Question 1 |
A three-phase, 50 Hz, 4-pole induction motor runs at no-load with a slip of 1%. With
full load, the slip increases to 5 %. The % speed regulation of the motor (rounded off
to 2 decimal places) is _________ .
4.21 | |
16.45 | |
2.21 | |
8.44 |
Question 1 Explanation:
\begin{aligned} \text{Synchronous speed, }\\ N_s&=\frac{120f}{P}\\ N_s&=\frac{120 \times 50}{4}rpm\\ N_s&=1500 \; rpm\\ \text{Speed at no-load,}\\ &=N_s(1-S_{nL})\\ &=1500(1-0.01)\\ &=1485 \; rpm\\ \text{Speed at full-load,}\\ &=N_s(1-S_{fL})\\ &=1500(1-0.05)\\ &=1425 \; rpm\\ \% \text{ speed regulation}&=\frac{N_{NL}-N_{FL}}{N_{FL}} \times 100\\ &=\frac{1485-1425}{1425} \times 100\\ &=4.21\% \end{aligned}
Question 2 |
The magnetic circuit shown below has uniform cross-sectional area and air gap of 0.2 cm. The mean path length of the core is 40 cm. Assume that leakage and fringing fluxes are negligible. When the core relative permeability is assumed to be infinite, the magnetic flux density computed in the air gap is 1 tesla. With same Ampere-turns, if the core relative permeability is assumed to be 1000 (linear), the flux density in tesla(round off to three decimal places) calculated in the air gap is __________
0.257 | |
0.456 | |
0.833 | |
0.658 |
Question 2 Explanation:
\begin{aligned} l_{ag} &=0.2 cm \\ l_m&= 40cm\\ B_0 &= 1 \text{Tesla}\\ \phi &= \frac{mmf}{\Re }\\ &\text{For the same mmf,} \\ \phi &\propto \frac{1}{\Re } \\ \text{In case-1:} &\\ \Re _1&=\frac{l_m}{\mu _0A} \\ &= \frac{l_{ag}}{\mu _0 A}+\frac{l_m}{\mu _0 \mu _r A}\\ \text{as} \;\;\mu _0=0& \\ \therefore \;\;\Re _1&= \frac{0.2 \times 10^{-2}}{\mu _0 A}\\ \text{In case-2:} &\\ \Re _2&= \frac{l_{ag}}{\mu _0 A}+\frac{l_m}{\mu _0 \mu _r A}\\ &=\frac{0.2 \times 10^{-2}}{\mu _0 A}+\frac{40 \times 10^{-2}}{1000\mu _0 A}\\ &=\frac{0.24\times 10^{-2}}{\mu _0 A} \end{aligned}
As flux \phi \propto B for uniform cross section area.
\begin{aligned} \therefore \; B_2&=\frac{B_1 \times \Re _1}{\Re _2} \\ B_2&= \frac{1 \times 0.2 \times 10^{-2}/\mu _0 A}{0.24 \times 10^{-2}/\mu _0A}\\ &= 0.833T \end{aligned}
As flux \phi \propto B for uniform cross section area.
\begin{aligned} \therefore \; B_2&=\frac{B_1 \times \Re _1}{\Re _2} \\ B_2&= \frac{1 \times 0.2 \times 10^{-2}/\mu _0 A}{0.24 \times 10^{-2}/\mu _0A}\\ &= 0.833T \end{aligned}
Question 3 |
The equivalent circuit of a single phase induction motor is shown in the figure, where the
parameters are R_{1}=R'_{2}=X_{i1}=X'_{i2}=12\Omega , X_{M}=240\Omega and s is the slip. At no-load, the motor speed can be approximated to be the synchronous speed. The no-load lagging power factor of the motor is___________ (up to 3 decimal places).
