A 200 volt motor is taking 10 amperes armature current, the armature resistance is 0.1 ohm. The Back- EMF under these conditions will be
- 201 volts.
- 199 volts.
- 1 volt.
Explanation from PART66Online
Back-EMF equals the applied voltage minus the armature volt-drop: E = V - Ia*Ra. Here E = 200 - (10 A * 0.1 ohm) = 200 - 1 = 199 V. The small armature-resistance drop accounts for the 1 volt difference between supply and back-EMF.
A
arifadli8 asked · 3 Jan 2014
How???
Community Comments (1)
T
tsukiyo92
31 Jul 2014
First, we need to understand the total current flowing over the armature is due to 'effective voltage' which is sum of applied voltage and back emf. (difference)
Given that we ave actual current of 10A flowing over armature of 0.1ohm :
V=IR
V= 10 * 0.1
V = 1V
With only 1V of effective voltage, that means back emf of 199V.
Given that we ave actual current of 10A flowing over armature of 0.1ohm :
V=IR
V= 10 * 0.1
V = 1V
With only 1V of effective voltage, that means back emf of 199V.
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