coupling capacitance and actual capacitance

albbg said:

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.

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Venkadesh_M said:

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?

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albbg said:

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.

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kripacharya said:

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

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albbg said:

I think you are totally rigth.

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