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RRB JE CBT 2 Electrical study materials
RRB JE CBT 2 Electrical study resources

RRB JE CBT 2 Electrical practice questions

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Question 1Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A 12 V source drives a 4 Ω resistor. What current flows through the resistor?

  1. 3 A Correct answer.
  2. 0.33 A
  3. 16 A
  4. 48 A

Explanation

Ohm's law states that current equals voltage divided by resistance, so I = V/R = 12/4 = 3 A. This is a direct application of the fundamental V = IR relationship for a purely resistive element.

Question 2Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

Two resistors of 6 Ω and 3 Ω are connected in parallel. What is the equivalent resistance of the combination?

  1. 4.5 Ω
  2. 9 Ω
  3. 2 Ω Correct answer.
  4. 18 Ω

Explanation

For two resistors in parallel the equivalent resistance is the product divided by the sum: (6×3)/(6+3) = 18/9 = 2 Ω. The result is always smaller than the smallest branch resistance, which confirms 2 Ω is reasonable.

Question 3Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

Kirchhoff's current law is a direct statement of which conservation principle at a circuit node?

  1. Conservation of momentum
  2. Conservation of flux linkage
  3. Conservation of charge Correct answer.
  4. Conservation of energy

Explanation

Kirchhoff's current law states that the algebraic sum of currents entering a node is zero, which expresses the fact that charge cannot accumulate at a junction. Energy conservation underlies Kirchhoff's voltage law instead.

Question 4Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A coil has an inductance of 0.5 H and the current through it changes at 4 A/s. What emf is induced in the coil?

  1. 8 V
  2. 2 V Correct answer.
  3. 4.5 V
  4. 0.125 V

Explanation

Faraday's law for self-induction gives e = L·(di/dt) = 0.5 × 4 = 2 V. The induced emf opposes the change in current according to Lenz's law.

Question 5Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

In a purely capacitive AC circuit, the current waveform has what phase relationship with the applied voltage?

  1. Current leads voltage by 90° Correct answer.
  2. Current lags voltage by 90°
  3. Current is in phase with voltage
  4. Current leads voltage by 45°

Explanation

In a capacitor the current is proportional to the rate of change of voltage, so the current reaches its peak a quarter cycle before the voltage. Hence current leads voltage by 90° in a purely capacitive circuit.

Question 6Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

Which network theorem replaces a complex linear two-terminal network with a single voltage source in series with a resistance?

  1. Superposition theorem
  2. Reciprocity theorem
  3. Thevenin's theorem Correct answer.
  4. Norton's theorem

Explanation

Thevenin's theorem states that any linear two-terminal network can be replaced by an equivalent voltage source Vth in series with an equivalent resistance Rth. Norton's theorem gives the dual current-source form instead.

Question 7Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

The RMS value of a sinusoidal voltage with a peak value of 100 V is approximately:

  1. 50 V
  2. 70.7 V Correct answer.
  3. 63.7 V
  4. 100 V

Explanation

For a sine wave the RMS value equals the peak value divided by the square root of 2, so 100/1.414 ≈ 70.7 V. The RMS value represents the equivalent DC value producing the same heating effect.

Question 8Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

In a series RLC circuit at resonance, the impedance of the circuit is:

  1. Infinite
  2. Equal to the resistance alone Correct answer.
  3. Purely capacitive
  4. Zero

Explanation

At resonance the inductive and capacitive reactances are equal in magnitude and cancel each other, leaving only the resistance in the impedance. The circuit current is therefore at its maximum and limited solely by R.

Question 9Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A 10 Ω resistor carries a current of 2 A. What power is dissipated in the resistor?

  1. 80 W
  2. 40 W Correct answer.
  3. 20 W
  4. 5 W

Explanation

Joule heating power equals I²R, so P = 2² × 10 = 4 × 10 = 40 W. Alternatively, P = V·I with V = IR = 20 V gives the same 40 W result.

Question 10Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A network has two independent sources. The superposition theorem allows the total response to be found by:

  1. Adding the source values and analysing once
  2. Replacing all sources by their internal resistances
  3. Multiplying the responses of each source acting alone
  4. Adding the responses of each source acting alone Correct answer.

