If the voltage

\$v_I = \frac{jωC1R3}{1 + jωC1R3}*Vs \$

does that mean the voltage across the capacitor would be equal to \$v_S - v_I\$ ?

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  • 2
    \$\begingroup\$ Welcome to EE.SE! Please pay attention that current is though. Voltage is across. \$\endgroup\$
    – winny
    Apr 8, 2020 at 6:46
  • \$\begingroup\$ The problem with though, through, thought, tough is that they all pass the spelling test. \$\endgroup\$
    – Oldfart
    Apr 8, 2020 at 6:51
  • \$\begingroup\$ Remember that in steady-state the average current through an ideal capacitor is always zero. And you can use a superposition to find the DC component and AC component separately. electronics.stackexchange.com/questions/301921/… \$\endgroup\$
    – G36
    Apr 8, 2020 at 7:30

1 Answer 1


The voltage across the cap (not through it) is, by definition vS - vI. So yes.

  • \$\begingroup\$ Because the capacitor is DC blocking does that mean its voltage value will be the real part of the complex voltage? \$\endgroup\$
    – Simba N
    Apr 8, 2020 at 7:04
  • \$\begingroup\$ The direct voltage across the capacitor will be the direct voltage present in \$\small v_s\$. The real part of \$\small v_s\$ is not the direct voltage, it's the in-phase component. \$\endgroup\$
    – Chu
    Apr 8, 2020 at 8:43

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