I understand that this is really basics of electronics, but I would really appreciate if somebody clearly explained following question.

So here is a small set of questions related to similar topic:

  1. What's the difference between 1 and 2 terminal source? As far as I understood one wire is the max-potential(+) and the second is low-potential(-). But what is the low-potential? Is it usually 0?

  2. I tried to simulate a simple circuit with capacitor charge/discharge function. In one case I tried to use 2 terminal source, assuming that low-potential ~ virual ground. The capacitor did not discharge. In the second case I used 1 terminal source and ground - capacitor successfully discharged. Why the capacitor did not discharge in the first case?

  3. Is 0V the same as ground?

I have added 2 images below here illustrating the problem.

2 terminals 1 terminal/grounded


The one-terminal source is referred to ground, which has a specific symbol that you also used.

  1. The 1-terminal supply is like a battery in which the negative terminal is automatically connected to ground. The 2-terminal is more like a battery: you can stack more or put it inside a circuit, so it simply creates a voltage. Then you have to reference it to ground if you want.

  2. In the first circuit, the capacitor is connected between the low terminal of the voltage source and ground, so basically it's not supplied by anything. You have to put a ground symbol in the negative terminal of the voltage source, to make the circuits equivalent.
    As I said in n. 1, since the 2-terminals supply is not referenced, you are not creating any potential across the capacitor.

  3. Especially in this case, yes, 0 V is ground.

P.S. It's not the best to put horizontal grounds, consider turning it.

This is the equivalent of your second circuit with a 2-terminal supply

  • \$\begingroup\$ Thanks. In the first picture with 2-terminal source. One terminal is positive. But what about the second terminal? Isn't it the same as ground (and 0V)? Then why do I need to add (another) ground there? Sorry, I just want to understand it completely. \$\endgroup\$ Apr 24 '12 at 15:35
  • \$\begingroup\$ @Arturs no; the supply simply creates a voltage between the terminals. You have to put the negative one to ground if you want, otherwise it's just floating. Consider it like a battery. \$\endgroup\$
    – clabacchio
    Apr 24 '12 at 15:37
  • \$\begingroup\$ Could you please explain, what do you mean by "floating"? I thought, if there is -any- potential difference, current will flow. \$\endgroup\$ Apr 24 '12 at 15:39
  • 1
    \$\begingroup\$ @Arturs the problem is that in the real world, they would have a relative potential; but in this case, it's like the supply has voltage X and X+5, but you don't know X and it doesn't close the circuit with GND. \$\endgroup\$
    – clabacchio
    Apr 24 '12 at 15:43
  • \$\begingroup\$ Right. It makes it clear. The last thing - What would happen with the Circuit Nr.1 in the real world? What would be X in the real world? Would it discharge the capacitor? \$\endgroup\$ Apr 24 '12 at 15:48

Your simulation problem is that you've defined GND separately from your negative battery terminal in the first case. So it's not clear what the relationship between those two nodes is. I think what you wanted to do in the first case is also connect the negative terminal to GND which should behave the same way as your "one-terminal" simulation.

  • \$\begingroup\$ Does it mean, that if I want to get -5V from the source, I just add GND to the positive terminal? \$\endgroup\$ Apr 24 '12 at 15:37
  • \$\begingroup\$ @Arturs exactly :) \$\endgroup\$
    – clabacchio
    Apr 24 '12 at 15:48
  • \$\begingroup\$ @Arturs - Basically yes. For example, you could connect the + os one 9v battery to the - of another, and call that point GND. Then you'd have ±9v. \$\endgroup\$ Apr 24 '12 at 15:49

In your first circuit, you have not connected the negative terminal of the voltage source to ground. It's only connected to the node that you drew it connected to.

Because of that, the capacitor is not connected to any current loop. There's no way to charge or discharge it, because there's no complete circuit in the design that includes the capacitor.

In fact, the only way the simulator is able to simulate this circuit at all is by adding an extra very-high-value resistor in parallel with the capacitor. Without this added resistor, none of the nodes in the circuit would have a defined voltage relationship to ground.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.