DC Machines

Question 1
A 280 V, separately excited DC motor with armature resistance of 1\Omega and constant field excitation drives a load. The load torque is proportional to the speed. The motor draws a current of 30 A when running at a speed of 1000 rpm. Neglect frictional losses in the motor. The speed, in rpm, at which the motor will run, if an additional resistance of value 10\Omega is connected in series with the armature, is _______. (round off to nearest integer)
A
482
B
254
C
365
D
625
GATE EE 2022   Electrical Machines
Question 1 Explanation: 
Back emf,
E_{b1}=280-30 \times 1 =250V
When additional 10\Omega resistance connected in series with armature.
E_{b2}=280-(10+1) I_{a_2} =280-11I_{a_2}
Given: \tau \propto N
We know, \tau \propto \phi I_a \text{ (Here, }\phi \rightarrow constant)
So,
\begin{aligned} \tau &\propto I_a \\ \therefore \; N &\propto I_a \\ or \; \frac{N_2}{N_1}&=\frac{I_{a_2}}{I_{a_1}}\\ I_{a_2}&=N_2 \times \frac{30}{1000}\\ \therefore \; E_{b_2}&=280-\left ( 11 \times \frac{30}{1000} \right )N_2\\ &\text{We also have,}\\ E_b &\propto\phi \omega \\ \therefore \frac{E_{b_2}}{E_{b_1}}&=\frac{N_2}{N_1}\\ &\frac{280-\left ( 11 \times \frac{30}{1000} \right )N_2}{250}=\frac{N_2}{1000}\\ \Rightarrow N_2&=482.758rpm \end{aligned}
Question 2
A belt-driven \text{DC} shunt generator running at 300\: \text{RPM} delivers 100 \:\text{kW} to a 200 \:V\: \text{DC} grid. It continues to run as a motor when the belt breaks, taking 10 \:\text{kW} from the \text{DC} grid. The armature resistance is 0.025 \Omega, field resistance is 50\:\Omega, and brush drop is 2 \:V. Ignoring armature reaction, the speed of the motor is _____ \text{RPM}. (Round off to 2 decimal places.)
A
275.18
B
254.12
C
362.24
D
148.44
GATE EE 2021   Electrical Machines
Question 2 Explanation: 
\begin{aligned} I_{L} &=\frac{100 \times 10^{3}}{200}=500 \mathrm{~A} \\ I_{S h} &=\frac{200}{50}=4 \mathrm{~A} \\ I_{a} &=504 \mathrm{~A} \\ E_{q} &=V+I_{a} R_{a}+\text { Brush drop } \\ &=200+504(0.025)+2 \mathrm{~V} \\ &=214.6 \mathrm{~V}\\ \text{In motoring case:} V I&=10 \mathrm{~kW}, V=200 \mathrm{~V}\\ \therefore \qquad \qquad I &=\frac{10000}{200}=50 A \\ I_{f} &=4 \mathrm{~A}, I_{a}=I_{L}-I_{f} \\ &=46 \mathrm{~A} \\ E_{b} &=V-I_{a} R_{a}-\text { Brush Drop } \\ &=200-46(0.025)-2=196.85 \mathrm{~V} \\ \frac{N_{m}}{N_{g}} &=\frac{E_{b}}{E_{g}} \\ N_{m} &=\frac{E_{b}}{E_{g}} \times N_{g} \\ \therefore \qquad \qquad N_{m} &=\frac{196.85}{214.6} \times 300=275.18 \mathrm{rpm} \end{aligned}
Question 3
Consider a permanent magnet dc (PMDC) motor which is initially at rest. At t=0, a dc voltage of 5 V is applied to the motor. Its speed monotonically increases from 0 rad/s to 6.32 rad/s in 0.5 s and finally settles to 10 rad/s. Assuming that the armature inductance of the motor is negligible, the transfer function for the motor is
A
\frac{10}{0.5s+1}
B
\frac{2}{0.5s+1}
C
\frac{2}{s+0.5}
D
\frac{10}{s+0.5}
GATE EE 2020   Electrical Machines
Question 3 Explanation: 
\begin{aligned} \text{Input} &= 5 V \\ \Rightarrow \; R(s)&=\frac{5}{s} \\ \frac{C(s)}{R(s)}&=T.F.=\frac{K}{1+Ts} \\ C(s)&=\frac{5K}{s(1+Ts)} \\ \lim_{s\rightarrow 0}&=5K=10\\ \Rightarrow \; K&=2\\ \text{Steady state speed}&=10 \text{ rad/sec.(Given)} \\ C(s)&=\frac{2}{1+0.5s}\times \frac{5}{s} \\ \Rightarrow \; \lim_{s\rightarrow 0}sC(s)&=10 \end{aligned}
Question 4
A 250 V dc shunt motor has an armature resistance of 0.2\Omega and a field resistance of 100\Omega. When the motor is operated on no-load at rated voltage. It draws an armature current of 5 A and runs at 1200 rpm. When a load is coupled to the motor, it draws total line current of 50 A at rated voltage, with a 5% reduction in the air-gap flux due to armature reaction. Voltage drop across the brushes can be taken as 1 V per brush under all operating conditions. The speed of the motor, in rpm, under this loaded condition, is closest to:
A
1200
B
1000
C
1220
D
900
GATE EE 2020   Electrical Machines
Question 4 Explanation: 
no load current 5 A

