A junction or a point where two (or more) network elements intersect is called as:
(a) Node ✓ (b) Branch (c) Loop (d) Mesh
The frequency of domestic power supply in India is
(a) 200 Hz (b) 100 Hz (c) 60 Hz (d) 50 Hz ✓
Which of the following is not a part of field magnets?
(a) Field winding (b) Armature ✓ (c) Pole shoe (d) Pole core
What is the working principle of a Transformer?
(a) Transformer works on the principle of self induction (b) Transformer works on the principle of mutual induction ✓ (c) Transformer works on the principle of ampere law (d) Transformer works on the principle of coulomb law
As a PN junction is forward biased
(a) Holes as well as electrons tend to drift away from the junction (b) The depletion region decreases ✓ (c) The barrier tends to breakdown (d) None of the above
A FET is a __________ controlled device whereas a bipolar transistor is a __________ controlled device.
(a) Voltage, Current ✓ (b) Current, Voltage (c) Voltage, Power (d) Gate, Current
The logic expression A + B can be implemented by giving inputs A and B to a two-input
(a) OR gate ✓ (b) NAND gate (c) X-OR gate (d) X-NOR gate
Simplify the expression using K-maps:
F(A, B, C, D) = Σ(1, 3, 5, 6, 7, 11, 13, 14)(a) AB + BC’D + A’B’C ✓ (b) BCD’ + A’C’D + BD’ (c) A’D + BCD + A’BC + AB’C’ (d) AC’D’ + BC + A’BD + C’D’
Which of the following instruments is suitable for both AC and DC measurements?
(a) Moving Coil (b) Moving Iron (c) Dynamometer ✓ (d) Both (a) and (b)
DSO stands for
(a) Data Signal Oscilloscope (b) Data Storage Oscilloscope (c) Digital Storage Oscilloscope ✓ (d) Digital Signal Oscilloscope
11.Let’s solve this circuit problem step by step?
1. Define Mesh Currents:
Assume mesh currents I1 and I2 flowing clockwise in the two loops as shown in the diagram.
2. Apply Kirchhoff’s Voltage Law (KVL) to each loop:
Loop 1:
Starting from the negative terminal of the 6V source and going clockwise:
-6V + 4Ω(I1) + 8Ω(I1 – I2) = 0
Simplifying: 12Ω(I1) – 8Ω(I2) = 6V
Loop 2:
Starting from the negative terminal of the 10V source and going clockwise:
-10V + 8Ω(I2 – I1) + 2Ω(I2) + 2V = 0
Simplifying: -8Ω(I1) + 10Ω(I2) = 8V
3. Solve the system of equations:
We now have two equations with two unknowns (I1 and I2):
12I1 – 8I2 = 6
-8I1 + 10I2 = 8
There are several ways to solve this system of equations (substitution, elimination, matrices). Let’s use elimination:
Multiply equation 1 by 5: 60I1 – 40I2 = 30
Multiply equation 2 by 6: -48I1 + 60I2 = 48
Add the two new equations: 12I1 + 20I2 = 78
Solve for I1: 12I1 = 78 – 20I2 => I1 = (78 – 20I2)/12
Substitute the expression for I1 into equation 2:
-8((78 – 20I2)/12) + 10I2 = 8
Simplify and solve for I2: I2 = 1.714 A (approximately)
Substitute the value of I2 back into the equation for I1:
I1 = (78 – 20 * 1.714)/12
I1 = 0.571 A (approximately)
4. Results:
I1 = 0.571 A (approximately) flowing clockwise in the left loop.
I2 = 1.714 A (approximately) flowing clockwise in the right loop.
Therefore, the mesh currents for the given circuit are approximately I1 = 0.571 A and I2 = 1.714 A.
Key Points for a 5-Mark Answer:
Clear Diagram:
Redraw the circuit with clearly labeled mesh currents.
KVL Equations:
Write the KVL equations correctly. This is the most crucial step.
Show your work:
Demonstrate how you solved the system of equations. Don’t just write the final answer.
Units:
Include the units (Amperes) in your final answer.
Accuracy:
Calculate accurately. If you make a small calculation error, it can affect the final result.
Conclusion:
Clearly state the values of the mesh currents.
11. (b) Define the following terms:
(i) RMS Value:
The effective value of a varying (AC) quantity. It is the DC equivalent value that would produce the same power dissipation in a resistive load. Expressed as Vrms = Vpeak / √2 (for a sine wave).
