# Analyzing a circuit with 1 point earthed

• Krushnaraj Pandya
In summary: The answer says +4V, also at point A, the answer says the potential is -6V. This is just a mirror to what you said, though. Can you explain how you got...The answer says +4V, also at point A, the answer says the potential is -6V. This is just a mirror to what you said, though. Can you explain how you got from -6V at point A to +4V at point B?The answer says +4V, also at point A, the answer says the potential is -6V. This is just a mirror to what you said, though. Can you explain how you got from -6V at point A to +

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## Homework Statement

Find potential of A and B in the given figure. (pardon poor quality, I only have a webcam)

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## The Attempt at a Solution

marking 1 to 4 from left to right. C1,C2,C3 are in series. C123=(2/3)C. which is in parallel with C4. C effective is 5C/3. What now? I don't know what to make of the point between them (lets call it D) being earthed.

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The voltage across a resistor at time t depends on the current at time t, however the voltage across a capacitor depends on past current so you need to make some assumptions about how the system starts off.

There is only one connection to Earth so what does that mean for the current flowing to/from earth?

CWatters said:
The voltage across a resistor at time t depends on the current at time t, however the voltage across a capacitor depends on past current so you need to make some assumptions about how the system starts off.
Usually with such problems in my textbook we are supposed to assume we just put the circuit in place.

Krushnaraj Pandya said:
Usually with such problems in my textbook we are supposed to assume we just put the circuit in place.
Ok so if the capacitor came out of it's box with some random voltage on it this problem would be impossible to solve. Try assuming they all start with 0V across each when the battery is connected.

CWatters said:
Ok so if the capacitor came out of it's box with some random voltage on it this problem would be impossible to solve. Try assuming they all start with 0V across each when the battery is connected.
I already have...I don't understand from there what I can do when a point is earthed after I find effective capacitance

Sorry I added this bit to the question above...

CWatters said:
There is only one connection to Earth so what does that mean for the current flowing to/from earth?

Krushnaraj Pandya said:
marking 1 to 4 from left to right. C1,C2,C3 are in series. C123=(2/3)C. which is in parallel with C4.

C1,2,3 are in series with each other and in series with C4.

What do you know about the current through components in series?

CWatters said:
C1,2,3 are in series with each other and in series with C4.
I understand C 1,2,3 are in series with each other but how are they in series with C4. Did you mean to say parallel, since they are connected across the same potentials of 10 and zero?

No I do mean series. For components to be in parallel they must share two nodes.

The equivalent capacitance of C1-3 is connected to Earth and the -ve of the battery.
C4 is connected to Earth and the +ve of the battery.

So they only share one node (earth).

Cwatters is trying to say that the current from the earthed point to the ground will always be zero, so essentially all the capacitors are in series.

BUT I have trouble myself understanding why that current will always be zero.

In order to make progress I'll give you two hints...

There is only one connection to Earth so no current flows through that connection to earth.

All 4 capacitors are in series with the battery so when the battery is connected the same amount of charge flows through all 4 capacitors. This continues until the sum of the voltages on each capacitor add up to the battery voltage (eg Kirchoff's voltage law).

CWatters said:
No I do mean series. For components to be in parallel they must share two nodes.

The equivalent capacitance of C1-3 is connected to Earth and the -ve of the battery.
C4 is connected to Earth and the +ve of the battery.

So they only share one node (earth).
is there a way I can develop this intuition? a link perhaps...
current is same across series. Therefore charge on capacitors will be the same. Adding C 1,2,3,4 in series, equivalent capacitance is 2C/5, using Q=CV, total charge on each capacitor is 4C. Correct?
I don't know how to proceed further, shouldn't potential at A be 10 V itself since there is no capacitor till that point

Alright, so I know the potentials across each capacitor now. From 1 to 4 they are 2,2,2 and 4 V

How do I derive potentials at A and B from this?

I haven't checked the voltages you calculated but...

If the voltage across C4 is 4v and the left hand side is connected to Earth (0V) then the right hand side of C4 must be at -4V.

Edit: ok I agree your voltages are correct.

CWatters said:
I haven't checked the voltages you calculated but...

If the voltage across C4 is 4v and the left hand side is connected to Earth (0V) then the right hand side of C4 must be at -4V.
The answer says +4V, also at point A, the answer says the potential is -6V. This is just a mirror to what you said, though. Can you explain how you got this?

maybe we need to remind you that the potential difference across any type of lumped element between points A and B is ##V_{AB}=V_A-V_B## where ##V_A## and ##V_B## the potentials at points A and B respectively. So you know all the potential differences and also that the potential at the Earth point (lets name it C) is ##V_C=0##. You also know ##V_{BC}## for example, so...

Sorry I got the battery voltage wrong way around. Book answer is correct.

Do we consider the negative terminal of the battery at -10V?

since there seems a gradual increase to zero till the earthed point. and then again from 0 till +10 to the positive terminal

can you explain again why no current flows into the earth?