\begin{aligned} B.R.D&=\text{1 V per brush}\\ \text{loaded, }I_{L}&=50 A \\R_{sh}&=100 \Omega\\ I_{sh}&=\frac{250}{100}=2.5\: A \\I_{a0}&=2.5\: A\\I_{aL}&=47.5\: A \\V&=E_{b}+I_{a}R_{a}+B.R.D\\ E_{b\, no\, load}&=V-I_{a0}R_{a}-B.R.D \\&=250-2.5(0.2)-1\times 2 \\&=247.5\, Volts \\E_{b\, load}&=250-47.5(0.2)-1\times 2 \\&=238.5\, volts\\ \frac{N_{2}}{N_{1}}&=\frac{E_{b_{2}}}{E_{b_{1}}}\times \frac{\phi_{1} }{\phi _{2}} \\ \frac{N_{2}}{1200}&=\frac{238.5}{247.5}\times \frac{\phi _{1}}{0.95\phi_{1}}\\ N_{2}&=1217.22\, rpm \end{aligned}
Question 5
A 220 V DC shunt motor takes 3 A at no-load. It draws 25 A when running at full-load at 1500 rpm. The armature and shunt resistances are 0.5 \Omega and 220 \Omega,respectively. The no-load speed in rpm (round off to two decimal places) is ________
A
4562.82
B
7896.28
C
1579.33
D
1583.44
GATE EE 2019   Electrical Machines
Question 5 Explanation: 


\begin{aligned} &\text{At full load condition,}\\ N_1&=1500 \; \text{rpm} \\ I_1 &=25A \\ I_f&= \frac{220V}{200\Omega }=1A\\ I_{a1}&=I_1-I_f=24A \\ E_{g1}&=V-I_{a1}R_a \\ &=220-24 \times 0.5=208V \\ &\text{At no-load condition,}\\ I_2&= 3A\\ I_f&= 1A\\ I_{a2}&=I_2-I_f=2A \\ E_{g2}&=220-2 \times 0.5=219V \\ E_g&\propto \phi N \\ & \text{Flux id constant,}\\ E_g&\propto N \\ \frac{E_{g2}}{E_{g1}}&=\frac{N_2}{N_1} \\ \frac{219}{208}&=\frac{N_2}{1500} \\ N_2&= 1579.33 \;\text{rpm} \end{aligned}
Question 6
A 200 V DC series motor, when operating from rated voltage while driving a certain load, draws 10 A current and runs at 1000 r.p.m. The total series resistance is 1 \Omega. The magnetic circuit is assumed to be linear. At the same supply voltage, the load torque is increased by 44%. The speed of the motor in r.p.m. (rounded to the nearest integer) is ________ .
A
253
B
825
C
365
D
412
GATE EE 2018   Electrical Machines
Question 6 Explanation: 
Form circuit diagram,

\begin{aligned} E_{b1} &=V-I_a(R_a+R_{se}) \\ &= 200-10(1)=190V\\ N_1&= 1000rpm \end{aligned}
Load torque increased by 44% (T\propto I_a^2)
\begin{aligned} \therefore \; \frac{T_2}{T_1} &=\left ( \frac{I_{a2}}{I_{a1}} \right )^2 \\ \Rightarrow \;\; \frac{1.44T_1}{T_1} &=\left ( \frac{I_{a2}}{10} \right )^2 \\ \therefore \;\; I_{a2}^2&= 144\\ I_{a2}&= 12A\\ &\text{At same voltage,}\\ E_{b2}&=V-I_{a2}(R_a+R_{se}) \\ \therefore \;\; E_{b2} &= 200-12(1)=188V\\ &\text{For series motor,}\\ N&\propto \frac{E_b}{\phi }\;\;(\phi \propto I_a) \\ \frac{N_2}{N_1}&=\frac{E_{b2}}{E_{b1}} \times \frac{I_{a1}}{I_{a2}}\\ \frac{N_2}{1000} &=\frac{188}{190}\times \frac{10}{12} \\ N_2&=824.56\; \text{rpm}\\ &\simeq 825\; \text{rpm} \end{aligned}
Question 7
A separately excited dc motor has an armature resistance R_{a}= 0.05\Omega. The field excitation is kept constant. At an armature voltage of 100 V, the motor produces a torque of 500 Nm at zero speed. Neglecting all mechanical losses, the no-load speed of the motor (in radian/s) for an armature voltage of 150 V is _____ (up to 2 decimal places).
A
50
B
500
C
600
D
850
GATE EE 2018   Electrical Machines
Question 7 Explanation: 
Given separately initiated DC motor,
Field excitaion is constant,