(ii) Real Power:
The actual power consumed by a load, measured in watts (W). It represents the power used to perform useful work. Calculated as P = VIcos(θ), where θ is the phase angle between voltage and current.
(iii) Reactive Power:
The power that oscillates between the source and the load without doing any real work, measured in volt-amperes reactive (VAR). It is associated with energy stored in inductive or capacitive elements. Calculated as Q = VIsin(θ).
(iv) Power Factor:
The ratio of real power to apparent power. It indicates how effectively power is being used. A power factor of 1 means all power is used for work. Calculated as PF = cos(θ).
(v) Apparent Power:
The product of voltage and current in an AC circuit, measured in volt-amperes (VA). It represents the total power flowing in the circuit, including both real and reactive power. Calculated as S = VI.
12. (a) Explain the working principle of a Transformer.
A transformer works on the principle of mutual induction. It consists of two or more coils (windings) linked by a common magnetic flux. When an alternating current flows through one coil (primary winding), it produces a time-varying magnetic flux. This changing flux induces a voltage in the other coil(s) (secondary winding). The voltage induced in the secondary can be higher (step-up transformer) or lower (step-down transformer) than the primary voltage, depending on the turns ratio of the windings.
12. (b) Derive the torque equation of a DC motor.
Force on a Conductor:
current, and L is the length of the conductor.
Torque:
The force acting on the conductor creates a torque, causing the armature to rotate. The torque is given by τ = rFsin(θ), where r is the radius of the armature and θ is the angle between the magnetic field and the conductor.
Torque Equation:
Combining these relationships and considering the number of conductors and magnetic poles, the torque equation for a DC motor is derived as: τ = (ΦZNP)/(2πA), where Φ is the flux per pole, Z is the number of conductors, N is the speed, P is the number of poles, and A is the number of parallel paths in the armature winding.
13. (a) Explain about the working of Zener Diode.
A Zener diode is a specially designed diode that operates in the reverse breakdown region. Unlike a normal diode, which blocks reverse current until a high voltage is reached, a Zener diode allows current to flow in the reverse direction when the reverse voltage exceeds its Zener voltage (Vz). This breakdown is non-destructive as long as the power dissipation is within the diode’s limits. Zener diodes are commonly used as voltage regulators to maintain a constant voltage across a load.
13. (b) Explain the characteristics and working of MOSFET.
A MOSFET (Metal-Oxide-Semiconductor Field-Effect Transistor) is a three-terminal (Gate, Source, Drain) semiconductor device that controls current flow between the source and drain by applying a voltage to the gate. It has a very high input impedance. There are two types:
Enhancement MOSFET:
Conducts current only when a gate voltage is applied.
Depletion MOSFET:
Conducts current with zero gate voltage and requires a gate voltage to stop conduction.
The gate voltage creates an electric field that controls the conductivity of a channel between the source and drain. MOSFETs are widely used in amplifiers, switches, and digital circuits.
14. (a) How the error can be detected using parity codes.
Parity codes are a simple form of error detection. A parity bit is added to a data word to make the total number of 1s either even (even parity) or odd (odd parity). When the data is transmitted, the receiver counts the number of 1s. If the parity doesn’t match the agreed-upon scheme (even or odd), an error is detected. Parity can detect single-bit errors but cannot detect multiple-bit errors or correct errors.
14. (b) Minimize the Boolean function using K-map: F(A,B,C) = Σm(0, 1, 4, 5, 7)
AB
00 01 11 10
C 0 1 1 1 1
1 0 0 1 0
Grouping the 1s:
Group 1: A’B’ (minterms 0, 1)
Group 2: AC (minterms 4, 5, 7)
Minimized function: F(A,B,C) = A’B’ + AC
15. (a) Draw the block diagram and explain about DSO.
A DSO (Digital Storage Oscilloscope) captures, stores, and displays waveforms digitally.
Block Diagram:
Input Amplifier: Amplifies the input signal.
Analog-to-Digital Converter (ADC): Converts the analog signal to digital.
Memory: Stores the digitized waveform data.
Processor: Processes the data for display and analysis.
Display: Shows the waveform on a screen.
Control Panel: Allows users to set parameters and control the DSO.
Explanation: The input signal is amplified and converted to digital form by the ADC. The digital data is stored in memory. The processor manipulates the data for display, performs measurements, and can provide advanced analysis. The waveform is displayed on the screen.
15. (b) Explain the principle of operation of attraction type MI meter.
