A fan operates at 200 volt (DC) consuming 1000 W when running at full speed. It's internal wiring has resistance 1 Ω Ω When the fan runs at full speed, its speed becomes constant. This is because the torque due to magnetic field inside the fan is balanced by the torque due to air resistance on the blades of the fan and torque due to friction between the fixed part and the shaft of the fan. The electrical power going into the fan is spent (i) in the internal resistance as heat, call it P 1 . (ii) in doing work against internal friction and air resistance producing heat, sound etc, call it P 2 . When the coil of fan rotates, an emf is also induced in the coil. This opposes the external emf applied to send the current into the fan. This emf is called back-emf, call it 'e'.
Answer the following questions when the fan is running at full speed.
(i) The current flowing into the fan and the value of back emf 'e' is -
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Ans.
(i)
Sol. The fan is operating at 200 V, consuming 1000 W, then I = 
= 5A
But as coil resistance is 1 Ω Ω then power dissipated by internal resistance heat is P 1 = I 2 R = 25 W
if V is net emf across coil then
= 25 W V = 5 volt
Net emf = source emf – back emf
V = V s – e ⇒ e = 195 C
The work done P 2 = 1000 – 25 = 975 W
(ii)
Sol. The fan is operating at 200 V, consuming 1000 W, then I = 
= 5A
But as coil resistance is 1 Ω Ω then power dissipated by internal resistance heat is P 1 = I 2 R = 25 W
if V is net emf across coil then
= 25 W V = 5 volt
Net emf = source emf – back emf
V = V s – e ⇒ e = 195 C
The work done P 2 = 1000 – 25 = 975 W
(iii)
Sol. The fan is operating at 200 V, consuming 1000 W, then I = 
= 5A
But as coil resistance is 1 Ω Ω then power dissipated by internal resistance heat is P 1 = I 2 R = 25 W
if V is net emf across coil then
= 25 W V = 5 volt
Net emf = source emf – back emf
V = V s – e ⇒ e = 195 C
The work done P 2 = 1000 – 25 = 975 W
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