Producing a torque of 500 N-m
\begin{aligned} &\text{At zero speed,}\\ \because \; N&= 0,\; E_b=0\\ E_b &= V-I_aR_a\\ 0&= 100-I_a(0.05)\\ \therefore \; I_{a1} &=\frac{100}{0.05} =2000A\\ T &= \frac{60}{2 \pi N}E_bI_a\\ &= \frac{60}{2 \pi N}\frac{\phi ZNP}{60A}\cdot I_a\\ \text{Let,}\;\; T_1&=\frac{1}{2 \pi}\frac{ZP}{A}\phi I_{a1}\;\text{N-m} \\ \text{Let,}\;\; T_1 &=kI_{a1} \\ \text{where,}\;\; k &= \frac{1}{2 \pi}\frac{ZP \phi}{A}\\ T_1 &= 500 \text{N-m}, \\ I_{a1}&=2000A\\ k&=\frac{500}{2000}=0.25 \end{aligned}
When motor runs on no-load given all mechanical losses neglected. No-load current is negligible and the voltage drop at no-load can be negligible.
\begin{aligned} \therefore \;\;E_b &\simeq V=150V \\ E_b&= \frac{\phi ZNP}{60A}\\ \text{Already solved,}\\ k&=\frac{1}{2\pi} \frac{ZP \phi }{A}\\ \therefore \;\;\frac{ZP \phi }{A} &=2 \pi (k) \\ \therefore \;\; E_b &= 2 \pi (k) \cdot \frac{N}{60}\\ \therefore \;\; E_b&= k \times \omega \\ (\omega : & \text{No-load speed in rad/sec})\\ \therefore \;\;\omega &= \frac{E_b}{k}\\ \therefore \;\;\omega &= \frac{150}{0.25}\\ \omega &= 600 \text{rad/sec} \end{aligned}
Question 8
A 120 V DC shunt motor takes 2 A at no load. It takes 7 A on full load while running at 1200 rpm. The armature resistance is 0.8 \Omega , and the shunt field resistance is 240\Omega . The no load speed, in rpm, is _______________.
A
1856
B
1479
C
3214
D
1242
GATE EE 2017-SET-2   Electrical Machines
Question 8 Explanation: 
Shunt motor take 7A on full load and runs at 1200 rpm.

\begin{aligned} \therefore \; E_{b1} &=120-65(0.8) \\ E_{b1} &= 114.8V\\ N_1&=1200\;\text{rpm} \end{aligned}
Shunt motor takes 2A at no load

\begin{aligned} E_{b2} &=120-1.5(0.8)=118.8 \\ & \text{For shunt motor,}\\ \frac{N_2}{N_1} &= \frac{E_{b2}}{E_{b1}}\\ \therefore \; N_2&=\frac{118.8}{114.8}\times 1200\\ &=1241.81 \text{rpm} \end{aligned}
Question 9
A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance R_a=0.02\Omega. When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply is
A
34.2A
B
30A
C
22A
D
4.84A
GATE EE 2017-SET-2   Electrical Machines
Question 9 Explanation: 
\begin{aligned} \text{Neglecting rotor loss,}&\\ E_bI_a &= 10kW\\ (V-I_aR_a)I_a &= 10kW\\ (220-I_a(0.02))I_a &= 10000\\ 220I_a-0.02I_a^2-10000 &=0 \\ 0.02I_a^2-220I_a+10000&=0\\ \text{By sloving,}\;\; & I_a=45.75A\\ \text{Torque corrosponding,}&\\ T&=\frac{60}{2\pi N}E_bI_a\\ \therefore \; T&=\frac{60}{2 \pi (900)}10000\\ &=106.15N-m \end{aligned}
Separately excited motor,
\therefore \;\;T \propto I_a
For a torque 106.15 N-m
I_a=45.75A
For a torque 70 N-m,
New I_a=\frac{70 \times 45.75}{106.15}=30.16A
Question 10
A separately excited DC generator supplies 150 A to a 145 V DC grid. The generator is running at 800 RPM. The armature resistance of the generator is 0.1 \Omega. If the speed of the generator is increased to 1000 RPM, the current in amperes supplied by the generator to the DC grid is _______.
A
200
B
55
C
550
D
145
GATE EE 2017-SET-1   Electrical Machines
Question 10 Explanation: 


Given, R_a=0.1\Omega
As, DC generator is connected with DC grid, the output voltage will be constant.
When generator is running at 800 rpm,
Generator emf,
\begin{aligned} E_{a1} &=V+I_{a1}R-a \\ &= 145+150 \times 0.1\\ &= 160V \end{aligned}
When generator is running at 1000 rpm,
Generator emf,
\begin{aligned} E_{a2} &=E_{a1} \times \frac{N_2}{N_1}\\ &= 160 \times \frac{1000}{800}\\ &= 200V \end{aligned}
So, new armature currrent will be,
\begin{aligned} I_{a2} &=\frac{E_{a2}-V}{R_a}\\ &= \frac{200-145}{0.1}\\ &= 550A \end{aligned}
There are 10 questions to complete.