Question

An electric device draws 6.50A at 240V.

If the voltage drops by 25% , what will be the current, assuming nothing else changes?

If the resistance of the device were reduced by 25% , what current would be drawn at 240V ?

Answer 1

The resistance of the device = 240/6.5 = 36.92 ohm
so, Vnew = 240-0.25*240=180

therefore current drawn by device after reducing voltage = 180/36.92=4.875 Ampere (Ans)

ii ) R new = 36.92-0.25*36.92 = 27.69 ohm
current drawn = 240 / 27.69=8.667 Ampere

Answer 2

I = 6.5 A
V = 240 V

V = IR
=> R = V/I = 240/6.5 = 36.92 ohm

Now voltage drops by 25 %
Thus
new Voltage = 180 V

Thus New I = V/R = 4.875 A

Now New Resistance = .75 * R

Thus R' = 27.69

Thus New I = 8.66 A