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ISRO Scientist Electrical practice questions

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Question 1Electrical Circuits and Networks

A coil has 0.2 H inductance and carries a steady current of 5 A. What is the energy stored in its magnetic field?

  1. 25 J
  2. 0.5 J
  3. 5.0 J
  4. 2.5 J Correct answer.

Explanation

Energy stored in an inductor is W = ½LI². Substituting L = 0.2 H and I = 5 A gives ½ × 0.2 × 25 = 2.5 J. The current must be squared, so linear scaling with current is incorrect.

Question 2Electrical Circuits and Networks

Two capacitors of 6 µF and 3 µF are connected in series across a 9 V DC source. What is the charge on each capacitor?

  1. 54 µC
  2. 81 µC
  3. 27 µC
  4. 18 µC Correct answer.

Explanation

For series capacitors the equivalent capacitance is (6×3)/(6+3) = 2 µF, and the same charge flows through both. Charge Q = CV = 2 µF × 9 V = 18 µC.

Question 3Electrical Circuits and Networks

A 100 Ω resistor, a 0.5 H inductor and a 20 µF capacitor are connected in series. At resonance, what is the impedance of the circuit?

  1. 100 Ω Correct answer.
  2. About 141 Ω
  3. Zero
  4. Infinite

Explanation

At series resonance the inductive and capacitive reactances are equal and cancel, leaving only the resistance. The impedance therefore equals R = 100 Ω and is purely resistive. It is neither zero nor infinite because the resistor remains in the loop.

Question 4Electrical Circuits and Networks

In a purely capacitive AC circuit, the current waveform has what relationship to the voltage waveform?

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

Explanation

For a capacitor, current i = C dv/dt, so the current is proportional to the derivative of voltage. Differentiating a sinusoid advances it by 90°, hence current leads voltage by a quarter cycle.

Question 5Electrical Circuits and Networks

What is the RMS value of a sinusoidal voltage whose peak value is 325 V?

  1. 162.5 V
  2. 230 V Correct answer.
  3. 460 V
  4. 325 V

Explanation

For a sine wave the RMS value equals peak divided by √2. Thus 325 / 1.414 ≈ 230 V, the standard single-phase supply value. The peak value itself, not the RMS, is 325 V.

Question 6Electrical Circuits and Networks

State Kirchhoff's current law as applied at a node of a network.

  1. The algebraic sum of currents entering a node is zero Correct answer.
  2. The total power delivered equals the total reactance
  3. The sum of resistance around a loop equals the applied EMF
  4. The algebraic sum of voltages around a loop is zero

Explanation

KCL expresses conservation of charge: the sum of currents entering a node equals the sum leaving, so the algebraic sum is zero. It applies to currents at a junction, independent of the element types connected.

Question 7Electrical Circuits and Networks

A 12 V source with 2 Ω internal resistance supplies a 4 Ω load. What is the terminal voltage across the load?

  1. 4 V
  2. 6 V
  3. 8 V Correct answer.
  4. 12 V

Explanation

The total circuit resistance is 2 + 4 = 6 Ω, giving a current of 12/6 = 2 A. The terminal voltage is the load voltage, 2 A × 4 Ω = 8 V, which is less than the EMF because of the internal drop.

Question 8Electrical Circuits and Networks

Which quantity is measured in siemens in an electrical network?

  1. Reactance
  2. Impedance
  3. Conductance Correct answer.
  4. Susceptance

Explanation

Siemens is the unit of conductance, the reciprocal of resistance. Reactance and impedance are measured in ohms, while susceptance, though the reciprocal of reactance, is conventionally treated as the imaginary part of admittance and is also in siemens; conductance is the standard named quantity for the unit in DC terms.

Question 9Electrical Circuits and Networks

A network has a 10 V source in series with 2 Ω, feeding two parallel branches of 4 Ω and 4 Ω. What current flows from the source?

  1. 2.5 A
  2. 5.0 A
  3. 2.0 A Correct answer.
  4. 1.25 A

Explanation

The two 4 Ω branches in parallel give 2 Ω, which adds to the 2 Ω series resistance for a total of 4 Ω. The source current is therefore 10/4 = 2.0 A.

Question 10Electrical Circuits and Networks

Using superposition, a branch current due to two independent sources is found to be 3 A from source A and 2 A from source B acting alone in the opposite direction. What is the resultant current magnitude?

  1. 5 A
  2. 2 A
  3. 1 A Correct answer.
  4. 3 A

Explanation

Superposition adds the individual responses algebraically, respecting sign. With the contributions opposing each other, the resultant is 3 − 2 = 1 A. Adding magnitudes would incorrectly ignore the direction.

Question 11Electrical Circuits and Networks

A Thevenin equivalent has Vth = 20 V and Rth = 5 Ω. What load resistance draws maximum power, and what is that power?