0.182 | |
0.412 | |
0.214 | |
0.106 |
Question 3 Explanation:
Simplifying the above circuit into a simple R-L circuit,
\begin{aligned} Z_{eq} &=\frac{(3+j6)(j120)}{(3+j126)}+(12+j32) \\ &=138.563\angle \\ &\text{Current drawn by motor is } \\ I&= \frac{V\angle 0}{Z\angle \theta } \end{aligned}
(\therefore \; \theta : impedance angle will be p.f. angle)
\therefore No laod lagging p.f. of motor is (\cos\theta ),
\cos(83.9)=0.106 lagging power factor
Question 4 |
A transformer with toroidal core of permeability \mu is shown in the figure. Assuming uniform flux density across the circular core cross-section of radius r\lt \lt R, and neglecting any leakage flux, the best estimate for the mean radius R is
\frac{\mu Vr^{2}{N^{2}}_{P}\omega }{I} | |
\frac{\mu VIr^{2}N_{P}N_{S}\omega }{V} | |
\frac{\mu Vr^{2}{N^{2}}_{P}\omega }{2I} | |
\frac{\mu Ir^{2}{N^{2}}_{P}\omega }{2V} |
Question 4 Explanation:
Since we know reluctance of core:
\Re =\frac{l_c}{\mu a}
Here, l_c= mean core length, a= area of core reluctance,
\begin{aligned} \Re &=\frac{2 \pi R}{\mu \times \pi r^2}\\ &=\frac{2R}{\mu r^2}\\ [\text{where} R &= \text{mean radius of core}] \\ \text{or}\;\; R&=\frac{\Re \mu r^2}{2} \\ \text{and}\;\; \Re &= \frac{mmf}{flux}=\frac{N_pI_p}{\phi }\\ \text{Here,}\;\; \phi &=-\frac{1}{N_p}\int_{0}^{t}e\;dt \\ [\text{where}\; e &\text{is primary generated voltage}] \\ \text{or}\;\; \phi &=-\frac{1}{N_p}\int_{0}^{t}-(V_p)\;dt \\ \phi &= \frac{1}{N_p}\int_{0}^{t}V \cos \omega t \;dt\\ &= \frac{V}{\omega N_p}\sin \omega t\\ &\text{Now reluctance} \\ \Re &= \frac{N_p I \sin \omega t}{\frac{V}{\omega N_p} \sin \omega t}\\ &= \frac{N_p^2\omega I}{V}\\ & \text{So mean radius,}\\ R&=\frac{N_p^2 \times \omega I \times \mu r^2}{2V} \end{aligned}
Question 5 |
A 375W, 230 V, 50 Hz capacitor start single-phase induction motor has the following constants
for the main and auxiliary windings (at starting): Z_{m}=(24.50+j12.75)\Omega (main winding), Z_{n}=(24.50+j12.75)\Omega (auxiliary winding). Neglecting the magnetizing branch the value of the capacitance (in \muF ) to be added in series with the auxiliary winding to obtain maximum torque at starting is _______.
38.43 | |
200 | |
150 | |
50 |
Question 5 Explanation:
Auxiliary winding impedance =(24.5+j12.75)\Omega
Let X_c be the reactance of the capacitor connected in auxiliary winding
\begin{aligned} Z_a &= (24.5+j12.75-jX_c)\Omega \\ X_a &=(X_c-12.75),\; R_a=24.5\Omega \\ \therefore \;\; \phi _a &=\tan^{-1}\frac{X_c-12.75}{24.5}\;...(i) \\ Z_m&= 12.5+j15.75\\ &= 20.10\angle 51.56^{\circ}\Omega \\ \phi _m&= 51.56^{\circ} \end{aligned}
For maximum torque at starting, condition on \phi _a and \phi _m is given by \left [ \phi _a=\frac{90^{\circ}-\phi _m}{2} \right ]
\phi _a=\frac{90-51.56}{2}=19.22^{\circ}}\;...(ii)
Comparing equation (i) and (ii),
\begin{aligned} \tan(19.22^{\circ}) &=\frac{X_c-12.75}{24.5} \\ 0.3486 &= \frac{X_c-12.75}{24.5}\\ X_c &=8.541+12.75 \\ \text{Capacitive reactance,}& \\ X_c &= 21.291\Omega \\ \frac{1}{2 \pi fc}&= 21.291\\ \Rightarrow \;C&= \frac{1}{2\pi \times 50 \times 21.291}\\ &= 149.51\mu F \end{aligned}