Explanation

Superposition states that in a linear bilateral network the total response equals the algebraic sum of the responses caused by each source acting alone. All other sources must be deactivated appropriately while each source is considered.

Question 11Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A coil of 200 turns links a flux that changes from 0.02 Wb to 0.06 Wb in 0.1 s. What is the average induced emf?

  1. 40 V
  2. 800 V
  3. 80 V Correct answer.
  4. 8 V

Explanation

By Faraday's law, e = N·(dΦ/dt) = 200 × (0.06 − 0.02)/0.1 = 200 × 0.4 = 80 V. The change in flux per second is 0.4 Wb/s and the turns multiply this value.

Question 12Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

The form factor of a pure sinusoidal waveform is:

  1. 1.57
  2. 1.11 Correct answer.
  3. 0.707
  4. 1.414

Explanation

The form factor is the RMS value divided by the average value, which for a sine wave is 0.707/0.637 ≈ 1.11. It measures how much the waveform deviates from a pure DC shape.

Question 13Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

In a balanced three-phase star-connected system with a line voltage of 400 V, the phase voltage is approximately:

  1. 230 V Correct answer.
  2. 692 V
  3. 200 V
  4. 400 V

Explanation

For a star connection the phase voltage equals the line voltage divided by the square root of 3, so 400/1.732 ≈ 231 V. This is why 400 V three-phase systems provide roughly 230 V single-phase supplies.

Question 14Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

The magnetic field strength H inside a long solenoid depends on which quantity?

  1. The resistance of the winding
  2. The current and turns per unit length Correct answer.
  3. The cross-sectional area of the coil
  4. The permeability of the core only

Explanation

For a long solenoid H = NI/l, so the field strength is set by the ampere-turns per unit length and is independent of the core material. The core permeability affects the flux density B, not the field strength H.

Question 15Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A capacitor of 100 μF is connected across a 50 Hz supply. What is its capacitive reactance approximately?

  1. 0.0318 Ω
  2. 31.8 Ω Correct answer.
  3. 3.18 Ω
  4. 318 Ω

Explanation

Capacitive reactance is Xc = 1/(2πfC) = 1/(2π × 50 × 100×10⁻⁶) ≈ 1/0.0314 ≈ 31.8 Ω. Reactance falls as frequency or capacitance increases.

Question 16Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

Two 10 Ω resistors are connected in series across a 20 V supply. What is the voltage across each resistor?

  1. 20 V
  2. 5 V
  3. 40 V
  4. 10 V Correct answer.

Explanation

The total resistance is 20 Ω, so the current is 20/20 = 1 A. Each 10 Ω resistor then drops V = IR = 1 × 10 = 10 V, which also matches the equal division of voltage between identical series resistors.

Question 17Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

Maximum power transfer from a source to a load occurs when the load resistance is:

  1. Zero
  2. Much smaller than the source resistance
  3. Equal to the source resistance Correct answer.
  4. Much larger than the source resistance

Explanation

The maximum power transfer theorem states that the load receives maximum power when its resistance equals the internal resistance of the source. At this condition the efficiency is only 50 percent, which is why power systems do not operate this way.

Question 18Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

The time constant of an RL circuit with R = 20 Ω and L = 4 H is:

  1. 0.2 s Correct answer.
  2. 0.05 s
  3. 80 s
  4. 5 s

Explanation

The time constant of an RL circuit is L/R = 4/20 = 0.2 s. It represents the time in which the current reaches about 63 percent of its final steady value.

Question 19Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A delta-connected load has a phase current of 10 A. What is the line current?

  1. 10 A
  2. 17.32 A Correct answer.
  3. 30 A
  4. 5.77 A

Explanation

In a delta connection the line current equals the phase current multiplied by the square root of 3, so 10 × 1.732 ≈ 17.32 A. This contrasts with a star connection, where line and phase currents are equal.

Question 20Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

Two coupled coils have a mutual inductance of 0.2 H. If the current in one coil changes at 10 A/s, what emf is induced in the other coil?

  1. 50 V
  2. 0.02 V
  3. 2 V Correct answer.
  4. 0.2 V

Explanation

Mutually induced emf is e = M·(di/dt) = 0.2 × 10 = 2 V. The coupling coefficient and winding geometry are already embedded in the given value of M.

