Why Voltage Becomes Zero During Short Circuit?

GA for you.

all kvs are kilo volt sized in intelligence . . .

No Sir, I am very simple man... self made.. but with the help of every one I met... so everyone is my Guru!

Let your intelligence be used at unity power-factor.I just mean to say have a nice day.Happy Engineer's Day.

Happy Engineer's Day to you also. Engineer's Day is celebrated on 15 September every year. The day is the birthday of Sir Mokshagundam Visvesvarayya an eminent Great Indian engineer.

Sir Do you mind looking in General Discussion.I posted something.

I have already in the first hour!

Doh!

Since V = I * R and with R=0, therefore V=0.

Duh!

That formula doesn't apply when R is close to 0, there is very small resistance so there is very small voltage needed to push the current thru the shorted connection. The resistance equals the resistance of the wires which is probably too small for our ohm meters to measure. You can't have any current flow without pressure in volts. This attractive force between the two polarities somehow gets converted from voltage to current during the short similar to a step-down transformer transduces voltage to current proportionally by the ratio of coils. All current thru a wire has an electromagnetic field and a magnetic field around it which may play a part in this change from voltage to current. Perhaps they provide the pressure to push the current and it is a different kind of pressure rather than electromotive force (voltage). Current comes from voltage applied bouncing the electrons around in the wires, not actually from the battery or AC Source. When too much current goes back into the battery, etc. it can damage the sulfur bridge internal resistance then the voltage goes down, this is why batteries need to be recharged. See my other reply.

That is why a short circuit causes damage to your power supply, batteries or components caught in this path of current flow. There should be a way to detect a short in a circuit so it is routed to a ground connection to avoid the damage to batteries or voltage sources. When the current gets too high it could trip the magnet in a relay which directs it to ground. You would need two of these relay circuits for AC and just one for DC. This would make circuits costlier, heavier, but safer with less repairs to be fixed. Good idea?

When there is no fault or healthy network:

Generator voltage = Terminal Voltage of Load + Load Current*closed Loop impedance (such as Transmission line, Transformer, cables etc.) of a close circuit.

Worse case fault is 3-Phase fault in network, During Fault:

Generator voltage = Zero terminal voltage of Load + Short circuit Current*closed Loop impedance (such as Transmission line, Transformer, cables etc.) of a close circuit.

Therefore, at fault, Terminal voltage is zero but Generator voltage remains same and Whole generator voltage is dropped in the close loop impedance of the network that causes very high current (in Kilo-Amps) which is called Short Circuit current.

At short circuit:

a. Generator Voltage remains same and equal to Isc*without load loop impedance

b. Load Terminal Voltage is ZERO

I thought generated emf = load voltage+IaRa.

Generated emf & and generator terminal voltage are the same. Therefore to put my statement in equations then as follows:

Before a load fault:

Vg = Vt + IL x Z; Z is Impedance of the close loop network (without load impedance) and neglecting generator internal impedance.

After a load fault:

Vg = Isc x Z; Z is Impedance of the close loop network (without load impedance) and neglecting generator internal impedance

Where:

Vg = Generator emf in Volts

Vt = Load terminal voltage in volts

IL = Current in Load

Z = Impedance of the close loop network (without load impedance) and neglecting generator internal impedance; in Ohm

Isc = 3-Phase short circuit current in Amps

Why Voltage Becomes Zero During Short Circuit? At load terminal voltage is zero because it is connected with an impedance in parallel whose impedance value is, very, very, very & very small which is in comparison to Network loop impedance.

I think this: for AC the hot wire is 180 degrees out of phase with the neutral wire so when the two are connected the voltage waveforms of both are cancelled out via destructive interference, but the current waveform adds somehow probably in proportion to what the input voltage was. Therefore if the circuit was using 1 amp, then the short might draw as much as 120 amps from the AC Source, but does not since the breaker switch is 60 amps or as low as 15 amps.

For a DC 12 volt battery, it could be the same - the positive pole has 12 volts, but the negative may be a negative 12 volts in comparison. During the shorting the two battery cables, then they too cancel out by destructive interference (phase doesn't apply in DC).

You could probably see all this canceling effect on a fast oscilloscope that records it, but try it with very low voltage source such as millivolts DC first.