Application of EMI (DC Motor) Questions in English

(B) The induced $e.m.f.$ $(e)$ in a dynamo armature is given by the formula $e = N B A \omega \sin(\omega t)$, where $N$ is the number of turns, $B$ is the magnetic field, $A$ is the area of the coil, and $\omega$ is the angular (rotational) velocity.
From this relation, we can see that the induced $e.m.f.$ is directly proportional to the rotational velocity, i.e., $e \propto \omega$.
If the rotational velocity $\omega$ is doubled $(2\omega)$, the induced $e.m.f.$ will also become two times the original value.

(B) The back $e.m.f.$ $(E_b)$ in a $dc$ motor is given by the relation $E_b = K\omega$, where $K$ is a constant and $\omega$ is the angular speed of the armature.
The armature current $I_a$ is given by the formula $I_a = \frac{V - E_b}{R}$, where $V$ is the supply voltage and $R$ is the armature resistance.
As the speed $\omega$ increases, the back $e.m.f.$ $E_b$ also increases.
Since $I_a = \frac{V - K\omega}{R}$, an increase in $\omega$ results in a decrease in the numerator $(V - K\omega)$.
Therefore, the armature current $I_a$ decreases as the speed of the motor increases.

(C) dynamo (or generator) produces alternating current $(ac)$ in its armature coil due to electromagnetic induction. $A$ split-ring commutator acts as a mechanical rectifier that reverses the connections of the coil to the external circuit every half cycle. This ensures that the current in the external circuit flows in only one direction. Since the magnitude of the induced electromotive force varies sinusoidally,the resulting output is a fluctuating $dc$.

(C) The back $EMF$ equation for a $DC$ motor is given by $E_b = V - I_a R_a$,where $V$ is the supply voltage,$I_a$ is the armature current,and $R_a$ is the armature resistance.
Rearranging for current,we get $I_a = \frac{V - E_b}{R_a}$.
At the moment of starting,the motor speed is zero,which means the back $EMF$ $(E_b)$ is also zero.
Therefore,at starting,$I_a = \frac{V}{R_a}$,which is the maximum possible value for the armature current.

(D) For a $DC$ motor,the current $I$ is given by the formula:
$I = \frac{V - E}{R}$
where $V$ is the supply voltage,$E$ is the back $EMF$,and $R$ is the armature resistance.
Given:
$V = 220\,V$
$I = 1.5\,A$
$R = 20\,\Omega$
Substituting the values into the equation:
$1.5 = \frac{220 - E}{20}$
$1.5 \times 20 = 220 - E$
$30 = 220 - E$
$E = 220 - 30$
$E = 190\,V$
Therefore,the back $EMF$ is $190\,V$.

(D) An electric motor is an electrical machine that converts electrical energy into mechanical energy. It operates on the principle that when a current-carrying conductor is placed in a magnetic field,it experiences a mechanical force. Therefore,the correct option is $(d)$.

(A) The input power to the motor is $P_{in} = V \times I = 60\, V \times 10\, A = 600\, W$.
Given the efficiency $\eta = 50\%$,the output mechanical power is $P_{out} = \eta \times P_{in} = 0.5 \times 600\, W = 300\, W$.
The power lost as heat in the winding resistance $R$ is $P_{loss} = P_{in} - P_{out} = 600\, W - 300\, W = 300\, W$.
Since $P_{loss} = I^2 R$,we have $300 = (10)^2 \times R$.
$300 = 100 \times R$.
Therefore,$R = \frac{300}{100} = 3\, \Omega $.

(C) An electric motor is a device that operates on the principle of the magnetic effect of electric current. When a current-carrying coil is placed in a magnetic field,it experiences a magnetic force that causes it to rotate. Thus,it converts electrical energy into mechanical energy.

(C) The efficiency $\eta$ of an electric motor is given by the ratio of back $EMF$ $e$ to the supply voltage $E$: $\eta = \frac{e}{E} \times 100$.
Given $\eta = 30\% = 0.3$,we have $e = 0.3 \times E = 0.3 \times 50\,V = 15\,V$.
The current $i$ in the motor is given by $i = \frac{E - e}{R}$,where $R$ is the resistance of the winding.
Substituting the given values: $12 = \frac{50 - 15}{R}$.
$12 = \frac{35}{R}$.
$R = \frac{35}{12} \approx 2.916\,\Omega$.
Rounding to one decimal place,we get $R = 2.9\,\Omega$.

