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I'm supposed to calculate the voltage drop (Ux). Where R=R=R=1kΩ, C=1µF, I=1mA, E= 2sin(ωt+pi/2), ω = 10^3 rad/s.

I know I have to use superposition method, but I dont know how to calculate impedance in both situations.

1](http[![enter image description here

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    \$\begingroup\$ Which voltage drop do you want to calculate? Where does the voltage Ux occur? \$\endgroup\$ – TomS Feb 10 '16 at 9:49
  • \$\begingroup\$ Oh god, I forgot, added to the picture. \$\endgroup\$ – Vendace Feb 10 '16 at 10:19
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  • 1 Substitute current source with it's output impedance. Ideally it is infinite, so leave it open.
  • 2 Calculate currents and voltages due to voltage source.
  • 3 Substitute voltage source with it's output impedance, Ideally, zero, so replace it with a short circuit to ground.
  • 4 Calculate currents and voltages due to current source. (Hint: it is DC, so what does that mean for the capacitor?)
  • 5 Add Corresponding currents and voltages from step 2 and 4.

Where is the top resistor connected to? It seems to float. If it is not connected at the other side, it has no function in this diagram.

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  • \$\begingroup\$ I get the steps. But that top resistor makes no sense to me, I just redrew it from the task as it was shown there. Do I just ignore it when counting impedance? \$\endgroup\$ – Vendace Feb 10 '16 at 12:14
  • \$\begingroup\$ If the other terminal is not connected to anything, it carries no current. And neither does it have a voltage drop. Then yes, ignore it. \$\endgroup\$ – Bart Feb 10 '16 at 12:16
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schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

Superpositioning seems definitely right.

1st part: - Treat the DC current source as open circuit and calculate the AC impedance of the capacitor. Then calculate the voltage on X2.

2nd part: - Treat the AC voltage source as shortcut and the capacitor as an open circuit. Then calculate the potential of the node between R1 and R3

Add both two voltages and you should have an answer.

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  • \$\begingroup\$ Your second diagram is wrong. It should have a current source. \$\endgroup\$ – Bart Feb 10 '16 at 21:43
  • \$\begingroup\$ I've changed both diagrams accordingly (The variant "current source" was already mentioned in the answer text, but now also the schematics are correct. I've also modified the answer to only address a current source. \$\endgroup\$ – TomS Feb 11 '16 at 7:59

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