Delta2
if you consider the negative terminal at -10V and the positive terminal at +10V then the potential difference across the source would be 10-(-10)=20V...

Delta² said:
if you consider the negative terminal at -10V and the positive terminal at +10V then the potential difference across the source would be 10-(-10)=20V...
this seems counter-intuitive but undeniably correct. How can I easily understand that the sign of potential at A is negative, and at B is positive. You mention that we know potential at Earth is 0 and we got the potential differences correctly too. Then what's to stop the point at B from having a potential -4 V?

Krushnaraj Pandya said:
this seems counter-intuitive but undeniably correct. How can I easily understand that the sign of potential at A is negative, and at B is positive.
You can understand or infer that, all you can infer is because B is connected at the positive terminal will have a higher potential than A which is connected at the negative terminal, it might as well both potential be negative with ##V_B=-1V >-11V=V_A##, ##V_{BA}=V_B-V_A=-1-(-11)=+10V.##
You mention that we know potential at Earth is 0 and we got the potential differences correctly too. Then what's to stop the point at B from having a potential -4 V?
Because it is ##V_{BC}=V_B-V_C=4V##

Delta² said:
You can understand or infer that, all you can infer is because B is connected at the positive terminal will have a higher potential than A which is connected at the negative terminal, it might as well both potential be negative with ##V_B=-1V >-11V=V_A##, ##V_{BA}=V_B-V_A=-1-(-11)=+10V.##

Because it is ##V_{BC}=V_B-V_C=4V##
Got it! Thank you, I just realized that I learned Kirchhoff's laws on PF even though they are in the next chapter hehe
Thank you very much @CWatters for your help (although I'd still like to know why no current flows to the earth)

Delta2
Krushnaraj Pandya said:
Got it! Thank you, I just realized that I learned Kirchhoff's laws on PF even though they are in the next chapter hehe
Thank you very much @CWatters for your help (although I'd still like to know why no current flows to the earth)

@CWatters I will also like to know that. I can ,sort of, rephrase the question , let's say that the voltage source in this example is 1000V that means that ##V_{BC}## will be 400V, so if we haven't earthed C point yet, and I go touch the C point, I will feel absolutely nothing, since the current through me will be zero?

The simple answer is no current flows in the Earth connection because it's not a complete circuit. There is no closed loop around which current can flow. In theory you could make a 1000v battery and hold onto the +ve terminal with one hand and Earth with the other. No current would flow because the -ve terminal of the battery isn't connected to anything.

The longer answer is yes you might feel it/get a shock. The reason is that in the real world there can be some parasitic capacitance between the circuit and Earth that completes the loop. This capacitance isn't shown on the circuit diagram and is typically quite small. It's this capacitance that could discharge through you giving you a shock.

If you happen to have 1000v battery don't try this at home.

Delta2
CWatters said:
The longer answer is yes you might feel it/get a shock. The reason is that in the real world there can be some parasitic capacitance between the circuit and Earth that completes the loop. This capacitance isn't shown on the circuit diagram and is typically quite small. It's this capacitance that could discharge through you giving you a shock.
I totally agree with this explanation. But tell me something else, assume a typical electrostatic scenario with a charged conductor at a potential V, and then we connect this conductor to ground with a wire, will the current that will flow through the wire will be , again, due to a hypothetical closed loop consisting from the conductor, the wire, the ground and a parasitic capacitance between the conductor and the ground?
If you happen to have 1000v battery don't try this at home.
Nope don't have such a battery, the biggest battery in house is the one from my smartphone

Krushnaraj Pandya
Yes, the wire shorts out the parasitic capacitance.

If we are talking about something like a small charged sphere some height above the ground I think I'm right in saying that the "self capacitance" of the sphere would be a reasonable approximation to the capacitance between sphere and earth. If the charged object was more like a flat sheet of metal closer to the ground the equation for a flat plate capacitor would be more accurate.

## 1. How does earthing affect circuit analysis?

Earthing a circuit provides a reference point for voltage measurements and helps to reduce noise and interference in the circuit. This can make it easier to analyze the behavior of the circuit.

## 2. What does it mean to have 1 point earthed in a circuit?

Having 1 point earthed means that one point in the circuit is connected to the ground or earth. This creates a reference point for voltage measurements and can help to stabilize the circuit.

## 3. Why is it important to earth a circuit?

Earthing a circuit helps to protect against electric shocks and can improve the overall safety of the circuit. It also helps to reduce noise and interference, making it easier to analyze the circuit's behavior.

## 4. What types of circuits benefit from earthing?

All types of circuits can benefit from earthing, but it is especially important for circuits that handle high voltages or carry sensitive signals. This includes power grids, telecommunication networks, and electronic devices.

## 5. How do you analyze a circuit with 1 point earthed?

To analyze a circuit with 1 point earthed, you can use the same techniques and equations as you would for any other circuit. However, you should take into account the effect of the earth connection on voltage measurements and noise reduction.