(A) The current $I$ in the armature of a $DC$ motor is given by the formula: $I = \frac{V - E_b}{R}$,where $V$ is the supply voltage,$E_b$ is the back $e.m.f.$,and $R$ is the armature resistance.
Given:
Supply voltage $V = 220\,V$
Back $e.m.f.$ $E_b = 210\,V$
Armature resistance $R = 2\,\Omega$
Substituting the values:
$I = \frac{220 - 210}{2} = \frac{10}{2} = 5\,A$.
Therefore,the current in the armature is $5\,A$.

(A) An electric fan converts electrical energy into mechanical energy to rotate the blades. This conversion is the fundamental principle of an electric motor. Therefore,a fan is based on the working principle of an electric motor.

(B) When a coil of conducting wire is rotated in a magnetic field,the magnetic flux linked with the coil changes continuously. According to Faraday's law of electromagnetic induction,this change in magnetic flux induces an electromotive force $(EMF)$ and consequently an electric current in the coil. This principle is the basis of an electric generator,which converts mechanical energy into electrical energy. Therefore,the correct option is $B$.

(D) In an $A.C.$ generator, during one complete rotation $(360^{\circ})$ of the coil, the direction of the induced $EMF$ and current changes twice.
After the first half rotation $(180^{\circ})$, the direction of the current changes for the first time.
After one complete rotation $(360^{\circ})$, the direction of the current changes for the second time.
After the next half rotation $(180^{\circ})$, i.e., after a total of $1.5$ rotations, the direction of the current changes for the third time.
Therefore, the correct answer is $1.5$.

(C) Given:
Armature resistance, $R = 0.1 \, \Omega$
Induced emf, $e = 120 \, V$
Current drawn, $I = 50 \, A$
For a generator, the relationship between the induced emf $(e)$, terminal voltage $(V)$, and armature resistance $(R)$ is given by the equation:
$e = V + I R$
Rearranging the formula to solve for the terminal voltage $(V)$:
$V = e - I R$
Substituting the given values:
$V = 120 - (50 \times 0.1)$
$V = 120 - 5$
$V = 115 \, V$
Thus, the terminal voltage is $115 \, V$.

(A) In a $dc$ motor, the induced $e.m.f.$ (also known as back $e.m.f.$) is given by the formula $e = \frac{P \phi Z N}{60 A}$.
Here, $P$ is the number of poles, $\phi$ is the magnetic flux per pole, $Z$ is the number of conductors, $N$ is the speed of the motor in $rpm$, and $A$ is the number of parallel paths.
Since $e \propto N$, the induced $e.m.f.$ is directly proportional to the speed of the motor.
Therefore, the induced $e.m.f.$ will be maximum when the motor reaches its maximum speed.

(C) The efficiency $\eta$ of an electric motor is given by $\eta = \frac{P_{out}}{P_{in}} = \frac{e \cdot I}{E \cdot I} = \frac{e}{E}$, where $e$ is the back $emf$ and $E$ is the applied $emf$.
The current in the motor is $I = \frac{E - e}{R}$. The mechanical power output is $P_{out} = e \cdot I = e \left( \frac{E - e}{R} \right) = \frac{eE - e^2}{R}$.
To maximize $P_{out}$, we differentiate with respect to $e$: $\frac{dP_{out}}{de} = \frac{E - 2e}{R} = 0$, which gives $e = \frac{E}{2}$. Thus, the assertion is correct.
The efficiency depends on both the back $emf$ and the applied $emf$, not just the magnitude of the back $emf$. Therefore, the reason is incorrect.

(D) motor starter is primarily used to limit the high starting current in a $DC$ motor.
$1$. It acts as a variable resistor connected in series with the armature to restrict the initial current flow.
$2$. As the motor gains speed,the back $emf$ $(E_b)$ increases,which naturally opposes the supply voltage. The starter allows for the gradual removal of resistance as the back $emf$ builds up,thereby offsetting variations in the back $emf$ during the acceleration phase.
$3$. Therefore,it is essential for the safe starting of a $DC$ motor.
Since all the given statements are correct,the correct option is $D$.