Question 21Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A coil with negligible resistance draws 5 A from a 230 V, 50 Hz supply and consumes 575 W. What is its inductance approximately?

  1. 0.073 H
  2. 0.293 H
  3. 1.46 H
  4. 0.146 H Correct answer.

Explanation

The impedance is Z = V/I = 230/5 = 46 Ω, and the resistance is R = P/I² = 575/25 = 23 Ω. The reactance is X = √(46² − 23²) ≈ 39.8 Ω, so L = X/(2πf) = 39.8/314 ≈ 0.127 H, closest to 0.146 H among the choices.

Question 22Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A 100 V source with internal resistance of 10 Ω feeds a load. What load resistance extracts maximum power, and what is that power?

  1. 5 Ω, 500 W
  2. 10 Ω, 1000 W
  3. 20 Ω, 250 W
  4. 10 Ω, 250 W Correct answer.

Explanation

Maximum power transfer requires the load to equal the source resistance, so Rload = 10 Ω. The current is then 100/20 = 5 A and the load power is I²R = 25 × 10 = 250 W.

Question 23Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

In a series RLC circuit with R = 10 Ω, L = 0.1 H and C = 100 μF, the resonant frequency is approximately:

  1. 15.9 Hz
  2. 50 Hz
  3. 159 Hz Correct answer.
  4. 100 Hz

Explanation

Resonance occurs at f = 1/(2π√(LC)) = 1/(2π√(0.1 × 100×10⁻⁶)) = 1/(2π × 0.00316) ≈ 50.3 Hz, so among the options the value nearest the computed resonance is 50 Hz; the 159 Hz figure corresponds to a different L·C product and is not the resonance here.

Question 24Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A 50 Hz transformer core experiences a peak flux density. If the supply frequency is raised to 100 Hz at the same voltage, the peak flux density:

  1. Remains unchanged
  2. Becomes four times larger
  3. Doubles
  4. Halves Correct answer.

Explanation

From the emf equation E = 4.44·f·N·Φm, the peak flux is inversely proportional to frequency for a fixed applied voltage. Doubling the frequency therefore halves the peak flux density in the core.

Question 25Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A network is reduced to a Thevenin equivalent of 24 V in series with 6 Ω. What current flows into a 2 Ω load?

  1. 3 A Correct answer.
  2. 6 A
  3. 12 A
  4. 4 A

Explanation

The total series resistance is 6 + 2 = 8 Ω, so the load current is 24/8 = 3 A. The load terminal voltage would then be 3 × 2 = 6 V.

Question 26Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

Two resistors of 6 Ω and 3 Ω are connected in parallel across a 12 V DC supply. What current flows through the 6 Ω resistor?

  1. 0.5 A
  2. 1.33 A
  3. 2 A Correct answer.
  4. 4 A

Explanation

In a parallel circuit each branch sees the full supply voltage, so the branch current follows Ohm's law independently of the other branch. Thus I = V/R = 12/6 = 2 A. The 3 Ω branch separately carries 4 A, giving a total supply current of 6 A.

Question 27Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A 230 V, 50 Hz supply is applied to a pure inductor of 0.5 H (negligible resistance). What is the RMS current drawn?

  1. 10.4 A
  2. 1.46 A Correct answer.
  3. 0.73 A
  4. 2.30 A

Explanation

Inductive reactance is XL = 2πfL = 2 × 3.1416 × 50 × 0.5 ≈ 157 Ω. The RMS current is then V/XL = 230/157 ≈ 1.46 A. In a purely inductive circuit the current lags the voltage by 90° but the magnitude is set purely by reactance.

Question 28Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A galvanometer of 50 Ω resistance gives full-scale deflection at 1 mA. What series resistance converts it into a voltmeter reading 10 V at full scale?

  1. 50 Ω
  2. 10000 Ω
  3. 950 Ω
  4. 9950 Ω Correct answer.

Explanation

At full scale the meter itself drops I×R = 0.001 × 50 = 0.05 V, so the series multiplier must drop 10 − 0.05 = 9.95 V at 1 mA. That gives R = 9.95/0.001 = 9950 Ω. The total voltmeter resistance is therefore 10 000 Ω, corresponding to 1000 Ω/V sensitivity.

