Electrical Drives

Question 1
A shunt-connected DC motor operates at its rated terminal voltage. Its no-load speed is 200 radian/second. At its rated torque of 500 Nm, its speed is 180 radian/second. The motor is used to directly drive a load whose load torque T _{L} depends on its rotational speed \omega _{r} (in radian/second), such that T _{L}=2.78 \times \omega _{r}. Neglecting rotational losses, the steady-state speed (in radian/second) of the motor, when it drives this load, is _______.
A
120
B
180
C
240
D
285
GATE EE 2015-SET-2   Power Electronics
Question 1 Explanation: 
T_L=2.78 \times \omega
After neglecting rotational losses,
(T_{sh}=T_a)=T_L
At rated load,
\begin{aligned} T_a&=500Nm=2.78 \times \omega \\ \omega _r&=179.85 \\ \omega _r&\simeq 180 \text{ rad/sec} \end{aligned}
Question 2
The separately excited dc motor in the figure below has a rated armature current of 20 A and a rated armature voltage of 150 V. An ideal chopper switching at 5 kHz is used to control the armature voltage. If L_a=0.1mH, R_a=1\Omega, neglecting armature reaction, the duty ratio of the chopper to obtain 50% of the rated torque at the rated speed and the rated field current is
A
0.4
B
0.5
C
0.6
D
0.7
GATE EE 2013   Power Electronics
Question 2 Explanation: 
\begin{aligned} V_a &=E_b+I_aR_a \\ 150&=E_b+20 \times 1 \\ E_b&= 130V\\ \text{For half} &\text{ the rated torque,} \\ I_a&=10A \\ V_a&=130+10 \times 1=140 \\ D&=\frac{140}{200}=0.7 \end{aligned}
Question 3
A 3-phase squirrel cage induction motor supplied from a balanced 3-phase source drives a mechanical load. The torque-speed characteristics of the motor(solid curve) and of the load(dotted curve) are shown. Of the two equilibrium points A and B, which of the following options correctly describes the stability of A and B ?
A
A is stable, B is unstable
B
A is unstable, B is stable
C
Both are stable
D
Both are unstable
GATE EE 2009   Power Electronics
Question 3 Explanation: 


Let torque developed by the motor =T_m
Load torque =T_L
Accelerating torque,T_a=T_m-T_L
At point A,
T_m=T_L
\Rightarrow \;\;T_a=T_m-T_L=0
So, no accelerating , hence the motor rums at speed N_A.
When a smal disturbance decreases speed to N_c
At point C,
T_m \gt T_L
\Rightarrow \;\;T_a \gt 0
So, motor will accelerate and comes to point A. When the disturbance increases speed to N_D.
At point D,
T_m \lt T_L
\Rightarrow \;\;T_a \lt 0
Motor will decelerate and comes at point A
So, motor is stable at point A.
When the motor is operating at point B.,
T_m = T_L
\Rightarrow \;\;T_a= 0
If a small disturbance decreases speed to N_E.
T_m \lt T_L
\Rightarrow \;\;T_a \lt 0
Motor will decelerate and motor speed will keep on decreasing.
If the disturbance increases speed to N_F.
At point F,
T_m \gt T_L
\Rightarrow \;\;T_a \gt 0
Motor will accelerate and motor speed will keep on increasing.
So, motor is unstable at point B.
Question 4
A single phase fully controlled converter bridge is used for electrical braking of a separately excited dc motor. The dc motor load is represented by an equivalent circuit as shown in the figure.

Assume that the load inductance is sufficient to ensure continuous and ripple free load current. The firing angle of the bridge for a load current of I_0=10 A will be
A
44^{\circ}
B
51^{\circ}
C
129^{\circ}
D
136^{\circ}
GATE EE 2008   Power Electronics
Question 4 Explanation: 
Average output voltage of the converter,
V_0=\frac{2V_m}{\pi}\cos \alpha
Load current =I_0=10A
Back emf =E_b=150V
Armature resistance=R_a=2\Omega
Applying KVL,
\begin{aligned} &V_0-2I_0+150=0 \\ \Rightarrow \;& V_0=-150+2 \times 10\\ &=-130V\\ \Rightarrow \; & \frac{2V_m}{\pi} \cos \alpha =-130\\ \Rightarrow \; &\frac{2 \times \sqrt{2} \times 230}{\pi}\cos \alpha=-130 \\ \Rightarrow \;&\alpha =129^{\circ} \end{aligned}
Question 5
A three-phase, 440 V, 50 Hz ac mains fed thyristor bridge is feeding a 440 V dc, 15 kW, 1500 rpm separately excited dc motor with a ripple free continuos current in the dc link under all operating conditions, Neglecting the losses, the power factor of the ac mains at half the rated speed is
A
0.354
B
0.372
C
0.9
D
0.955
GATE EE 2007   Power Electronics
Question 5 Explanation: 


