Difference Between DC Motor and Induction Motor Feature DC Motor Induction Motor Power Supply Direct Current (DC) Alternating Current (AC) Working Principle Magnetic force on a current-carrying conductor Electromagnetic induction Rotor Current Supplied directly Induced automatically Brushes Required (except BLDC motors) Not required Commutator Present Not present Maintenance Higher Lower Efficiency High Very High Speed Control Easy and precise Requires Variable Frequency Drive (VFD) Starting Torque High Moderate to High Cost Usually higher Generally lower Reliability Good Excellent Common Applications Electric vehicles, cranes, robotics, conveyors Pumps, fans, compressors, industrial machinery Advantages of DC Motors Excellent speed control. High starting torque. Quick acceleration and deceleration. Suitable for battery-powered applications. Ideal for precision control systems. Advantages of Induction Motors Simple construction. Low maintenance. No brushes or commutator. Long se...

Electric motors are essential components in industrial machines, household appliances, electric vehicles, robotics, and automation systems. Among the most widely used motors are the DC motor and the induction motor . Although both convert electrical energy into mechanical energy, they differ significantly in their operating principles, construction, and applications. What Is a DC Motor? A Direct Current (DC) motor is an electrical machine that converts direct current (DC) electrical energy into mechanical rotation. It operates based on the interaction between a magnetic field and a current-carrying conductor. How Does a DC Motor Work? The working principle of a DC motor is based on Lorentz Force , which states that when a current-carrying conductor is placed in a magnetic field, it experiences a force. This force creates torque that rotates the motor shaft. Working Process DC voltage is supplied to the armature winding. The armature current produces ...

  A traditional ceiling fan is an example of an induction motor , specifically a single-phase induction motor . Traditional ceiling fan: usually a single-phase induction motor , commonly a capacitor-run motor . The capacitor creates a phase difference between the main and auxiliary windings, producing a rotating magnetic field.  A single AC wire is enough to produce a magnetic field, but not a naturally rotating magnetic field.  Suppose we connect one coil to AC:  So the field simply gets stronger, weaker, and reverses direction. It does not continuously rotate around the motor.  Because of this, a single-phase induction motor has zero starting torque when stationary (in the ideal case). Role of a Capacitor in a Ceiling Fan  A typical AC fan has two windings:  - Main winding → connected to AC - Auxiliary winding → AC through a capacitor  The capacitor shifts the current in the auxiliary winding, creating a phase difference between the two magne...

  The key point is that the magnetic field is continuously changing inside the stator. In an AC induction motor: AC current changes direction every half cycle. Therefore, the magnetic field also changes direction . In a 3-phase induction motor , the three phase currents are 120° apart, so their combined magnetic field produces a continuously rotating magnetic field . The rotor experiences this continuously changing magnetic field, so current is induced in the rotor. That induced rotor current produces torque , keeping the rotor rotating. So even though the AC current becomes zero momentarily , the motor doesn't stop. The rotating magnetic field has inertia/momentum in the rotor, and the field immediately changes direction and continues rotating. But in a 3-phase motor,  the magnetic field is changing continuously . That's why a 3-phase induction motor runs continuously rather than starting and stopping 50 times per second. A single-phase induction motor is differe...

  If V r e f V_{ref} is the output reference value and V f s V_{fs} is the ideal full-scale output voltage when all digital inputs are 1. Which of the following equations is correct? A. V r e f = V f s ( 1 − 1 2 n ) V_{ref} = V_{fs}\left(1-\frac{1}{2^n}\right) B. V f s = V r e f ( 1 2 n − 1 ) V_{fs} = V_{ref}\left(\frac{1}{2^n}-1\right) C. V f s = V r e f ( 1 − 1 2 n ) V_{fs} = V_{ref}\left(1-\frac{1}{2^n}\right) D. V r e f = V f s ( 1 2 n − 1 ) V_{ref} = V_{fs}\left(\frac{1}{2^n}-1\right) The correct answer is C : V f s = V r e f ( 1 − 1 2 n ) \boxed{V_{fs}=V_{ref}\left(1-\frac{1}{2^n}\right)} Why? For an n-bit DAC , the largest digital input is: 111 … 111 = 2 n − 1 111\ldots111 = 2^n-1 The DAC output is: V o = V r e f Digital input 2 n V_o=V_{ref}\frac{\text{Digital input}}{2^n} At the maximum input: V f s = V r e f 2 n − 1 2 n V_{fs}=V_{ref}\frac{2^n-1}{2^n} Separate the fraction: V f s = V r e f ( 2 n 2 n − 1 2 n ) V_{fs}=V_{ref}\left(\frac{2^n}{2^n}-\frac{1}{2^n}\r...

  Q. The final code after encoding data bits 1101 into 7-bit even parity Hamming Code is: A. 1110101 B. 1011101 C. 1010101 D. 0110101 Hamming (7,4) – Step by Step Data bits: D = [1,1,0,1] Step 1: Parity-Check Matrix H H = [1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 0 1 1 1 1] Parity bits are at positions 1,2,4; data bits at positions 3,5,6,7. Step 2: Extract P from H Using standard form H = [P^T | I], we get: P^T (columns 3,5,6,7) = [1 1 0 1 1 0 1 1 0 1 1 1] Transpose to get P : P = [1 1 0 1 0 1 0 1 1 1 1 1] Step 3: Generator Matrix G = [I | P] G = [1 0 0 0 | 1 1 0 0 1 0 0 | 1 0 1 0 0 1 0 | 0 1 1 0 0 0 1 | 1 1 1] Step 4: Encode Data → Codeword Multiply D × G (mod 2): D = [1 1 0 1] C = D × G = [1 1 0 1 1 0 0] In the above the sequence is D1 D2 D3 D4 P1 P2 P3   (But Transmitted Codeword should be = [P1 P2 D1 P3 D2 D3 D4] or 1010101 ) Answer: C. 1010101 Another Approach Data bits: 1101 1101 For a 7-bit Hamming code , parity bi...

  Q. The A/O gates in which an additional variable or a combination of variables can be included in the logic operation are called: Options: A. AOI Gates B. Expandable Gates C. Variable Gates D. Scalable Gates The correct answer is: B. Expandable Gates \boxed{\text{B. Expandable Gates}} Why? Expandable gates allow additional variables/inputs to be incorporated into the logic operation. Final answer C. 1010101   Browse All Solved Papers (2012 - 2025) → UGC-NET : Electronics Science Study Material (Subject: 088) → Further Reading: GATE EC Previous Year Papers with Solutions

Home / UGC NET PYQ / June 2020 Solved UGC NET Electronic Science June 2020 Question Paper with Answer Key and Full Explanations 📥 Download Question Paper (PDF) 2025 2024 2023 2022 2021 2020 Explanations 1.  Answer: Option (2) For Varactor Diode, m = ln(C2/C1) / ln{(1+V2/phi) / (1+V1/phi)} 2.  Answer: Option (4) 3.  Answer: Option (2) 4.  Answer: Option (2) 5.  Answer: Option (4) 6.  Answer: Option (4)  L= (5*10^-3) / P Or, P = (5*10^-3) / (5*10^7) = 10^-10 7.  Answer: Option (4) 8.  Answer: Option (3) 9.  Answer: Option (3) 10.  Answer: Option (1) 11.  Answer: Option (2) 12.  Answer: Option (2) 13.  Answer: Option (1) Slew Rate Slew rate tells us how fast the output voltage of an amplifier/op-amp can change . It is defined as: S R = max ⁡ ∣ d...