Question 29Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A coil of 500 turns is wound on a closed iron ring of mean length 0.5 m and cross-section 4 × 10⁻⁴ m². Taking the relative permeability of the iron as 1000, what is the inductance?

  1. 1.26 H
  2. 50.3 mH
  3. 0.503 H Correct answer.
  4. 0.503 mH

Explanation

For a uniform magnetic core, L = μ0μrN²A/l = (4π×10⁻⁷ × 1000 × 500² × 4×10⁻⁴)/0.5. The numerator is about 0.2513 and dividing by 0.5 gives L ≈ 0.503 H. The result scales with N², so turn count dominates the inductance.

Question 30Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

Which statement correctly describes the behaviour of a practical capacitor in a DC circuit after the switch has been closed for a long time?

  1. It behaves as an inductance and opposes changes in current
  2. It behaves as a short circuit and passes steady current
  3. It behaves as an open circuit and blocks steady current Correct answer.
  4. It behaves as a fixed resistance equal to its reactance

Explanation

Once fully charged, the voltage across a capacitor equals the applied DC voltage and no further charge flows, so the steady-state current is zero. The capacitor therefore looks like an open circuit to DC. Its reactance 1/(2πfC) is infinite at zero frequency, which is the same conclusion.

Question 31Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

In a two-wire DC distributor, the sending-end voltage is 240 V and the receiving-end voltage is 230 V while the load current is 50 A. What is the total resistance of the go-and-return conductor path?

  1. 0.4 Ω
  2. 0.1 Ω
  3. 2.0 Ω
  4. 0.2 Ω Correct answer.

Explanation

The voltage drop along the conductors is 240 − 230 = 10 V, and this drop occurs across the total loop resistance carrying the 50 A load current. Hence R_total = 10/50 = 0.2 Ω, which is the sum of the outgoing and return conductor resistances.

Question 32Basic Electrical Engineering (circuit laws, network theorems, AC/DC fundamentals, electromagnetic induction)

A moving-coil ammeter reads 5 A on a circuit whose current contains a significant alternating ripple superimposed on the DC component. Why does the instrument fail to indicate the ripple?

  1. Its coil inductance blocks the alternating component from entering the instrument
  2. Its permanent magnet cancels any alternating flux produced by the coil
  3. Its deflection responds to the average value of current, and the alternating component averages to zero Correct answer.
  4. Its scale is calibrated only for sinusoidal waveforms

Explanation

A moving-coil instrument produces torque proportional to the instantaneous coil current, and mechanical inertia makes it respond to the mean value over a cycle. A symmetrical alternating component has zero mean, so only the DC component moves the pointer. This is why moving-coil meters indicate average values and are used with rectifiers for AC measurement.

Question 33Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

In a DC shunt generator running at rated speed with no load, the terminal voltage is 230 V. If the field circuit resistance is 115 ohms, what is the shunt field current?

  1. 0.5 A
  2. 2 A Correct answer.
  3. 4 A
  4. 1 A

Explanation

The shunt field winding is connected directly across the armature terminals, so the field current equals terminal voltage divided by field resistance. Here 230 V divided by 115 ohms gives 2 A. This current flows through the field even at no load.

Question 34Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

In a transformer, the open-circuit test is normally performed on the low-voltage side with the high-voltage side open. What quantity does this test primarily determine?

  1. Core losses and magnetizing branch parameters Correct answer.
  2. Copper losses and equivalent resistance
  3. Voltage regulation at full load
  4. Winding temperature rise under load

Explanation

With the secondary open, only a small magnetizing current flows, so the input power measured is essentially the no-load core loss. The measured voltage, current and power give the shunt branch parameters of the equivalent circuit. Copper loss is negligible because rated current does not flow.

Question 35Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

A three-phase squirrel-cage induction motor is connected to a 50 Hz supply and runs at 1440 rpm at full load. What is the percentage slip?

  1. 2 percent
  2. 6 percent
  3. 3 percent
  4. 4 percent Correct answer.

Explanation

For a 4-pole machine the synchronous speed is 120f/p = 120 times 50 divided by 4, which equals 1500 rpm. Slip equals (1500 minus 1440) divided by 1500, giving 0.04 or 4 percent. The rotor always runs slightly below synchronous speed in motoring mode.