For a separately excited dc motor
back emf =E_a=V_0-I_aR_a
Since, losses are neglected R_a can be neglected
So,
\begin{aligned} E_a&=V_0 \\ V_0&=E_a =k_a\phi N \;\;...(i)\\ V_0&\propto N \end{aligned}
At rated voltage V_0=440V and N=1500rpm, so, at half the rated speed \left ( \frac{N}{2} =750 \;rpm\right ) output voltage of the bridge (V_0) is 220V.
if I_a is the average value of armature current rms value of supply current will be
I_s=I_a\sqrt{\frac{2}{3}}
Power delivered to the motor
P_0=V_0I_a
Input VA to the thyristor bridge
S_{in}=\sqrt{3}V_sI_s
Input power factor
=\frac{P_0}{S_{in}}=\frac{V_0I_a}{\sqrt{3}V_sI_s}=\frac{220 \times I_a}{\sqrt{3} \times 440 \times I_a\sqrt{\frac{2}{3}}}=0.345
Question 6
A solar cell of 350 V is feeding power to an ac supply of 440 V, 50 Hz through a 3-phase fully controlled bridge converter. A large inductance is connected in the dc circuit to maintain the dc current at 20 A. If the solar cell resistance is 0.5 \Omega, then each thyristor will be reverse biased for a period of
A
125^{\circ}
B
120^{\circ}
C
60^{\circ}
D
55^{\circ}
GATE EE 2006   Power Electronics
Question 6 Explanation: 
Solar cell emf = 350V
DC current I_{dc}=20A
Solar cell resistance
\begin{aligned} R_{cell}&=0.5\Omega \\ V_0&=\text{Voltage across inverter} \\ &=-(E-I_{dc}R_{cell}) \\ &=-(350-20 \times 0.5)=-340V \end{aligned}
The bridge acts as inverter,
Output voltage of 3-\phi fully controlled bridge
\begin{aligned} V_0&=\frac{3V_{ml} \cos \alpha }{\pi}\\ -340&=\frac{3V_{ml} \cos \alpha }{\pi}\\ -340&=\frac{3 \times 440\sqrt{2} \cos \alpha }{\pi}\\ \Rightarrow \; \alpha &=125^{\circ} \end{aligned}
Therefore, each thyristor will be reverse biased for a period of 55^{\circ}.
Question 7
The speed of a 3-phase, 440 V, 50 Hz induction motor is to be controlled over a wide range from zero speed to 1.5 time the rated speed using a 3-phase voltage source inverter. It is desired to keep the flux in the machine constant in the constant torque region by controlling the terminal voltage as the frequency changes. The inverter output voltage vs frequency characteristic should be
A
A
B
B
C
C
D
D
GATE EE 2006   Power Electronics
Question 7 Explanation: 
For a 3-\phi induction motor, stator voltage per phase is given by
\begin{aligned} V_1 &=\sqrt{2}\pi f_1 N_{pn_1}\phi K_{\omega 1} \\ \phi &\propto \frac{V_1}{f_1} \end{aligned}
To keep the \phi constant, constant V/f control is applied and voltage (V_1) varies linearly with frequency (f_1). For frequency above 50 Hz, such control is not possible because voltage can not be increased above rated voltage. Therefore, for frequency above 50 Hz, voltage is kept constant, During this control induction motor is said to be working in field weakling mode.
Question 8
An electric motor, developing a starting torque of 15 Nm, starts with a load torque of 7 Nm on its shaft. If the acceleration at start is 2 rad/sec^{2} , the moment of inertia of the system must be (neglecting viscous and coulomb friction)
A
0.25 kgm^{2}
B
0.25 Nm^{2}
C
4 kgm^{2}
D
4 Nm^{2}
GATE EE 2005   Power Electronics
Question 8 Explanation: 
T_s= Starting torque developed by the motor =15 N-m
T_L= Load torque =7 N-m
T_a= Accelerating torque =T_s-T_l= 15-7=8 N-m
\alpha =Acceleration = 2 \; rad/sec^2
T_a=I\alpha
I= Moment of inertia =\frac{T_a}{\alpha }=\frac{8}{2}=4kgm^2
Question 9
A variable speed drive rated for 1500 rpm, 40 Nm is reversing under no load. Figure shows the reversing torque and the speed during the transient. The moment of inertia of the drive is
A
0.048 kg m^{2}
B
0.064 kg m^{2}
C
0.096 kg m^{2}
D
0.128 kg m^{2}
GATE EE 2004   Power Electronics
Question 9 Explanation: 
Speed changes from -1500 rpm to 500 rpm in 0.5 sec
So angular acceleration,
\begin{aligned} \alpha &=\frac{500-(-1500)}{0.5}\times \frac{2 \pi}{60}\\ &=418.88 \;rad/sec^2\\ \text{Torque}&=T=20N-m\\ T&=I\alpha \\ &\text{Moment of inertia}\\ I&=\frac{T}{\alpha }=\frac{20}{418.88}=0.048kgm^2 \end{aligned}
Question 10
A single-phase half-controlled rectifier is driving a separately excited dc motor. The dc motor has a back emf constant of 0.5 V/rpm. The armature current is 5A without any ripple. The armature resistance is 2\Omega. The converter is working from a 230V, single-phase ac source with a firing angle of 30^{\circ}. Under this operating condition, the speed of the motor will be
A
339 rpm
B
359 rpm
C
366 rpm
D
386 rpm
GATE EE 2004   Power Electronics
Question 10 Explanation: 


\begin{aligned} \text{Back emf }&=E_a=k\phi N\\ E_a&=k_bN \end{aligned}
where,
k_b= Back-emf constant =0.25 V/rpm
Average output voltage of 1-\phi half controlled rectifier =V
\begin{aligned} V &=\frac{V_m}{\pi} (1+ \cos \alpha ) \\ &= \frac{230\sqrt{2}}{2 \pi} (1+ \cos 30^{\circ})\\ \Rightarrow \; V &=96.6V \\ E_a&=V-I_aR_a \\ &= 96.6-5 \times 2=86.6V\\ \text{Speed}&=N=\frac{E_a}{k_b}=\frac{86.6}{0.25}=346.4V \end{aligned}
There are 10 questions to complete.
Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.