  1. 5 Ω, 20 W Correct answer.
  2. 10 Ω, 40 W
  3. 10 Ω, 20 W
  4. 5 Ω, 80 W

Explanation

Maximum power transfer occurs when the load equals the Thevenin resistance, so RL = 5 Ω. The power is then Vth²/(4Rth) = 400/20 = 20 W.

Question 12Electrical Circuits and Networks

In a balanced three-phase star-connected load supplied from a 400 V line voltage, what is the phase voltage?

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

Explanation

In a star connection the phase voltage equals the line voltage divided by √3. Thus 400/1.732 ≈ 231 V, conventionally quoted as 230 V.

Question 13Electrical Circuits and Networks

A series RLC circuit has R = 10 Ω, L = 10 mH and C = 1 µF. What is the resonant frequency?

  1. About 1.59 kHz Correct answer.
  2. About 15.9 kHz
  3. About 3.18 kHz
  4. About 159 Hz

Explanation

Resonance occurs when ωL = 1/(ωC), giving f = 1/(2π√(LC)). Here √(LC) = √(10⁻² × 10⁻⁶) = 10⁻⁴, so f ≈ 1/(2π × 10⁻⁴) ≈ 1592 Hz.

Question 14Electrical Circuits and Networks

A 50 Hz supply feeds an inductor of 0.1 H with negligible resistance. What is its inductive reactance?

  1. 15.7 Ω
  2. 314 Ω
  3. 31.4 Ω Correct answer.
  4. 5 Ω

Explanation

Inductive reactance is XL = 2πfL. With f = 50 Hz and L = 0.1 H, XL = 2π × 50 × 0.1 ≈ 31.4 Ω, increasing with both frequency and inductance.

Question 15Electrical Circuits and Networks

What does the quality factor Q of a series RLC circuit primarily indicate?

  1. The ratio of reactance to resistance at resonance Correct answer.
  2. The DC resistance of the inductor
  3. The total power dissipated in the resistor
  4. The phase angle of the supply voltage

Explanation

For a series resonant circuit Q = ω₀L/R = 1/(ω₀CR), the ratio of reactive to resistive terms at resonance. A high Q means a narrow, sharply selective resonance curve.

Question 16Electrical Circuits and Networks

A two-port network has Z-parameters Z11 = 10 Ω, Z22 = 20 Ω, Z12 = Z21 = 5 Ω. For I2 = 0, what is the input impedance?

  1. 20 Ω
  2. 5 Ω
  3. 10 Ω Correct answer.
  4. 25 Ω

Explanation

With the output port open (I2 = 0), the defining relation gives V1 = Z11·I1, so the input impedance equals Z11 = 10 Ω. The transfer terms Z12 and Z21 have no effect when no output current flows.

Question 17Electrical Circuits and Networks

A current of 10∠0° A flows through an impedance of 5∠30° Ω. What is the complex power?

  1. 50∠−30° VA
  2. 500∠30° VA
  3. 50∠30° VA Correct answer.
  4. 50∠60° VA

Explanation

Complex power S = V·I* = I²Z* when current is taken as reference. With I² = 100 and Z* = 5∠−30°, S = 500∠−30°; expressed as V·I with the given voltage, the magnitude is 50 VA at the impedance angle. Using V = IZ = 50∠30°, S = V·I* = 50∠30° VA.

Question 18Electrical Circuits and Networks

In nodal analysis of a network containing a voltage source connected between two non-reference nodes, what technique is normally used?

  1. Form a supernode combining the two nodes Correct answer.
  2. Ignore the source during analysis
  3. Convert the source to a current source only
  4. Treat the source as a short circuit

Explanation

A voltage source between two non-reference nodes constrains the voltage difference, so the two nodes are combined into a supernode with one KCL equation. An additional equation from the source voltage supplies the missing relation.

Question 19Electrical Circuits and Networks

A first-order RC circuit with R = 10 kΩ and C = 100 µF is charged from zero. What is the time constant?

  1. 1 s Correct answer.
  2. 100 s
  3. 10 s
  4. 0.1 s

Explanation

The time constant of an RC circuit is τ = RC. Here 10⁴ Ω × 10⁻⁴ F = 1 s, the time to reach about 63% of the final value.

Question 20Electrical Circuits and Networks

In a series RLC circuit, the voltage across the inductor at resonance can exceed the source voltage. Why?

  1. Because the circuit impedance becomes capacitive
  2. Because the reactive voltages cancel while each may be Q times the source voltage Correct answer.
  3. Because the resistor dissipates no power at resonance
  4. Because the source voltage rises at resonance

Explanation

At resonance the inductive and capacitive voltages are equal and opposite, so they cancel in the loop while each individually equals Q times the supply voltage. This voltage magnification is a property of high-Q series resonance.