Question 36Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

Which condition must be satisfied before a DC shunt generator can build up its terminal voltage from residual magnetism?

  1. The field resistance must be below the critical field resistance at that speed Correct answer.
  2. The load must be connected across the armature
  3. The series field must aid the shunt field
  4. The armature must be driven in the reverse direction

Explanation

Voltage build-up requires residual flux to induce a small armature emf, which drives field current that must strengthen that flux. This self-excitation succeeds only if the field resistance line intersects the magnetization curve, that is, field resistance is below the critical value. The machine must also turn in the correct direction and be on no load or light load.

Question 37Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

For a transformer, the emf per turn is the same in both primary and secondary windings. This statement follows directly from which relationship?

  1. The winding resistance drop being equal
  2. The leakage reactance being equal
  3. The ampere-turn balance under load
  4. The common mutual flux linking both windings Correct answer.

Explanation

Both windings link the same alternating mutual flux, so the flux changes at the same rate per turn in each winding. Faraday's law then gives identical induced emf per turn in primary and secondary. Turns ratio therefore sets the voltage ratio.

Question 38Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

A 6-pole, 50 Hz three-phase induction motor has a full-load slip of 0.05. At what frequency do the rotor currents alternate?

  1. 50 Hz
  2. 0.5 Hz
  3. 2.5 Hz Correct answer.
  4. 5 Hz

Explanation

Rotor current frequency equals slip times supply frequency, that is, s times f. Here 0.05 times 50 Hz gives 2.5 Hz. At standstill slip is one and rotor frequency equals supply frequency; at synchronous speed it falls to zero.

Question 39Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

In a salient-pole synchronous machine, the direct-axis synchronous reactance is considerably larger than the quadrature-axis reactance. What is the main reason?

  1. The armature reaction is purely magnetizing on the quadrature axis
  2. The damper winding is more effective on the quadrature axis
  3. The field winding resistance is lower on the direct axis
  4. The air gap along the pole axis is much smaller than along the interpolar axis Correct answer.

Explanation

In a salient-pole rotor the iron path along the pole axis presents a short air gap, so direct-axis flux meets low reluctance. Along the interpolar axis the gap is long, giving higher reluctance and lower quadrature-axis reactance. This saliency produces reluctance torque and requires the two-reaction theory.

Question 40Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

A 3-phase, 400 V, 50 Hz star-connected induction motor has a per-phase stator resistance of 0.5 ohm. In the no-load test the input power is 600 W and the current is 5 A per phase. What is the approximate no-load rotational loss?

  1. 37.5 W
  2. 562.5 W Correct answer.
  3. 600 W
  4. 187.5 W

Explanation

No-load input power covers stator copper loss plus rotational (friction, windage and core) loss. Stator copper loss is 3 times I squared R, that is 3 times 25 times 0.5, equal to 37.5 W. Subtracting from 600 W gives about 562.5 W of rotational loss.

Question 41Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

A 100 kVA transformer has a full-load copper loss of 1.6 kW and a core loss of 0.4 kW. At what fraction of full load does maximum efficiency occur?

  1. 0.75
  2. 1.0
  3. 0.25
  4. 0.5 Correct answer.

Explanation

Maximum efficiency occurs when variable copper loss equals constant core loss. Since copper loss varies with the square of load fraction x, we set x squared times 1.6 kW equal to 0.4 kW, giving x squared equal to 0.25 and x equal to 0.5. So peak efficiency is at half load.

Question 42Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

Which starting method is used for a large synchronous motor to avoid excessive inrush and mechanical stress?

  1. Reduced-voltage starting with the field shorted through a resistor, then field applied near synchronous speed Correct answer.
  2. Direct-on-line starting with full field applied
  3. Starting with the field winding open-circuited
  4. Starting as a DC series motor

Explanation

A synchronous motor has no starting torque of its own, so it is usually started as a squirrel-cage induction motor using damper windings. The field winding is closed through a discharge resistor to limit induced high voltage during starting, and DC excitation is applied near synchronous speed to pull the rotor into synchronism.

Question 43Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

A separately excited DC motor draws 20 A from a 220 V supply and runs at 1000 rpm. Its armature resistance is 0.5 ohm. What is the back emf?

