Im trying to calculate the voltage through C2 in this circuit but i am unsure how to.

e(t) = 0.3 sin(ωt + 18°)V , C1= 10μF , C2= 5μF , R = 5 Ohm , ω=3*10^3 rad/s

enter image description here

This is what i got so far: frequency = 477.5 Hz

I put 0.3 sin(ωt + 18) in my calculator and got that one period takes 0.1057s so t = 0.1057s

then i put that in e(0.1057) = 0.3 sin(3000*0-1057 + 18) = -0.1263V

I calculated Xc1 and Xc2 and got them to be: Xc1 = 33.32 ohm and Xc2 = 66.67 ohm.

I then used these and calculated the impedance of R and C2 (4.9853ohm) and got the total resistance to be RT = 34.068 with 81.7 degrees.

I then used that to calculate the voltage in Uc2 with the following equation: (4.9853/34.068)*-0.1263 = -0.018482V

I did not get the correct answer though. I didnt know how to calculate this before i started so i gave it a shot, what ive done is probably completely wrong. All help would be appreciated.

  • \$\begingroup\$ Current flows through and voltage is "across" \$\endgroup\$
    – Andy aka
    Aug 3, 2016 at 12:52

1 Answer 1


I think the issue is with the parallel calculation of R and ZC2.

I agree that ZC2 = 0-j66.667, but putting that in parallel with 5+j0 yields: Z2 = 1/(1/R+1/ZC2) = 4.972 - j0.3729 or 4.986 at an angle of -4.3 degrees

From there, just multiply your input 0.3 @ 18 (radians?) by the impedance divider Z2 / ( ZC1 + Z2 )

I get 0.0439 at an angle of 77.32 degrees. Of course you still have to add back in the phase of 18 (by default such an equation would use radians - though 18 is an awful lot..., but degrees are usually specifically called out if used)

  • \$\begingroup\$ I forgot to put the degree sign after 18, sorry about that. When you say multiply your input 0.3 @ 18, what do you mean? \$\endgroup\$
    – Joo223
    Aug 3, 2016 at 13:00
  • \$\begingroup\$ no worries. So the 0.3 gets multiplied by the absolute value of the impedance divider expression to give you about 0.044. @ 18 degrees, means that you then add the 77.32 degrees to 18 for 95.32 degrees \$\endgroup\$
    – MikeP
    Aug 3, 2016 at 13:16
  • \$\begingroup\$ Ah okay, Thanks alot. Another thing im wondering is if there is a different approach to this problem? Tryding to find the period on my calculator feels really awkward. Is there an easier way? \$\endgroup\$
    – Joo223
    Aug 3, 2016 at 13:21
  • \$\begingroup\$ Nope, sorry, I don't know any easier ways. The only other way is to write the differential equations at each capacitor, and that's definitely the harder way! \$\endgroup\$
    – MikeP
    Aug 3, 2016 at 13:40
  • \$\begingroup\$ Okay thanks for the help, ill keep doing what im doing then! \$\endgroup\$
    – Joo223
    Aug 3, 2016 at 13:46

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