Electrical Engineer

  Derivation the value of K in induction motor  torque equation  In case of 3-phase induction motor , there occur copper losses in rotor. These rotor copper losses are expressed as P c = 3I 2 2 R 2 We know that rotor current, Substitute this value of I 2 in the equation of rotor copper losses, P c . So, we get The ratio of P 2 : P c : P m = 1 : s : (1 – s) Where, P 2 is the rotor input, P c is the rotor copper losses, P m is the mechanical power developed. Substitute the value of Pc in above equation we get, On simplifying we get, The mechanical power developed P m = Tω, Substituting the value of P m We know that the rotor speed N = N s (1 – s) Substituting this value of rotor speed in above equation we get, N s is speed in revolution per minute (rpm) and n s is speed in revolution per sec (rps) and the relation between the two is Substitute this value of N s in above equation and simplifying it we get Comparing both the equations, we get, constant   ...

  Firstly the magnitude of rotor current, secondly the flux which interact with the rotor of three phase induction motor and is responsible for producing emf in the rotor part of the induction motor  lastly the power factor  of rotor of the three phase induction motor. Combining all these factors, we get the equation of torque as- Where, T is the torque produced by the induction motor, φ is flux responsible for producing induced emf, I 2 is rotor current, cosθ 2 is the power factor of rotor circuit. The flux φ produced by the stator is proportional to stator emf E 1 . i.e φ ∝ E 1 We know that transformation ratio K is defined  as the ratio of secondary voltage (rotor voltage) to that of primary voltage (stator voltage). Rotor current I 2 is defined as the ratio of rotor-induced emf under running condition , sE 2 to total impedance, Z 2 of rotor side, and total impedance Z 2 on the rotor side is given by , Putting this value in above equation we get, s...

  Working of Diode In the N-type region, electrons are the majority charge carriers and holes are the minority charge carriers. In the P-type region, the holes are the majority of charge carrier and the electrons are negative charge carriers. Because of the concentration difference, the majority charge carriers diffuse and recombine with the opposite charge. It makes a positive or negative ion. These ions are collected at the junction. And this region is known as the depletion region. When anode terminal of diode is connected with a negative terminal and cathode is connected with the positive terminal of a battery, the diode is said to be connected in reverse bias. Similarly, when anode terminal is connected with a positive terminal and cathode is connected with the negative terminal of the battery, the diode is said to be connected in forward bias. Operation of Diode in Reverse Bias Condition The diode is connected in reverse bias. In this condition, free electrons diffusing into ...

  What is a Diode? “Di “= Two , and “ Ode “= Electrodes  i.e a device or component having two electrodes viz Anode “+” (P) and Cathode “-” (N). A diode is a two-terminal unidirectional power electronics device. The  semiconductor  diode is the first invention in a family of semiconductor electronics devices. After that many  types of diodes  are invented. But today also the most commonly used diode is a semiconductor diode. Generally, silicon is used to make a diode. But another semiconductor material like germanium or germanium arsenide is also used. A diode allows current to flow only in one direction and it blocks the current in another direction. It offers low resistance (ideally zero) in one direction and it offers a high resistance (ideally infinite) in another direction. Symbol of Diode

  Conductor Material Required in Underground 1-Phase 2-Wire AC System with Mid-Point Earthed Figure-2 shows the underground single-phase two-wire AC system with mid-point earthed. Here, M a x i m u m v o l t a g e b e t w e e n l i n e w i r e s = V m R M S v a l u e o f v o l t a g e = V m 2 √ P o w e r f a c t o r o f t h e l o a d = c o s ϕ Thus, the load current is given by, I 2 = P V m 2 √ c o s ϕ = 2 – √ P V m c o s ϕ If 𝑎 2  is the cross-section area per conductor and R 2  is the resistance per conductor. Then, the total power loss in the line conductors is W = 2 I 2 2 R 2 = 2 × ( 2 – √ P V m c o s ϕ ) 2 × ( ρ l a 2 ) ⇒ W = 4 P 2 ρ l V 2 m c o s 2 ϕ a 2 ∴ A r e a o f c r o s s s e c t i o n , a 2 = 4 P 2 ρ l W V 2 m c o s 2 ϕ Hence, the volume of conductor material required in the underground single-phase two-wire AC system with mid-point earthed, say K 1 , is given by, K 1 = 2 a 2 l = 2 × ( 4 P 2 ρ l W V 2 m c o s 2 ϕ ) × l ∴ K 1 = 8 P 2 ρ l 2 W V 2 m c o s 2 ϕ ⋅...