Say we have a low pass filter circuit as the one that follows:

enter image description here

If I want to study how the charging/discharging of the capacitor works, I need to apply the universal equation for voltage. However, this implies that apart from knowing the final voltage that we are trying to reach (10 V in the example), I also need to know the initial voltage drop across the capacitor. Generally, I would say it is 0 V at t = 0, but I realize this is wrong because my Professor said that in this case, it would be -10 V at t = 0. Does this have to do with the fact that there is no ground in the circuit? I am having a hard time visualizing this.

I appreciate any help.

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    \$\begingroup\$ this circuit won't act as LPF if you feed it with DC voltage. You are right and your professor is wrong \$\endgroup\$ – Long Pham Mar 14 '18 at 7:53
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    \$\begingroup\$ Ground is a relative term, it is a reference to which all other circuit voltages are compared. In this case your ground is on the negative terminal of the voltage source (You have 10 V where? Between the top wire of the source and the bottom). \$\endgroup\$ – A.S. Mar 14 '18 at 7:59
  • \$\begingroup\$ The voltage across the capacitor at any given time will be Vcap = Vsource - Vresistor. Vc = Vs - IR \$\endgroup\$ – A.S. Mar 14 '18 at 8:00

I assume with ground, you mean a reference. All voltages are measured between two points. However, to make things easier we might want to choose one fixed point we refer all other measurements to. This is what we then often call "ground"

You see, when we talk about ground in a circuit, it's not some special point that is determined by some laws of physics. It is what we choose it to be. We say "This is what I will say is zero, and I will refer my other voltages to this point". Of course, there are often points that make sense to call ground: For example, the middle between two supplies, to give you a \$ \pm 12 \ V\$ supply. But we can equally call the most negative point the ground, and then we have a supply that gives us 12 and 24 V.

Hence, the "absence" of a ground does not force any voltages.

In your example. I assume there is a switch somewhere that closes at \$ t = 0 \$. Indeed, then we must look at the voltage equations to see how the voltage on each node behaves. It would make sense to define the node that is shared as the ground - but we don't have to. Then indeed, in an ideal world there might be some voltage across the capacitor already that can't escape. This would set our starting conditions.

When your Professor says "it is \$-10\ V\$, does he mean that in a "assume the voltage across the capacitor is \$-10\ V\$ for \$t \leq 0 \$?

  • \$\begingroup\$ His professor violates the law of Conservation of Energy too ;) \$\endgroup\$ – Long Pham Mar 14 '18 at 8:04
  • \$\begingroup\$ @LongPham there is nothing that stops us from defining a capacitor to be at that voltage at t=0. Actually building a implementation that does this would of course require some more bits as shown on the schematic, but it is possible - charge-pump based DC/DC converters do this all the time. \$\endgroup\$ – Joren Vaes Mar 14 '18 at 8:06
  • \$\begingroup\$ If you attach a voltage source to a capacitor, will you get A volt across the capacitor instantly ? No, because there is always the existence of resistance and you need infinite current to charge the capacitor to A volt \$\endgroup\$ – Long Pham Mar 14 '18 at 8:14
  • \$\begingroup\$ Oh, I do know how a charge pump works. It's about ground-shifting. \$\endgroup\$ – Long Pham Mar 14 '18 at 8:16
  • \$\begingroup\$ @LongPham again, we can always define a circuit to have a certain starting state. It doesn't tell us how to get there - but it is possible and does not break physics. You can charge a capacitor to A volt - you need a current source and a stop watch to do so. Just because the circuit shown can't, doesn't mean it is not possible. Exercises like this, which seem to be entry-level electronics, always assume things about starting conditions. Being a know-it-all and saying "but that is not possible" does not help anyone understand the problem the exercise wants to illustrate and make us understand. \$\endgroup\$ – Joren Vaes Mar 14 '18 at 8:22

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