Let X_c be the reactance of the capacitor connected in auxiliary winding
\begin{aligned} Z_a &= (24.5+j12.75-jX_c)\Omega \\ X_a &=(X_c-12.75),\; R_a=24.5\Omega \\ \therefore \;\; \phi _a &=\tan^{-1}\frac{X_c-12.75}{24.5}\;...(i) \\ Z_m&= 12.5+j15.75\\ &= 20.10\angle 51.56^{\circ}\Omega \\ \phi _m&= 51.56^{\circ} \end{aligned}
For maximum torque at starting, condition on \phi _a and \phi _m is given by \left [ \phi _a=\frac{90^{\circ}-\phi _m}{2} \right ]
\phi _a=\frac{90-51.56}{2}=19.22^{\circ}}\;...(ii)
Comparing equation (i) and (ii),
\begin{aligned} \tan(19.22^{\circ}) &=\frac{X_c-12.75}{24.5} \\ 0.3486 &= \frac{X_c-12.75}{24.5}\\ X_c &=8.541+12.75 \\ \text{Capacitive reactance,}& \\ X_c &= 21.291\Omega \\ \frac{1}{2 \pi fc}&= 21.291\\ \Rightarrow \;C&= \frac{1}{2\pi \times 50 \times 21.291}\\ &= 149.51\mu F \end{aligned}
Question 6 |
The flux linkage (\lambda) and current (i) relation for an electromagnetic system is \lambda = (\sqrt{i})/g. When i=2A and g (air-gap length)=10cm, the magnitude of mechanical force on the moving part, in N, is ________.
128.52 | |
165.65 | |
188.56 | |
214.25 |
Question 6 Explanation:
Energy in magnetic system,
\begin{aligned} W_f &= \int_{0}^{\lambda } i(\lambda )\; d\lambda \\ \because \;\; i &=\lambda ^2g^2 \\ W_f &= \int_{0}^{\lambda } \lambda ^2g^2 \; d\lambda \\ &=g^2\frac{\lambda ^3}{2} \end{aligned}
Now, mechanical force,
\begin{aligned} F_F &=-\frac{\partial W_f (\lambda ,g)}{\partial g} \\ &=-\frac{\partial }{\partial g} \left ( \frac{g^2\lambda ^3}{3} \right )\\ &= -\frac{2}{3}\lambda ^3 g\\ \because \;\; i=2, \;g&=10cm ,\; \lambda =10\sqrt{2} \\ \text{Hence,}\;\;|F_F|&=\frac{2}{3} \times (10\sqrt{2})^3 \times 0.1\\ &=188.56N \end{aligned}
\begin{aligned} W_f &= \int_{0}^{\lambda } i(\lambda )\; d\lambda \\ \because \;\; i &=\lambda ^2g^2 \\ W_f &= \int_{0}^{\lambda } \lambda ^2g^2 \; d\lambda \\ &=g^2\frac{\lambda ^3}{2} \end{aligned}
Now, mechanical force,
\begin{aligned} F_F &=-\frac{\partial W_f (\lambda ,g)}{\partial g} \\ &=-\frac{\partial }{\partial g} \left ( \frac{g^2\lambda ^3}{3} \right )\\ &= -\frac{2}{3}\lambda ^3 g\\ \because \;\; i=2, \;g&=10cm ,\; \lambda =10\sqrt{2} \\ \text{Hence,}\;\;|F_F|&=\frac{2}{3} \times (10\sqrt{2})^3 \times 0.1\\ &=188.56N \end{aligned}
Question 7 |
The direction of rotation of a single-phase capacitor run induction motor is reversed by
interchanging the terminals of the AC supply | |
interchanging the terminals of the capacitor | |
interchanging the terminals of the auxiliary winding | |
interchanging the terminals of both the windings |
Question 7 Explanation:
Inter changing the terminals of the auxiliary winding.
Question 8 |
A 220 V, 3-phase, 4-pole, 50 Hz inductor motor of wound rotor type is supplied at rated voltage and frequency. The stator resistance, magnetizing reactance, and core loss are negligible. The maximum torque produced by the rotor is 225 % of full load torque and it occurs at 15% slip. The actual rotor resistance is 0.03 \Omega/phase. The value of external resistance (in Ohm) which must be inserted in a rotor phase if the maximum torque is to occur at start is ______.
