While studying for my end-term exam through past year papers(of my college). I came across this question in which we need to do AC Analysis.

Need to find out the values of Resistor (R) and Capacitor (C), also the phase angle between Vs and Vb, as well as Vb and I.

Info given:

  • |Va|related to|Vb|
  • Phase difference ∠Vb -∠Va

I did try researching on the Internet but I do not have the correct keywords I guess. Couldn't even find anything here on EE.SE either.(Internet links as answer would be most appreciated)

My Attempt

After converting to phasors(omega=100π rad/s),the first thing I can think of is to use Voltage Division Rule as Vs is given[in terms of 240sin(100πt)] then equate according to |Va| =|Vb|/3--but I get an equation in 2 variables(R and C).

How do I use the phase difference given? I can easily find other phase differences once I figure out R and C.

Edit: The exact Question is uploaded on Chegg--sadly I don't have a Chegg account.

  • \$\begingroup\$ If 90 deg then impedances must form a right angle triangle. \$\endgroup\$ Commented Jun 4, 2021 at 18:29
  • \$\begingroup\$ What do you know about the current through each element? What does that tell you about the relative impedances and their angles? We don't hand out homework solutions here; we expect you to do a fair amount of work. \$\endgroup\$ Commented Jun 4, 2021 at 18:33
  • \$\begingroup\$ It's not exactly homework but okay. (1) V = iZ (Thanks) ; (2) I'm still trying to figure out wdym. \$\endgroup\$
    – Elrond
    Commented Jun 4, 2021 at 19:08
  • \$\begingroup\$ It's the other way around. Using phase difference, you have to get the second equation. Two unknown variables, two equations..solve.. \$\endgroup\$
    – Mitu Raj
    Commented Jun 4, 2021 at 19:13
  • \$\begingroup\$ Just try making the RC impedance 90 degrees out of phase with the LC impedance and the adjust to get the right magnitude. \$\endgroup\$
    – ErikR
    Commented Jun 4, 2021 at 20:02

1 Answer 1


Let \$Z_{LR}\$ and \$Z_{RC}\$ be the impedances. Then

$$ V_b = \frac{Z_{LR}}{Z_{LR}+Z_{RC}}V_{in} \quad\text{and}\quad V_a = \frac{Z_{RC}}{Z_{LR}+Z_{RC}}V_{in} $$ so $$ \frac{V_b}{V_a} = \frac{Z_{LR}}{Z_{RC}} = 3i $$ (or maybe it's \$-3i\$.) Now just solve for \$Z_{RC}\$.

  • \$\begingroup\$ Thank you my good sir. I was actually trying to figure out what you said in your comment for the past 20 mins. Btw, here LC refers to RL I believe and i is iota(usually represented by 'j' in EE) \$\endgroup\$
    – Elrond
    Commented Jun 4, 2021 at 20:32
  • \$\begingroup\$ You're right about LC vs. LR -- I'll change it! \$\endgroup\$
    – ErikR
    Commented Jun 4, 2021 at 20:33

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