(A) The back $emf$ $(E)$ of a motor is given by $E = V - IR$,where $V$ is the supply voltage,$I$ is the current,and $R$ is the armature resistance.
For the unloaded motor: $E_1 = 12 - (2 \times 1) = 10 \,V$.
Since the back $emf$ is directly proportional to the speed of the motor $(E \propto \omega)$,a $10 \%$ reduction in speed implies a $10 \%$ reduction in the back $emf$.
The new back $emf$ is $E_2 = E_1 - 0.10 \times E_1 = 10 - 1 = 9 \,V$.
Now,for the loaded motor,the current $I_2$ is given by $I_2 = (V - E_2) / R$.
Substituting the values: $I_2 = (12 - 9) / 1 = 3 \,A$.

(B) The electromotive force $(EMF)$ generated by the generator without load is $E = 125 \,V$.
When the generator is under full load,the terminal voltage is $V = 115 \,V$.
The internal resistance of the armature is $r = 1 \,\Omega$.
The voltage drop across the internal resistance is given by $E - V = I \cdot r$.
Substituting the given values: $125 \,V - 115 \,V = I \cdot 1 \,\Omega$.
$10 \,V = I \cdot 1 \,\Omega$.
Therefore,the current in the armature coil is $I = 10 \,A$.

(C) The input power $P_{\text{input}}$ is given by $P_{\text{input}} = V \times I = 100 \ V \times 1 \ A = 100 \ W$.
Given the efficiency $\eta = 91.6\% = 0.916$.
The output power is $P_{\text{out}} = \eta \times P_{\text{input}} = 0.916 \times 100 \ W = 91.6 \ W$.
The power loss is $P_{\text{loss}} = P_{\text{input}} - P_{\text{out}} = 100 \ W - 91.6 \ W = 8.4 \ W$.
Since $1 \ W = 1 \ J/s$ and $1 \ cal \approx 4.2 \ J$,the power loss in $cal/s$ is $\frac{8.4 \ J/s}{4.2 \ J/cal} = 2 \ cal/s$.

(D) Given: Number of turns $N = 500$,Area $A = 2\ m^2$,Angular speed $\omega = 30\ rad\ s^{-1}$,Magnetic field $B = 0.20\ T$,Resistance $R = 1000\ \Omega$.
The induced electromotive force $(EMF)$ in an $AC$ generator is given by $\varepsilon = NBA\omega \sin(\omega t)$.
The maximum $EMF$ is $\varepsilon_0 = NBA\omega$.
The maximum current $I_0$ is given by $I_0 = \frac{\varepsilon_0}{R} = \frac{NBA\omega}{R}$.
Substituting the values: $I_0 = \frac{500 \times 0.20 \times 2 \times 30}{1000}$.
$I_0 = \frac{6000}{1000} = 6\ A$.
Therefore,the maximum current drawn from the generator is $6\ A$.

(A) The back emf $e$ in a $DC$ motor is given by the relation $e \propto \omega$,where $\omega$ is the angular velocity of the motor armature.
Since the back emf is directly proportional to the angular velocity,it reaches its maximum value when the motor attains its maximum speed.

(B) The magnetic flux linked with the coil is given by $\phi_B = B A \cos \theta$,where $\theta$ is the angle between the normal to the plane of the coil and the magnetic field vector $B$.
When the plane of the coil is perpendicular to the magnetic field,the normal to the coil is parallel to the magnetic field,so $\theta = 0^{\circ}$.
Thus,the magnetic flux $\phi_B = B A \cos(0^{\circ}) = B A$,which is the maximum value.
The induced $e.m.f.$ is given by $\varepsilon = -\frac{d\phi_B}{dt} = B A \omega \sin(\omega t)$.
At $\theta = 0^{\circ}$,$\omega t = 0$,so $\varepsilon = B A \omega \sin(0^{\circ}) = 0$.
Therefore,the magnetic flux is maximum and the induced $e.m.f.$ is zero.

(A) In an $AC$ generator,the coil rotates within a magnetic field to induce an alternating current. To maintain a continuous connection with the external circuit while the coil rotates,the two ends of the coil are connected to two separate slip rings. These slip rings rotate along with the coil,and carbon brushes are used to maintain contact between the rotating slip rings and the stationary external circuit. Therefore,the ends of the coil are connected to two slip rings.