6. An instrument is rated at 250 W if it is connected across a 120-V dc power supply.
(a) What current does it draw under normal operation?
(b) What is its resistance?
(c) How many kilowatt-hours does it use in a day if it is left on all the time?

466.67 s or 7.78 min

Explanation:

Applying,

E = LdI/dtEquation 1

Where E = Energized E.m.f in the solenoid, dI = change in current, dt = change in time.

make dt the subject of the equation,

dt = Ldi/E Equation 2

Given: L = 5.6 H, dI = 100-0 = 100 A, E = 1.2 V

Substitute into equation 2

dt = 5.6(100)/1.2

dt = 560/1.2

dt = 466.67 s

dt = (466.67/60) min = 7.78 min

Hence, it takes 466.67 s or 7.78 min for the solenoid to reach its operating current.