GATE EE 2004

Atresonance, the circuit should be in unity power factor.
Hence, 'Z' should be capacitive.
Admittance of the parallel circuit.
Y=\frac{1}{jL\omega }+\frac{1}{1/jC\omega }=0
\frac{-1}{L\omega }+C\omega =0
\therefore \;\; C=\frac{1}{L\omega ^2}=\frac{1}{2(2 \pi \times 500)^2} =0.05\mu F

When two capacitor formed by two layer of dielectics are connected in series, then equivalent capacitance
\begin{aligned} C_{eq} &= \frac{C_1 \times C_2}{C_1+C_2}\\ &=\frac{\frac{\varepsilon _{r1} \varepsilon _0 A}{d_1} \times \frac{\varepsilon _{r2} \varepsilon _0 A}{d_2}}{\frac{\varepsilon _{r1} \varepsilon _0 A}{d_1} + \frac{\varepsilon _{r2} \varepsilon _0 A}{d_2}} \\ &= \frac{(8.85 \times 10^{-12})(400 \times 10^{-3})^2 \times 4 \times 2}{4 \times 8 \times 10^{-3}+6 \times 2 \times 10^{-3}} \\ & =257 \; \text{pF} \end{aligned}

Let, B=\mu _0nI
Then, \phi =B.S=\mu _0nSI
Where 'S' is the cross sectional area of solenoid flux linkage,
a'=n\phi =\mu _0n^2SI
Hence, Inductance length =\mu _0n^2S
For,
l=1000mm=1m
\therefore \;\;L=4 \pi \times 10^{-7} \times 3000^2 \times \pi \times (30 \times 10^{-3})^2
\;\;\;=32mH

Impedance of the load
=Z_L=R+j\omega L=|Z|\angle \theta _L
where, \theta _L=\tan ^{-1}\left ( \frac{\omega L}{R} \right )
Voltages of 3-\phi supply
\begin{aligned} V_a&=V_m \sin \omega t \\ V_b&=V_m \sin (\omega t-120^{\circ}) \\ V_c &=V_m \sin (\omega t+120^{\circ}) \\ I_a&=\frac{V_a}{Z_L}=\frac{V_m \sin \omega t}{|Z|\angle \theta _L} \\ &= I_M \sin (\omega t-\theta _L)\\ \text{where, } I_m&=\frac{V_m}{|Z|}\\ \text{Similarly,}&\\ I_b&=I_m \sin (\omega t-120-\theta _L)\\ I_c&=I_m \sin (\omega t+120-\theta _L)\\ & \text{Instantaneous power}\\ &=P=V_aI_a+V_bI_b+V_cI_c\\ P&=V_mI_m[\sin \omega t\cdot \sin (\omega t-\theta _L)\\ &+\sin (\omega t-120^{\circ}) \cdot \sin (\omega t-120-\theta _L)\\ &+\sin (\omega t+120^{\circ})\cdot \sin (\omega t+120-\theta _L)]\\ &=\frac{V_mI_m}{2}[(\cos \theta _L -\cos(2\omega t-\theta _L) ) \\ &+(\cos \theta _L -\cos(2\omega t-240-\theta _L) )\\ &+(\cos \theta _L -\cos(2\omega t+240-\theta _L) )]\\ P&=\frac{3V_mI_m}{2}\cos \phi = \text{constant} \end{aligned}

Copper loass at any load
\begin{aligned} P_{cu} &=x^2 \times [\text{Full load copper loss}] \\ &= x^2 \times P_{fl,cu}\\ \text{where,}\;\; x&= \text{fraction of laod} \end{aligned}
Maximum occurs when
Copper loss-Iron loss
x^2 \times P_{fl,cu}=P_i
x^2 \times 600=300 x=0.707 =70.7\%

Detent torque: It is amount of torque that the motor produces when it is not energized. No current is slowing through the winding.
Holding Torque: It is amount of torque that the motor produces when it has rated current flowing through the winding but motor is at rest..
Pull-in Torque: Shows the maximum value of torque at given speeds that the motor can start, stop, or reverse in synchronism with the input pulses.
Pull-out Torque: Shows the maximum value of torque at given speeds that the motor can generate while running in synchronism. If the motor is run outside of this curve, it will stall.

Brushless dc motor definitely has a permanent magnet.

Capacitor start type motor has high staring torque and therefore is used for hard starting loads such as compressor, conveyors, pumps et.
Capacitor-run type motor has better running power factor and efficiency and a quiter and smoother operation. It is used for both easy and hard to start loads. Moder applications of such motors are ceiling fans, blower etc.

To reverse the direction of rotation, phase sequence of the supply has to be reversed. For clockwise direction, the phase sequence was A-B-C. For counter-clockwise direction, the phase sequence has to be A-C-B.

Where, \lambda =N\phi =Flux linkage
Field energy is the energy absorbed by the magnetic system to establish flux \phi.
For a linear electromagnetic circuit
Field energy = Co-energy =\frac{1}{2}\lambda i