An electrical device draws 5.44 A at 110 V. If voltage drops by 10.0 percent, what will be the current, assuming nothing else changes?
ANSWER :
GIven data: V (initial) = 110 V ; Current (initial) = 5.44 A; Voltage drop by 10%
From Ohms law , we know V= IR
Since Voltage droped by 10 % and nothing else has changed. SO , current also will drop by 10 % ( 100-10 = 90/100 = 0.9)
Therefore, Current (final) = 0.9 * Current (initial)
Current (final) = 0.9 * 5.44
So answer is Current (final) = 4.896 A
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