Question 1 |
The four stator conductor \left(A, A^{\prime}, B\right. and \left.B^{\prime}\right) of a rotating machine are carrying DC currents of the same value, the directions of which are shown in figure (i). The rotor coils a-a^{\prime} and b-b^{\prime} are formed by connecting the back ends of conductor 'a' and 'a^{\prime}' and 'b' and 'b^{\prime}', respectively, as shown in figure (ii). The e.m.f. induced in coil a-a^{\prime} and coil b-b^{\prime} are denoted by E_{a-a^{\prime}} and E_{b-b^{\prime}}; respectively. If the rotor is rotated at uniform angular speed \omega \mathrm{rad} / \mathrm{s} in the clockwise direction then which of the following correctly describes the \mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}} and \mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}} ?
E_{a-a^{\prime}} and E_{b-b^{\prime}} have finite magnitudes and are in the same phase | |
E_{a-a^{\prime}} and E_{b-b^{\prime}} have finite magnitudes with \mathrm{E}_{\mathrm{b}-b^{\prime}} leading \mathrm{E}_{\mathrm{a-a}} | |
\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}} and \mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}} have finite magnitudes with E_{a-a^{\prime}} leading E_{b-b^{\prime}} | |
\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}=\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}=0 |
Question 1 Explanation:
Since, a-a' and b-b' are placed at GNA (Geometric Neutral Axis) therefore,
\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}=\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}=0
\mathrm{E}_{\mathrm{a}-\mathrm{a}^{\prime}}=\mathrm{E}_{\mathrm{b}-\mathrm{b}^{\prime}}=0
Question 2 |
A 10-pole, 50 \mathrm{~Hz}, 240 \mathrm{~V}, single phase induction motor runs at 540 RPM while driving rated load. The frequency of induced rotor currents due to backward field is
100 \mathrm{~Hz} | |
95 \mathrm{~Hz} | |
10 \mathrm{~Hz} | |
5 \mathrm{~Hz} |
Question 2 Explanation:
Synchronous speed,
\begin{aligned} N_{s} & =\frac{120 f}{P} \\ & =\frac{120 \times 50}{10}=600 \mathrm{rpm} \end{aligned}
Now, forward slip,
\begin{aligned} S & =\frac{N_{\mathrm{s}}-N}{N_{\mathrm{s}}} \\ & =\frac{600-540}{600}=0.1 \end{aligned}
\therefore Backward slip =2-\mathrm{S}=1.9
Thus, required frequency =1.9 \times 50=95 \mathrm{~Hz}
\begin{aligned} N_{s} & =\frac{120 f}{P} \\ & =\frac{120 \times 50}{10}=600 \mathrm{rpm} \end{aligned}
Now, forward slip,
\begin{aligned} S & =\frac{N_{\mathrm{s}}-N}{N_{\mathrm{s}}} \\ & =\frac{600-540}{600}=0.1 \end{aligned}
\therefore Backward slip =2-\mathrm{S}=1.9
Thus, required frequency =1.9 \times 50=95 \mathrm{~Hz}
Question 3 |
The following columns present various modes of induction machine operation and the ranges of slip
\begin{array}{ll} \textbf{A (Mode of operation)}& \textbf{B (Range of slip)}\\\\ \text{a. Running in generator mode}&\text{p) From 0.0 to 1.0}\\\\ \text{b. Running in motor mode} & \text{q) From 1.0 to 2.0}\\\\ \text{c. Plugging in motor mode} & \text{r) From - 1.0 to 0.0} \end{array}
The correct matching between the elements in column A with those of column B is
\begin{array}{ll} \textbf{A (Mode of operation)}& \textbf{B (Range of slip)}\\\\ \text{a. Running in generator mode}&\text{p) From 0.0 to 1.0}\\\\ \text{b. Running in motor mode} & \text{q) From 1.0 to 2.0}\\\\ \text{c. Plugging in motor mode} & \text{r) From - 1.0 to 0.0} \end{array}
The correct matching between the elements in column A with those of column B is
a-r, b-p, and c-q | |
a-r, b-q, and c-p | |
a-p, b-r, and c-q | |
a-q, b-p, and c-r |
Question 3 Explanation:
Torque speed characteristic of 3-\phi IM :
\mathrm{S} \gt 1 \Rightarrow Plugging mode
0 \lt \mathrm{S} \lt 1 \Rightarrow Motoring mode
\mathrm{S} \lt 0 \Rightarrow Generating Mode
\mathrm{S} \gt 1 \Rightarrow Plugging mode
0 \lt \mathrm{S} \lt 1 \Rightarrow Motoring mode
\mathrm{S} \lt 0 \Rightarrow Generating Mode
Question 4 |
A 4-pole induction motor with inertia of 0.1 kg-m^2 drives a constant load torque of
2 Nm. The speed of the motor is increased linearly from 1000 rpm to 1500 rpm in
4 seconds as shown in the figure below. Neglect losses in the motor. The energy, in
joules, consumed by the motor during the speed change is ____________. (round
off to nearest integer)
1732 | |
2534 | |
1245 | |
3251 |
Question 4 Explanation:
We have
\begin{aligned} J\frac{d\omega }{dt}&=\tau _e-\tau _L\\ J\omega \frac{d\omega }{dt}&=P_e-P_L \;\;(\because P=\tau \omega )\\ \therefore P_e&=J\omega \frac{d\omega }{dt}+P_L\\ E&=\int P_edt\\ R&=J\int_{1000}^{1500}\omega d\omega +\int_{1000}^{1500}P_Ldt\\ &=\frac{J}{2}\left ( \frac{2 \pi}{60} \right )^2[1500^2-1000^2]+\frac{2 \pi \tau }{60}\int_{1000}^{1500}Ndt\\ &=\frac{0.1}{2}\left ( \frac{2 \pi}{60} \right )^2\times 125 \times 10^4+\frac{2 \pi}{60} \times 2\times [1000 \times 4+\frac{1}{2} \times 500 \times 4]\\ &=1732.586J \end{aligned}
\begin{aligned} J\frac{d\omega }{dt}&=\tau _e-\tau _L\\ J\omega \frac{d\omega }{dt}&=P_e-P_L \;\;(\because P=\tau \omega )\\ \therefore P_e&=J\omega \frac{d\omega }{dt}+P_L\\ E&=\int P_edt\\ R&=J\int_{1000}^{1500}\omega d\omega +\int_{1000}^{1500}P_Ldt\\ &=\frac{J}{2}\left ( \frac{2 \pi}{60} \right )^2[1500^2-1000^2]+\frac{2 \pi \tau }{60}\int_{1000}^{1500}Ndt\\ &=\frac{0.1}{2}\left ( \frac{2 \pi}{60} \right )^2\times 125 \times 10^4+\frac{2 \pi}{60} \times 2\times [1000 \times 4+\frac{1}{2} \times 500 \times 4]\\ &=1732.586J \end{aligned}
Question 5 |
The type of single-phase induction motor, expected to have the maximum power
factor during steady state running condition, is
split phase (resistance start). | |
shaded pole. | |
capacitor start. | |
capacitor start, capacitor run. |
There are 5 questions to complete.