Grounding of secondary side transformer

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Grounding the secondary side of a transformer affects current flow and circuit protection. In the case of an earth fault, grounding at L2 results in equal voltage across the secondary windings, preventing current flow. Conversely, grounding at L1 can allow current to bypass protective measures, potentially starting a motor without blowing a fuse. The discussion highlights the importance of proper relay placement to avoid undetected faults that could lead to overload conditions. Understanding these grounding scenarios is crucial for safe electrical system design.
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Grounding of secondary side of transformer
Hello,

I am a mech engineer teaching myself electrical engineering so please forgive my ignorance. Could someone explain to me how the coil stays open when grounded in the figure 2-12b.

With my limited understanding I can see that in 2-13a the if the side of the circuit on the left of the coil contacted ground (earth fault) then because of grounding at L2 both sides of the secondary windings would be at the same voltage so no current. In 2-13b grounding at L1 means that because of the Earth fault ground is now hot, current would pass through grounding at L1, bypassing the stop and into the coil? Is that correct?

many thanks,

will

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In (a), grounding any point between L1 and the motor, would blow the fuse.
In (b), grounding any point between L1 and the motor, could start the motor without blowing the fuse.
 
Figure 2-13A is not ideal either. Overload relay labeled OL should not be in the low side. A fault to ground between M and OL can exist for years without anyone realizing it until one day due to an overload condition on the motor causes the overload relay to trip and it has been bypassed due to an undetected fault between M and OL.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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