# DC Machines

 Question 1
A separately excited DC motor rated $400 \mathrm{~V}, 15A, 1500 RPM$ drives a constant torque load at rated speed operating from $400 \mathrm{~V}$ DC supply drawing rated current. The armature resistance is $1.2 \Omega$. If the supply voltage drops by $10 \%$ with field current unaltered then the resultant speed of the motor in RPM is ___ (Round off to the nearest integer).
 A 1343 B 2541 C 1145 D 1852
GATE EE 2023   Electrical Machines
Question 1 Explanation:
Seperately excited motor : Back emf, $\mathrm{E}_{\mathrm{b}_{1}}=400-15 \times 1.2=382 \mathrm{~V}$
Given: $\quad \tau=$ constant
$\because \quad \tau \propto \phi I_{a}$
where $\phi \rightarrow$ constant
$\therefore \mathrm{I}_{\mathrm{a}} \rightarrow$ consant
If supply voltage is drops by $10 \%$, then new value of supply voltage $=0.9 \times 400=360 \mathrm{~V}$
Now, back emf, $\mathrm{E}_{\mathrm{b}_{2}}=360-15 \times 1.2=342 \mathrm{~V}$

We have,
\begin{aligned} \mathrm{E}_{\mathrm{b}} & =\phi \omega_{\mathrm{m}} \\ \therefore \quad \frac{\mathrm{E}_{\mathrm{b}_{2}}}{\mathrm{E}_{\mathrm{b}_{1}}} & =\frac{\mathrm{N}_{2}}{\mathrm{~N}_{1}} \\ \Rightarrow \quad \frac{342}{382} & =\frac{\mathrm{N}_{2}}{1500} \\ \Rightarrow \quad \mathrm{N}_{2} & =342.93 \mathrm{rpm} \\ &\approx 1343 \mathrm{rpm} \end{aligned}
 Question 2
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 2 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 3
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 3 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 4
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 4 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 5
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 5 Explanation:
\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}