  1. 230 V
  2. 220 V
  3. 200 V
  4. 210 V Correct answer.

Explanation

For a motor, supply voltage equals back emf plus armature resistance drop. The drop is 20 A times 0.5 ohm, equal to 10 V. Therefore back emf is 220 minus 10, which is 210 V.

Question 44Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

A transformer has 500 primary turns and 100 secondary turns. If the primary is connected to a 250 V, 50 Hz supply, what is the approximate peak value of the mutual flux in the core?

  1. 0.45 Wb
  2. 0.0113 Wb
  3. 0.00225 Wb Correct answer.
  4. 0.0045 Wb

Explanation

The emf per turn is 250 V divided by 500 turns, equal to 0.5 V per turn. Using E = 4.44 f N phi, the flux per turn is 0.5 divided by (4.44 times 50), giving about 0.00225 Wb peak. This flux is common to both windings.

Question 45Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

Why is a starter necessary when starting a DC shunt motor?

  1. To reverse the direction of rotation
  2. To reduce the field current at start
  3. To limit the large starting armature current because back emf is zero at standstill Correct answer.
  4. To increase the supply voltage during acceleration

Explanation

At standstill the armature is stationary, so back emf is zero and the armature current is limited only by its small resistance. This current could be many times rated value and damage the commutator. A starter inserts external resistance that is cut out progressively as the motor speeds up and back emf rises.

Question 46Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

A 50 kVA, 2500/250 V single-phase transformer has a per-unit resistance of 0.02 and per-unit reactance of 0.05. What is the approximate percentage voltage regulation at full load and 0.8 power factor lagging?

  1. 2.0 percent
  2. 4.6 percent Correct answer.
  3. 6.0 percent
  4. 3.6 percent

Explanation

Regulation is approximately r cos phi plus x sin phi, expressed per unit. With r equal to 0.02, x equal to 0.05, cos phi 0.8 and sin phi 0.6, the result is 0.016 plus 0.030, equal to 0.046, or 4.6 percent. Lagging power factor increases regulation.

Question 47Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

In a three-phase synchronous generator, what is the effect of increasing the DC field excitation while the real power output is held constant?

  1. The armature current decreases to a minimum at unity power factor and then rises leading Correct answer.
  2. The speed of the prime mover increases
  3. The power factor becomes more lagging
  4. The real power delivered increases

Explanation

With real power fixed, varying excitation moves the machine along V-curves of armature current. Overexcitation gives a lagging power factor and underexcitation gives a leading one, with minimum current at unity power factor. Excitation controls reactive power, not real power.

Question 48Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

Which loss in a transformer is essentially independent of the load current?

  1. Hysteresis and eddy current loss in the core Correct answer.
  2. Stray load loss in the tank walls
  3. Winding eddy loss due to load current
  4. Copper loss in the windings

Explanation

Core loss consists of hysteresis and eddy current loss, and both depend on the maximum flux density and frequency. Since the applied voltage and frequency are nearly constant, the flux is constant and the core loss stays almost fixed from no load to full load. Copper and stray losses vary with load current.

Question 49Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

A 4-pole wave-wound DC armature has 400 conductors and a flux per pole of 0.03 Wb. If it runs at 1000 rpm, what is the generated emf?

  1. 300 V
  2. 400 V
  3. 200 V Correct answer.
  4. 100 V

Explanation

Generated emf equals phi times Z times N times P divided by 60 times A. For a wave winding the number of parallel paths A is 2, so emf is 0.03 times 400 times 1000 times 4 divided by (60 times 2), giving 200 V. Using lap winding with A equal to 4 would halve this value.

Question 50Electrical Machines (DC machines, transformers, induction machines, synchronous machines)

In a single-phase transformer, why is the core laminated?

  1. To reduce hysteresis loss by changing the flux density
  2. To increase the mechanical strength of the core
  3. To reduce eddy current loss by restricting circulating current paths Correct answer.
  4. To increase the magnetizing current

Explanation

Laminations are thin sheets insulated from each other, which breaks up the large eddy current loops that would flow in a solid core. This sharply reduces eddy current loss, which varies with the square of lamination thickness. Hysteresis loss is reduced by choosing suitable core material, not by laminating.