An attraction-type Moving Iron (MI) meter measures AC or DC current or voltage. It operates on the principle that a piece of iron placed in a magnetic field is attracted towards the stronger part of the field. The instrument consists of a moving iron piece attached to a pointer and placed near a coil carrying the current to be measured. The current in the coil produces a magnetic field, which attracts the moving iron. The deflection of the pointer is proportional to the square of the current, making the scale non-linear. A control spring provides a counter-torque to balance the magnetic force and allows the pointer to reach a stable position.
16. (a) Find the current in the 5Ω resistance for the given network by using Nodal Analysis.
Identify Nodes: Label the nodes in the circuit. Let’s call the top node V1 and the bottom node (reference) 0V.
Apply Kirchhoff’s Current Law (KCL) at Node V1: The sum of currents entering the node must equal the sum of currents leaving the node.
Current through 4A source: 4A (entering)
Current through 2Ω resistor: (V1 – 0)/2 = V1/2 (leaving)
Current through 5Ω resistor: (V1 – V2)/5 (leaving), where V2 is the voltage at the node between the 5Ω and 2Ω resistors.
Current through 2Ω resistor: (V2 – 8)/2 (leaving)
KCL equation at V1: 4 = V1/2 + (V1 – V2)/5
Current through 5Ω resistor: (V1 – V2)/5 (entering)
Current through 2Ω resistor: (V2 – 8)/2 (leaving)
KCL equation at V2: (V1 – V2)/5 = (V2 – 8)/2
Solve the system of equations: We have two equations with two unknowns (V1 and V2):
4 = V1/2 + (V1 – V2)/5
(V1 – V2)/5 = (V2 – 8)/2
Solve these equations simultaneously to find the values of V1 and V2.
Calculate the current through the 5Ω resistor:
Current = (V1 – V2)/5
Substitute the values of V1 and V2 obtained in step 4 to calculate the current.
16. (b) (i) Differentiate Star and Delta connection of AC circuit.
Feature Star Connection Delta Connection
Phase Voltage Vph = VL / √3 Vph = VL
Phase Current Iph = IL Iph = IL / √3
Line Voltage VL = √3 * Vph VL = Vph
Line Current IL = Iph IL = √3 * Iph
Neutral Wire Requires a neutral wire No neutral wire required
Power P = √3 * VL * IL * cos(θ) P = 3 * Vph * Iph * cos(θ)
16. (b) (ii) Write about voltage division and current division rule.
Voltage Division Rule: In a series circuit, the voltage across each resistor is proportional to its resistance value relative to the total resistance.
V1 = Vtotal * (R1 / (R1 + R2 + … + Rn))
Current Division Rule: In a parallel circuit, the current through each branch is inversely proportional to its resistance value relative to the total equivalent resistance.
I1 = Itotal * (Req / R1)
17. (a) Explain the construction and working of a three-phase alternator.
Construction:
Stator: Contains three-phase windings (armature windings) spaced 120 degrees apart.
Rotor: Rotating magnetic field produced by permanent magnets or electromagnets.
Slip Rings: Provide a means for connecting the rotor winding to an external DC source (for excitation).
Brushes: Contact the slip rings to provide the DC excitation current.
Working: The rotor is rotated by a prime mover (turbine, engine). The rotating magnetic field cuts the stator windings, inducing an EMF according to Faraday’s law of electromagnetic induction. Since the stator windings are spaced 120 degrees apart, the induced EMFs are also 120 degrees out of phase, generating a three-phase voltage.
17. (b) Explain the construction and working of a DC generator.
Construction:
Field Magnets: Provide the magnetic field.
Armature: Rotating coil where EMF is induced.
Commutator: Rectifies the AC induced in the armature to DC.
Brushes: Contact the commutator to provide a path for the DC current to flow to the external circuit.
Working: The armature is rotated by a prime mover. The rotating coil cuts the magnetic field, inducing an EMF. The commutator and brushes act as a rectifier, converting the AC induced in the armature to DC at the output terminals.
18. (a) Draw the circuit diagram of a half-wave rectifier and explain its working.
Circuit Diagram: A half-wave rectifier consists of a single diode connected in series with an AC source and a load resistor.
Working: During the positive half-cycle of the AC input, the diode is forward biased and conducts current, allowing the positive half of the waveform to pass through to the load resistor. During the negative half-cycle, the diode is reverse biased and does not conduct, blocking the negative half of the waveform. The output across the load resistor is a pulsating DC waveform, containing only the positive half cycles of the input.
