During the normal operation within a building and with the motor running, there were 350 volts at the panelboard and 300 volts at the motor. Determine the percent voltage drop for this circuit.
What will be an ideal response?
Percent Voltage Drop =[(VoltageSource – Voltage Load )/VoltageSource] x 100
=[(350-300)/350]x 100
= 14.3%
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