0 | |
0.18 | |
0.24 | |
0.28 |
Question 9 |
A 200 V, 50 Hz, single-phase induction motor has the following connection
diagram and winding orientations as shown. MM' is the axis of the main stator
winding(M_1M_2) and AA' is that of the auxiliary winding(A_1A_2). Directions of the
winding axis indicate direction of flux when currents in the windings are in the
directions shown. Parameters of each winding are indicated. When switch S is
closed, the motor
rotates clockwise | |
rotates anti-clockwise | |
does not rotate | |
rotates momentarily and comes to a halt |
Question 9 Explanation:
f=50Hz
Impedance of main winding
\begin{aligned} Z_m &= r_m+j2 \pi f L_m\\ &=0.1+ j 2 \pi \times 20 \times \frac{0.1}{\pi} \\ Z_m&=0.1+j10\Omega \end{aligned}
Impedance of auxiliary winding,
\begin{aligned} Z_a &=r_a+j2 \pi fL_a \\ &= 1+j 2 \pi \times 50 \times \frac{10}{\pi}\\ Z_a&= 1+j1000\Omega \end{aligned}
Current through main winding,
I_m=\frac{V_s}{Z_m}=\frac{220}{0.1+j10}
I_m=200\angle -89.427^{\circ}A
Current through auxiliary winding
I_a=\frac{V_s}{Z_a}
\;\;\;=\frac{220}{1+j1000}=0.22\angle -89.942
Thaing V_s as the reference
I_m leads I_a the field crreated by the two current also have same difference thereby constituting an unbalanced field system, The result is the production of the starting torque.
Space orientation of the field
The motor rotates in the direction of leading phase to lagging phase.In this case, the motor rotates anti-clockwise.
Impedance of main winding
\begin{aligned} Z_m &= r_m+j2 \pi f L_m\\ &=0.1+ j 2 \pi \times 20 \times \frac{0.1}{\pi} \\ Z_m&=0.1+j10\Omega \end{aligned}
Impedance of auxiliary winding,
\begin{aligned} Z_a &=r_a+j2 \pi fL_a \\ &= 1+j 2 \pi \times 50 \times \frac{10}{\pi}\\ Z_a&= 1+j1000\Omega \end{aligned}
Current through main winding,
I_m=\frac{V_s}{Z_m}=\frac{220}{0.1+j10}
I_m=200\angle -89.427^{\circ}A
Current through auxiliary winding
I_a=\frac{V_s}{Z_a}
\;\;\;=\frac{220}{1+j1000}=0.22\angle -89.942
Thaing V_s as the reference
I_m leads I_a the field crreated by the two current also have same difference thereby constituting an unbalanced field system, The result is the production of the starting torque.
Space orientation of the field
The motor rotates in the direction of leading phase to lagging phase.In this case, the motor rotates anti-clockwise.
Question 10 |
A 230 V, 50 Hz, 4-pole, single-phase induction motor is rotating in the clockwise
(forward) direction at a speed of 1425 rpm. If the rotor resistance at standstill is
7.8 \Omega, then the effective rotor resistance in the backward branch of the equivalent circuit will be
2\Omega | |
4\Omega | |
78\Omega | |
156\Omega |
Question 10 Explanation:
Rotor resistance at stand still, R=7.8\Omega
Synchronous speed,
N_s=\frac{120f}{P}
\;\;=\frac{120 \times 50}{4}=1500 \text{rpm}
The slip(s) of rotor with respect to forward field
s=\frac{N_s-N_r}{N_s}
\;\;=\frac{1500-1425}{1500}=0.05
The slip of rotor with respect to backward field
=2-s=2-0.05=1.95
Effective rotor resistance
\;\;=\frac{R_2}{2(2-s)}=7.8
\;\;=\frac{7.8}{2 \times 1.95}=2\Omega
Synchronous speed,
N_s=\frac{120f}{P}
\;\;=\frac{120 \times 50}{4}=1500 \text{rpm}
The slip(s) of rotor with respect to forward field
s=\frac{N_s-N_r}{N_s}
\;\;=\frac{1500-1425}{1500}=0.05
The slip of rotor with respect to backward field
=2-s=2-0.05=1.95
Effective rotor resistance
\;\;=\frac{R_2}{2(2-s)}=7.8
\;\;=\frac{7.8}{2 \times 1.95}=2\Omega
There are 10 questions to complete.