to time domain? \$\endgroup\$
    – Vetenskap
    May 18, 2020 at 17:44
  • \$\begingroup\$ I would really appreciate if you could answer me. :) \$\endgroup\$
    – Vetenskap
    May 18, 2020 at 18:28
  • \$\begingroup\$ @Vetenskap I edited my answer, hope it helps. \$\endgroup\$
    – Ali Nategh
    May 18, 2020 at 18:47
  • \$\begingroup\$ For the capacitor I got it to be, 2j V < -90 degrees. However, I'm a little confused if I should consider the answer to be Vc(t) = 2cos(wt -90)? \$\endgroup\$
    – Vetenskap
    May 18, 2020 at 18:52
  • \$\begingroup\$ @Vetenskap Yes if you convert -2j (which is polar form) to rectangular form( which i used in my answer) it will be 2<270 degrees OR 2<-90 degrees. So yes Vc(t)= 2 cos(wt-90) is correct. \$\endgroup\$
    – Ali Nategh
    May 18, 2020 at 21:27

As far as I see, you don't need to use complex calucalations here

The Voltages on a Resistor, a Capacitor and an Inductive are defined as follows in the time-domain:

$$ V_R(t)=R*i(t) $$ $$ V_C(t)=\frac{1}{C}*\int_{}{}{i(t)}{dt} $$ $$ V_L(t)=L*\frac{di(t)}{dt} $$

In your example, all components are in series and so the current through each component is the same: $$i(t) = 2cos(\omega t) $$

So it should be done by simple inserting \$i(t)\$ in the formulas above:

$$ V_R(t)=2R*cos(\omega t) $$ $$ V_C(t)=\frac{2sin(\omega t)}{\omega C}$$ $$ V_L(t)=-2\omega Lsin(\omega t) $$

  • \$\begingroup\$ For Vc(t) = 2sin(wt)/wC, when you say insert i(t), do you mean 2 mA? \$\endgroup\$
    – Vetenskap
    May 18, 2020 at 17:40
  • \$\begingroup\$ No. You have Vc(t)=1/C * integral I(t)dt and I(t) = 2cos(wt). So by inserting I(t) in the first formula, you get: Vc(t)=1/C * integral 2cos(wt)dt By solving the integral, you get Vc(t)=2sin(wt)/wC \$\endgroup\$
    – Feuerlink
    May 18, 2020 at 17:45
  • \$\begingroup\$ I'm thinking that I got -2j V to be the voltage over the capacitor. However, you're telling me that Vc(t) = 2sin(wt)/wC, so does that mean that I have to insert the known C and w to calculate? I've got a little trouble understanding due to me being new to this chapter in electronics. \$\endgroup\$
    – Vetenskap
    May 18, 2020 at 18:01
  • \$\begingroup\$ Here, you do not need to calculate anything with complex-integers. -2jV is complex (because of the j) You can do that, but it is not necessary here, because in this exercise you should calculate V(t) (and not V(jw) of every component. \$\endgroup\$
    – Feuerlink
    May 18, 2020 at 18:07

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