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 '20 at 6:46
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\$\begingroup\$ The problem with though, through, thought, tough is that they all pass the spelling test. \$\endgroup\$ – Oldfart Apr 8 '20 at 6:51
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\$\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 '20 at 7:30
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The voltage across the cap (not through it) is, by definition vS - vI. So yes.
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\$\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 '20 at 7:04
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\$\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 '20 at 8:43
