Dc Machines Practice Questions

Given, $I_a=8A$ No-load speed, $N_0=1400 \; \text{rpm},$ $V=230\text{volt}$ $\therefore \;\;\omega _0=\frac{2 \pi}{60}\times N_0$ $\;\;= \frac{2 \pi}{60}\times 1400=146.6 \;red/s$ Power developed in the motor, $P=E_aI_a$ Also Power, $P=\text{Torque} \times \omega _0$ or, $\text{Torque} \;\; T=\frac{P}{\omega _0}=\frac{E_aI_a}{\omega _0}$ Since, armature resistance drop and brush drop are neglected therefore.

Given, $R_a=0.4\Omega ,$ $R_f=230\Omega ,$ $V=230 \;\text{volt}$ At no load, speed, $N_0=1400 \;\text{rpm},$ $I_{L_0}=5A$ Field current, $I_f=\frac{V}{R_f}=\frac{230}{230}=1A$ No load armature current, $I_{a_0}=5-1=4A$ Induced emf at no-load, $E_{a_0}=V-I_{a_0}R_a$ $\;\; \;\;=230-4 \times 0.4=228.4 V$ Also, at full load, $I_{L_{fl}}=70A$ $I_{a_{fl}}=I_{L_{fl}}-I_f=70-1=69A$ Induced emf at full load, $E_{a_{fl}}=V-I_{a{fl}}R_a$ $E_{a_{fl}}=230 -69 \times 0.4=202.4V$ we know that,

Therefore, full-load speed of motor = 1240.63 rpm

Also, full load armature current, $I_{a_f}=15A$ $\therefore$ induced emf in armature at load,

Let the speed of motor during loaded condition be N. We know that:

Given, $V=250V,$ $R_a=0.6\Omega ,$ $R_f=125\Omega ,$ $I_L=50A$ both for motor as well as generator action. Field current, $I_f=\frac{V}{R_f}=\frac{250}{125}=2A$ $\text{For motor action:}$

$\text{For generator action:}$

For a DC shunt machine, $E_a\propto \phi \omega _n$ or $\;\; E_a\propto N \;\text{(speed)}$ $(\because \;\phi =constant)$ $\therefore \;\;\frac{E_{a_g}}{E_{a_m}}=\frac{N_g}{N_m}$ $\frac{N_g}{N_m}=\frac{281.2}{221.2}=1.271$ Therefore, ratio of the speed as a generator to the speed of motor=1.271

i.e. field flux reduces Therefore, % change in flux $=\frac{\phi _2-\phi _1}{\phi _1}=-36.36\%$ $\Rightarrow$ Decrease in flux.

CASE-(i)

CASE-(ii) Back emf,

As speed is constant

as torque is constant

4-point starter is used to control the speed of shunt motor in field weakening region, i.e. above rates speeds. In field weakening region field current will reduce in 3-point starter holding coil unable to hold the plunger in ON position.

For, $N=1400\; \text{rpm}, \; E_a=186.67V$ Required $V=E_a+I_aR_a$ $\;\;\;=186.67+I_a(3.39)$ where $I_a$ is function of torque, [/latex] To develope 5 N-m it requires 3.925A

DC machine is acting as DC motor. When the brush is shifted in the direction of rotation in DC motor, field gets magnetized. Due to armature reaction, leading tip of N-pole and S-pole gets more magnetized and trailing tip of N-pole and S-pole gets demagnetized. Butincrease in flux density is more than the decreases in flux density. Due to magetizing effect of armature reaction waveform of $V_{A'B'}$ will be as given option (A).

Currently no explanation available

Currently no explanation available

Rated armature current= 12A Armature current during braking $I_b=1.25 \times 12=15A$ Total armature resistance including external resistance during braking

Speed-current characteristics of DC motors It can seen from the characteristics, speed regulation can be zero at full load in the case of differential compound dc motor.

Compensating winding is placed in slots cut out in pole faces such that the axis of this winding coincides with the brish axis. The compensating winding neutralizes the armature mmf directly under the pole while in the interpolar region, there is incomplete neutralization. To speed up the commutation process, the reactance voltage must be neutralized by injecting a suitable polarity dynamical (speed) voltage into the commutating coil. In rder that this injection is restricted to commutating coils, narrow iterpolar are provided in the interpolar region.

Magnetic poles revolves around stationary coils in clockwise direction. As the motion is relative, it can be assumed that, magnetic poles are stationary and coils are rotating in counter-clockwise. The direction of magnetic flux is from N-pole to S-pole. Now consider coils $C_1$ and $C_i$ The dynamically induced emf in conductor $C_1$ can be given by flamig's right hand rule $e=|\vec{V}|\times |\vec{B}|l=B/V \sin \theta$ The direction of emf $\vec{V}\times \vec{B}$ Therefore, $\odot \text { \in } C_1$ Now, consider, coils $C_2$ and $C'_2$ At this instant, B and V are antiparallel i.e. $\theta =180^{\circ}$ So, $\sin 180^{\circ}=0$ Hence,no emf is induced in $C_2$ .

As we know, (i) Armature emf, $E=k_a\phi \omega$ where $k_a$ is constant so, E depends on $\phi$ and $\omega$ only. (ii) Developed torque, $T=k_a\phi I_a$ T depends on $\phi$ and $I_a$ only. (ii) Developed power, $P=E I_a=(k_a\phi \omega )I_a$ or, $\; P=T\omega = (k_a\phi I_a)\omega$ So, P depends on $\phi ,\omega$ and $I_a$ .

Armature control provides constant torque drive.In the shunt motor case by keeping the filed current at maximum value full load torque can be obtaineda at full load armaturecurrent at all speeds

(output power delivered at half the rated speed by armature control). Filed control provides constant-power drive. Since the speed is inversely proportional to the flux/pole $\left ( \omega \propto \frac{1}{\phi } \right )$ while the torque is direclty proportional to flux/pole $(T\propto \phi )$ for a given armature current. Output power $=T\omega =\left ( \frac{k_1}{\phi } \right )(k_2\phi )$ $\;\;=k_1k_2=constant$ Output power delivered by the motor remains constant i.e. 50kW