18. (b) Explain the characteristics and working of SCR.
SCR (Silicon Controlled Rectifier): A four-layer (PNPN) semiconductor device that acts as a unidirectional switch. It has three terminals: Anode, Cathode, and Gate.
Characteristics:
Forward Blocking Region: The SCR blocks current flow in the forward direction until the gate current exceeds a certain value (trigger current).
Forward Conduction Region: Once triggered, the SCR conducts heavily in the forward direction, acting like a closed switch.
Reverse Blocking Region: The SCR blocks current flow in the reverse direction, similar to a diode.
Working: The SCR is turned on (triggered) by applying a sufficient current pulse to the gate terminal. Once triggered, the SCR continues to conduct even if the gate signal is removed, as long as the anode current remains above the holding current. The SCR can be turned off by reducing the anode current below the holding current.
19. (a) Explain and verify the truth table of logic gates.
Logic Gates: Basic building blocks of digital circuits that perform logical operations on binary inputs.
Common Logic Gates:
AND Gate: Output is 1 only if all inputs are 1.
OR Gate: Output is 1 if at least one input is 1.
NOT Gate: Output is the inverse of the input.
NAND Gate: Output is 0 only if all inputs are 1.
NOR Gate: Output is 1 only if all inputs are 0.
XOR Gate: Output is 1 if the inputs are different.
Truth Table: A table that shows all possible combinations of inputs and their corresponding outputs for a logic gate or circuit.
Verification: To verify the truth table, you can build the circuit using logic gate ICs and observe the output for different input combinations.
19. (b) Design and implement a full adder circuit.
Full Adder: A combinational circuit that adds two binary digits (A and B) and a carry-in (Cin) from a previous stage, producing a sum (S) and a carry-out (Cout).
Logic Equations:
S = A ⊕ B ⊕ Cin
Cout = AB + (A ⊕ B)Cin
Implementation: A full adder can be implemented using logic gates:
Two XOR gates for the sum (S).
Two AND gates and one OR gate for the carry-out (Cout).
Circuit Diagram: Draw the circuit diagram showing the interconnection of logic gates to implement the full adder logic equations.
Truth Table: Write the truth table showing all possible combinations of A, B, Cin, and their corresponding S and Cout outputs.
20. (a) Explain in detail about the instrument transformers.
Instrument transformers are used to extend the range of measuring instruments (ammeters, voltmeters, wattmeters) for measuring high currents and voltages in power systems. They provide isolation between the high-voltage circuit and the measuring instrument, ensuring safety. Two main types exist:
Current Transformers (CTs):
Used to measure high currents. The primary winding of a CT is connected in series with the line carrying the current. The current in the secondary winding is proportional to the primary current, but at a much lower and safer level.
Potential Transformers (PTs) or Voltage Transformers (VTs):
Used to measure high voltages. The primary winding is connected in parallel with the line. The secondary voltage is a scaled-down replica of the primary voltage.
Key Aspects:
Turns Ratio: The ratio of the number of turns in the primary winding to the number of turns in the secondary winding determines the transformation ratio.
Burden: The load connected to the secondary winding of the instrument transformer.
Accuracy: Instrument transformers are designed for high accuracy to ensure reliable measurements.
Applications: Used in power substations, transmission lines, and distribution systems for metering and protection purposes.
20. (b) Explain the principle of operation of a single-phase energy meter.
A single-phase energy meter measures the electrical energy consumed in a single-phase AC circuit. It operates on the principle of electromagnetic induction. It consists of:
Driving Element (Current Coil): A coil connected in series with the load, carrying the load current. It produces a magnetic flux proportional to the load current.
Shunt Element (Pressure Coil): A coil connected in parallel with the load, carrying a current proportional to the load voltage. Its magnetic flux is proportional to the load voltage.
Rotating Disc: Made of aluminum, placed in the magnetic field produced by both coils. The interaction of the two fluxes induces eddy currents in the disc.
Braking Magnet: A permanent magnet that applies a braking torque on the rotating disc, proportional to the disc’s speed.
Registering Mechanism: Counts the revolutions of the disc, which is proportional to the energy consumed, and displays it in kilowatt-hours (kWh).
Operation:
The magnetic fields produced by the current and voltage coils interact, inducing a torque on the aluminum disc. The disc rotates at a speed proportional to the power being consumed by the load. The braking magnet provides an opposing torque, stabilizing the disc’s speed. The revolutions of the disc are counted by the registering mechanism, providing a measure of the energy consumed.