Question

A fan operates at 200V (dc) consuming 1000W when running at full speed. Its internal wiring has resistance 1$\Omega$. When the fan runs at full speed, its speed becomes constant. This is because the torque due to the magnetic field inside the fan is balanced by the torque due to air resistance on the blades of the fan and the torque due to friction between the fixed part and the shaft of the fan. The electric 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 to the fan. This emf is called back emf; call it e. Answer the following questions when the fan is running at full speed. The current flowing into the fan and the value of back emf e is
A) 200 A, 5V
B) 5 A, 200V
C) 5 A, 195V
D) 1 A, 0V

Answer 1

In order to find the current flowing into the fan we have the power supplied to it and it is operating in dc voltage of 200V. To find the back emf e, we know that the net E.M.F is equal to the addition of the back emf with the voltage supply to it. This voltage is according to back emf which is the current flowing through fan and resistance of $1\Omega$

Step by step solution:
Step 1:
We are given:
A fan operates at 200V (dc) = E
It is consuming 200W power when running at full speed. So P=200W
The fan has its internal wiring has resistance 1$\Omega$. This implies R=1$\Omega$
We need to find the current flowing into the fan and the value of back emf e
Let us know first about the back current first:
Back current: - Counter-electromotive force (counter EMF, CEMF), also known as back electromotive force (back EMF), is the electromotive force or "voltage" that opposes the change in current which induced it.
Step 2:
According to the formula of power, it states that power supplied to any system is equal to the product of voltage and current supplied to it.
$\Rightarrow P = V \times I$……. (1) Where, V is in volts, and I is in amperes.
Substituting the value of V and P to get current supplied, I=$\dfrac{P}{V}$
This is equal to$\dfrac{{1000}}{{200}}$
Therefore the current I is 5 amperes. ……… (2)
Now, given the resistance of the coil is $1\Omega$
Equation 1 can be rewritten as, $P = {I^2}R$, where I is the current and R is the resistance of the coil.
Then the power dissipated by the internal resistance is ${P_1}$
$\rightarrow$ ${P_1} = {I^2}R$, Substituting the value we get${P_1}$ =${\left( 5 \right)^2} \times 1$. This is equal to 25W
Net E.M.F is equal to the addition of the back emf e and the voltage.
Here voltage V=IR.
Then, E=e+IR. On rearranging the equation for the value of back emf we get, e=E−IR
Substituting the value, e=200V−(5)×1
On solving it we will get the back emf e is equal to 195V ……. (3)
From 2 and 3 we can say that the current flowing into the fan and the value of back emf e is 5Ampere and 195 volt.

Hence option C is the correct answer.

Note: The back emf e cannot be prevented in the coils but it can be controlled. In suppressing the back EMF the objective is to prevent the very high voltages and dissipate the stored energy in a controlled way. If there is no back end then large current flows in starting of motors because initial speed is zero and back emf is zero thus winding gets damaged.