13) An amplifier has power gain of 800. Its decibel power gain is:  19 dB [Option ID = 2881] 32 dB [Option ID = 2882] 28 dB [Option ID = 2883] 29 dB [Option ID = 2884] Answer: D   Solution 10*log[10]800 = 29 Previous yr Question papers with Full Explanations → Electronics and Communiaction Study Materials →

How a Boost Converter Increases 20V to 40V Without Breaking Energy Conservation Ever wondered how a circuit can convert 20V into 40V without violating the law of energy conservation ? The answer lies in a powerful electronic device called a Boost Converter . What is a Boost Converter? A boost converter is a type of DC-DC converter that increases (boosts) input voltage to a higher output voltage using smart energy transfer techniques. Main Components: Inductor Switch (Transistor) Diode Capacitor It does NOT create energy . Instead, it converts low voltage & high current into high voltage & low current. Step-by-Step Working of a Boost Converter Step 1: Switch ON (Energy Storage Phase) The transistor closes. Current flows through the inductor. The inductor stores energy in its magnetic field. Energy stored in inductor: E = ½ × L × I² During this phase, the diode blocks current to the output. Step 2: Swi...

12) The roots of a system having a transfer function G(s) = 6(s + 2) / [(s + 3)(s + 4)] will be:  either -3 or -4 [Option ID = 2877] either -2 or -4 [Option ID = 2878] either -3 or -2 [Option ID = 2879] either 3 or 4 [Option ID = 2880] Answer: A Solution The Fundamental Concept: In Control Systems, the "Roots of a System" refer to the roots of the Characteristic Equation , which is the denominator of the Transfer Function. These are also known as Poles . Zeros: Roots of the Numerator (top). Poles (Roots): Roots of the Denominator (bottom). 1. Identify the Transfer Function G(s) = 6(s + 2) (s + 3)(s + 4) 2. Set the Denominator to Zero To find the system roots (poles), we solve for s in...

  11) Consider the network shown in the following figure. The state equation of the system will be: Answer: A Solution (State Equations) In state-space analysis of circuits, we typically choose the voltage across the capacitor ( V c ) and the current through the inductor ( i L ) as our state variables. Step 1: Apply KCL at the top node The input current I splits into the capacitor current (i c ) and the inductor branch current (i L ): I = i c + i L Since i c = C(dV c /dt): I = C(dV c /dt) + i L Rearranging for the derivative of the first state variable: dV c /dt = (0)V c - (1/C)i L + (1/C)I   --- (Eq. 1) Step 2: Apply KVL to the right-hand branch The voltage across the capacitor (V c ) is equal to the voltage across the series combination of L and R: V c = V L + V R Since V L = L(di L /dt) and V R = i L ...

10) A differential equation is given as x(n+2) + 3x(n+1) + 2x(n) = 0; x(0) = 0, x(1) = 1. The solution of this equation will be:  x(n) = -(-1) n + (2) n [Option ID = 2869] x(n) = -(-2) n - (1) n [Option ID = 2870] x(n) = -(-1) n + (-2) n [Option ID = 2871] x(n) = (-1) n - (-2) n [Option ID = 2872] Correct Answer: x(n) = (-1) n - (-2) n [Option ID = 2872] Answer: D Solution Note: This is a Linear Difference Equation (Recurrence Relation). It is solved similarly to a second-order linear differential equation by finding characteristic roots. 1. Identify the Equation & Conditions x(n+2) + 3x(n+1) + 2x(n) = 0 Initial Conditions: x(0) = 0, x(1) = 1 2. Form the Characteristic Equation Assume a solution of the form x(n) = r n . Substituting this into the equation gives: ...

9) The circuit shown in the following figure is initially under steady-state condition. The switch is moved from position 1 to position 2 at t = 0. The current after switching will be: 2e -5t A [Option ID - 2865] 1 - 2e -5t A [Option ID - 2866] 1 + 2e -5t A [Option ID - 2867] 2e 5t A [Option ID - 2868] Answer: A Solution Step 1: Initial State (t < 0) Before the switch moves, it is at position 1. In a DC steady-state condition, an inductor acts as a short circuit . The current flows only through the voltage source (20V) and resistor R 1 (10Ω). i(0 - ) = V / R 1 = 20V / 10Ω = 2 A Step 2: The Switching Property A...

  8) If C s and C l are the equilibrium concentration of impurities in the solid and liquid near the interface respectively, then for dilute solutions encountered in silicon growth, an equilibrium segregation coefficient k 0 may be defined as: C s / C l C l / C s C s - C l (C s - C l ) / C l Answer: A Solution When making silicon wafers for computer chips, we start with a vat of molten (liquid) silicon. We then slowly "grow" a solid crystal from this melt. To make the silicon conduct electricity, we add "impurities" (dopants like Boron or Phosphorus). As the liquid silicon turns into solid silicon at the interface (the boundary where they meet), the impurities don't distribute themselves equally. Some impurities "prefer...

Advanced Thermodynamic Refrigeration Simulator Cycle Controls Evaporator Pressure (P low ) 200 kPa Adjusting this sets the boiling temperature of the refrigerant. Condenser Pressure (P high ) 1000 kPa Determines the discharge temperature at the back of the fridge. Compressor Efficiency 85 % Fridge Cabinet Temp 4.0 °C System COP (Efficiency) 4.21 Evap Temp (T1) -10 °C Cooling Capacity 150 kJ/kg Compressor Work 35 kJ/kg Discharge Temp 65 °C COMPRESSOR CONDENSER EXP. VALV...