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Suppose you are given a 115 V motor that draws 6.5 A at its normal (steady-state) speed, but 21.5 A when it first starts. What is the back end of the motor in V?
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Added Sun, 26 Jul '15
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R= V/I = 115/21.5 = 5.34 ohm. This is because initially the motor is not turning and no back emf is being generated, so the system behaves like a simple resistor (ignoring effects due to inductance of the coils).

When the motor is at normal speed, it also acts as a generator producing a voltage called the back emf (Vb); this cancels out some of the applied voltage (V) so the effective voltage = (V - Vb)
6.5 = (V-Vb) / R
6.5 = (115-Vb) / 5.34
Vb = 80.23 V
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Added Sun, 26 Jul '15
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