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At the circuit above I built a simulation that shows the voltages on both sides of the capacitor(in red and green). However the voltage measurement shows that the voltage on the 3u capacitor is only 1[V] I couldn't understand why is that (and not 6[V])?

Important fact - it is a steady state condition.

Thanks a lot.


Your nodal analysis is incorrect. For steady-state resistive analysis, we can remove all of the capacitors from the circuit. This quickly shows us that the green node is at 4V with respect to the negative terminal of the 12V supply. This bold part is important.

Now, on the left hand side, we effectively have a 15V supply into three series 5k resistors - the 2k and the 3k combine in series and the capacitor has no effect. This puts the red node at 10V with respect to the negative terminal of the 15V supply. Notice that our reference point for the potential is different for the red and green node - so the voltages can not be used in the same equation just yet.

For that, we have to pick a common reference point (which if I were doing this analysis I would mark using a signal ground symbol to help me think). For convenience, let's make the negative terminal of the 12V supply the reference point - which means the green node voltage needs no transformation.

The red node does, however. We know that 5V is dropped across each of the 5k resistors from Kirchoff's laws, so across the 5k resistor between the red node and our newly chosen reference is 5V. So now we can see that with 5V wrt our reference at the red node and 4V wrt the same reference at the green node, we have 5V - 4V = 1V across the 3uF capacitor.


You need to use the same reference for both measurements (e.g. circuit ground) I suspect you are measuring from the bottom of each voltage source, rather than from circuit ground (wherever that is) Assuming circuit ground is at the bottom of the 12V source, then you will read 5V at the 3k/5k resistor junction, making 1V across the 3uF capacitor correct.


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