potenial with respect to earth

As per the law of electromagnetic induction whenever the magnetic linkage of a coil changes a emf is induced in the coil.The emf denotes the the potential difference between two ends of the coil.what i am confused about is the value of potential of the ends of the coil with respect to earth.Symbolically let the ends of the coil placed in a changing flux be denoted by a and b.The potential of end a of coil with respect to earth be Vae and that of end b of coil be Vbe. The emf induced in the coil due to changing flux be denoted by Vemf.Then Vemf=Vae-Vbe=n*rate of change of flux, which is given by a well known fromula. But how to evaluate the value of Vae and Vbe? i.e if we put a voltmeter between a end of coil and earth what will it read. pls give your valuable guidance. thanks for ur time.

Reply to
sanjay664lal
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----------------------- It could read anything. There is no relationship between the voltage induced in a coil and the potential to ground. Capacitance to ground, overhead clouds, and a variety of other sources will have an effect (and these will not be related to the rate of change of flux in the coil).

Reply to
Don Kelly

So if the potential of a particular end of a coil (placed in changing flux) with respect to earth cannot be determined by any formula. Then will it be true to say that in case of a three phase generator with coils of three phases connected in star (the star point not being connected to earth), the potential of star point of coils with respect to earth cannot be determined by any fromula. i.e if we connect a voltmeter in between the star point of generator and the earth then what it will read some value depending on the condition of the site(ground capacitance, overhead cloud) but the value can not be determined by any formula. Is it true?

If the phase voltage of the coils are 230 v and star point (which is not connected to earth) of coil has some potential say 100 V w.r.t earth. Now if we connect the star point to earth what will be the scenario?i.e what will be the value of potential of star point w..r.t earth ,value of phase voltages. If the potential of star point w.r.t. earth now becomes zero then where does 100 V go? (w.r.t.= with respect to) I am looking forward for your useful suggestions. Thanks for your time

Reply to
sanjay664lal

If the star point has a potential of 100V with respect to the earth with the system floating, it is primarily a result of capacitance between the various points of the system and the earth, as well as the currents that are flowing in each phase. It is easier to understand, however, if you assume all currents are zero.When you connect the star point to the earth, you bypass that capacitance and you force the potential at that point to be zero. The generator determines the potential between the star point and the phases, and therefore between earth and the phases.

Just to get your creative juices flowing, what do you think would happen if you connect one of the phases to earth, instead of the star point?

Reply to
Ben Miller

Reply to
Don Kelly

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Addition: I missed a very important scenario: If there is a low impedance fault to ground somewhere along a winding -then there can be a part of the phase voltage appearing between the ground and star point. In that case a high current can flow if the star point is grounded. This is serious.

Reply to
Don Kelly

This question can be simplified to What is the voltage between the positive terminal of a 9V battery (not connected to anything) and ground?

Reply to
peter

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You've put it neatly in a nutshell! Thank you

Reply to
Don Kelly

Thanks all of you for your suggestions. From the discussion what I have understood is that the potential of any point to the earth depends upon the charge of the point, distance of the point from the ground (ground capacitance!!). Thus to evaluate the potential difference (p.d.) between the floating star point and the ground we must be able to evaluate the charge of the star point. DK described that the p.d. depends on the ground capacitance, overhead cloud. This is because it affects the charge of the star point, am I right DK? But I still don=92t know how to evaluate the charge? When the star point is connected to the earth, we are simply putting a conductor between two plates of the capacitor i.e. earth and the star point which makes the p.d between them zero and all the charge at star point will be drained to the earth. Pls give ur suggestions.

Reply to
sanjay664lal

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