Here is the circuit which caused a long sleepless night and I would like to explain why:

enter image description here

I have seen similar schema in this question (How to calculate time constant for RC circuit with more than one resistor), however, the author was concerned about the Thévenin resistance. My question is about the Thévenin voltage. It is known that the capacitor can be represented as open circuit in DC conditions. I remove the capacitor from the circuit and obtain two voltage sources in series and two resistors in series. Then the voltage on the capacitor terminals is

V = (15 + 7.5) - (15 + 7.5) * 2/(2+6) = 16.875 V

However, the correct answer is 9.375 V which is calculated in the following manner. We assume that the ground is in the bottom of the schema. Then we assume that the capacitor node is still there and there are 3 currents: from 15 V source, fvrom 7.5 source and into capacitor. And then we write the equation for the capacitor node:

(Vc - 15)/2 + (Vc + 7.5)/6 = 0    ; 0 because no current flows through the capacitor

Vc = 9.375 V

Here is the simulation in LTSpice:

enter image description here

And here is the question: why should I assume that the ground is in the bottom of the schema? If I will ground the circuit just after the second resistor, I will got the value I calculated from the very beginning:

enter image description here


1 Answer 1


The currents are all the same through the components. The relative voltages between each node are the same. The difference is the absolute voltages are not the same. No matter where you place the ground, the nodes will have the same voltages relative to one another. The ground is a point that we decide to call 0V an reference all voltages to that node.

The relative voltages are the same V1 still has 15V across it in both simulations, R2 always has 5.625V (at DC deferentially) and R1 always has 16.875V deferentially across it, no matter where you place the ground

It does not matter where you place ground if there is no currents flowing into it (like if you have two grounds in the circuit. If there is one ground, you can place it anywhere and the circuit will function the same.

If you don't have a ground, the spice simulation will have an error. That is because the spice simulator makes equations from the circuit (like a human would) and then solves them. A human nor computer can solve these equations without a ground,because the solution for the system of equations will be indeterminate.

  • \$\begingroup\$ But as you see, the position of the ground does play the role. In the first case I have two parallel branches, where each branch contains a voltage source and a resistance in series. In the second case it is simple connection of Voltage sources and resistances in series. \$\endgroup\$
    – tenghiz
    Aug 1, 2019 at 4:22
  • \$\begingroup\$ @tenghiz: the ground symbol just indicates the point in the circuit that we choose to call "zero volts", and use as a reference when measuring voltages elsewhere in the circuit. \$\endgroup\$ Aug 1, 2019 at 5:04
  • \$\begingroup\$ The relative voltages are the same V1 still has 15V, R2 always has 5.625V at DC and R1 always has 16.875V across it, no matter where you place the ground \$\endgroup\$
    – Voltage Spike
    Aug 1, 2019 at 5:06
  • \$\begingroup\$ I see, LTSpice also shows that there is the same current across all the elements. However, why should I use different approach to calculate a voltage in each case? \$\endgroup\$
    – tenghiz
    Aug 1, 2019 at 5:22
  • \$\begingroup\$ @Peter Bennett, that's why I ask: how should I know where does the author of the problem connects the ground? Is there any convention? Basically I can attach ground to any point of the circuit. \$\endgroup\$
    – tenghiz
    Aug 1, 2019 at 6:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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