coupling capacitance and actual capacitance

[albbg]

OK. Take two resistors in series (same schematic of post #10 but with capacitors replaced by resistors). To measure the value of the series I can take a floating generator (like that is in a multimeter); the voltage is known then I measure the current and I have R=V/I.

1. let's suppose R1=R2=0.5 ohm and V=1 V. I decide my ground is A, I will have Vac=0.5 V (voltage divider) and Vab=1V the current is (single mesh) I=1 A. The resistance between A and B is 1 ohm.
2. let's suppose now the ground is in C. We measure Vca=0.5V, Vbc=0.5V then applying KVL Vab=1V, the current is then always 1A and then the resistance is 1 ohm.

I can connect C to any point, floating with respect to A and B and my result will not change. If the generator is floating with respect to the earth (the handheld multimeter, for instance) I can connect C to earth without any change in the measured result.

 

[Venkadesh_M]

Think it if you just decide any one (only one) point as ground then there will be never going to be any current, but talking about a circuit which already involves a ground and we are connecting some other part to it will surely going to make any changes in the potential Isn't it?

 

[albbg]

Yes, it is. But the original circuit didn't foresee to keep in touch the metal C with A or B. Their are floating then they are the only two nodes that can change the capacitance if in contact with C. Any other point connected to C does have any effect.

 

[kripacharya]

OK. Take two resistors in series (same schematic of post #10 but with capacitors replaced by resistors). To measure the value of the series I can take a floating generator (like that is in a multimeter); the voltage is known then I measure the current and I have R=V/I.

1. let's suppose R1=R2=0.5 ohm and V=1 V. I decide my ground is A, I will have Vac=0.5 V (voltage divider) and Vab=1V the current is (single mesh) I=1 A. The resistance between A and B is 1 ohm.
2. let's suppose now the ground is in C. We measure Vca=0.5V, Vbc=0.5V then applying KVL Vab=1V, the current is then always 1A and then the resistance is 1 ohm.

I can connect C to any point, floating with respect to A and B and my result will not change. If the generator is floating with respect to the earth (the handheld multimeter, for instance) I can connect C to earth without any change in the measured result.

Think it if you just decide any one (only one) point as ground then there will be never going to be any current, but talking about a circuit which already involves a ground and we are connecting some other part to it will surely going to make any changes in the potential Isn't it?

Yes, it is. But the original circuit didn't foresee to keep in touch the metal C with A or B. Their are floating then they are the only two nodes that can change the capacitance if in contact with C. Any other point connected to C does have any effect.

Both of you are correct even though your conclusions are totally different.

This is because neither of you is paying attention to the ports which are implicitly being used by the other. And since both of you are using different ports for driving voltage/ measurement, hence you are getting different (both correct tho) results.

I think the OP has long ago got his homework answer, so lets close this now ?

 

[albbg]

I think the OP has long ago got his homework answer, so lets close this now ?

I think you are totally rigth.

 

[Venkadesh_M]

I think you are totally rigth.

Yeah, me too think the same....