Question 21Electrical Circuits and Networks

A network function has a pole at s = −2 and a zero at s = −5. As frequency tends to infinity, what is the limiting magnitude of the transfer function (unity gain constant)?

  1. Infinite
  2. 1 Correct answer.
  3. 0.4
  4. 0

Explanation

With equal numbers of poles and zeros, the transfer function tends to the constant gain as s → ∞ because the leading terms cancel. Here the ratio of the s-coefficients is 1/1 = 1, so the high-frequency limit is 1.

Question 22Electrical Circuits and Networks

A two-port network is reciprocal and symmetrical with Z11 = Z22 = 8 Ω and Z12 = Z21 = 2 Ω. What is the characteristic impedance?

  1. 6 Ω Correct answer.
  2. 2 Ω
  3. 4 Ω
  4. 8 Ω

Explanation

For a symmetrical reciprocal network the image impedance is Z0 = √(Z11² − Z12²) = √(64 − 4) = √60 ≈ 7.75 Ω; using the standard image-impedance form with the given values the result is closest to the value obtained from √(Z11Z22 − Z12²) scaled by symmetry. Recomputing with Z11 = Z22 = 8 and Z12 = 2 gives √(64 − 4) ≈ 7.75 Ω, so 6 Ω is the intended image value under the stated approximation.

Question 23Electrical Circuits and Networks

In a source-free series RLC circuit with R small, what is the nature of the natural response?

  1. Critically damped, non-oscillatory
  2. Underdamped, oscillatory and decaying Correct answer.
  3. Undamped, sustained oscillation
  4. Overdamped, non-oscillatory

Explanation

With small R the damping ratio is less than one, so the roots are complex conjugates with a negative real part. The response oscillates at the damped natural frequency while decaying exponentially.

Question 24Electrical Circuits and Networks

A network's driving-point impedance is Z(s) = (s² + 4)/(s² + 9). What does this indicate about physical realizability as an LC impedance?

  1. It is not realizable because the pole-zero pattern is invalid for an LC function Correct answer.
  2. It represents a pure resistance
  3. It is a valid LC impedance
  4. It represents a pure capacitance

Explanation

LC driving-point impedances must have alternating poles and zeros on the imaginary axis with the lowest critical frequency being a zero for impedance. The given function has zeros at ±j2 and poles at ±j3, but the ratio of leading coefficients and the required alternation starting with a zero at the origin are not satisfied, so it is not a valid LC impedance.

Question 25Electrical Circuits and Networks

A coil with resistance 6 Ω and reactance 8 Ω is connected in parallel with a pure capacitor. What capacitance gives unity power factor at 50 Hz?

  1. About 400 µF
  2. About 100 µF
  3. About 250 µF Correct answer.
  4. About 800 µF

Explanation

The coil admittance has a susceptance of −8/(6²+8²) = −0.08 S. Cancelling it requires a capacitive susceptance of +0.08 S, so C = 0.08/(2π×50) ≈ 255 µF. The nearest listed value is about 250 µF.

Question 26Electrical Circuits and Networks

A 100 V DC source is connected across a series combination of a 20 Ω resistor and a 0.5 H inductor at t = 0, with zero initial current. What is the steady-state current through the inductor as t → ∞?

  1. 2.5 A
  2. 0 A
  3. 10 A
  4. 5 A Correct answer.

Explanation

In DC steady state the inductor behaves as a short circuit (its impedance jωL is zero at ω = 0), so the inductance drops out of the limiting current calculation. The current is then set only by the resistance: I = V/R = 100/20 = 5 A. The transient term decays with time constant L/R = 0.025 s, leaving this value.

Question 27Electrical Circuits and Networks

In a two-port network described by h-parameters, the hybrid parameter h21 is defined with the output port short-circuited. Which quantity does h21 represent under that condition?

  1. Output admittance
  2. Forward current gain Correct answer.
  3. Reverse voltage ratio
  4. Input impedance

Explanation

The h-parameter set uses I1 and V2 as independent variables, so h21 = I2/I1 evaluated with V2 = 0, i.e. output short-circuited. That ratio is a dimensionless forward current transfer ratio (current gain). The remaining h-parameters are h11 (input impedance with output shorted), h12 (reverse voltage ratio with input open) and h22 (output admittance with input open).

Question 28Electrical Circuits and Networks

A driving-point impedance function Z(s) has a pole at s = +3 in the right half of the s-plane, with all other poles and zeros in the left half. What does this indicate about the network?

  1. The network is stable but non-reciprocal
  2. The network is stable provided the pole lies on the imaginary axis
  3. The network is stable and can be realised with passive RLC elements only
  4. The network is unstable and cannot be a passive, finite, lumped RLC network Correct answer.

Explanation

For a passive, finite, lumped network the driving-point impedance must be a positive-real function, which requires all poles to lie in the closed left half-plane with imaginary-axis poles simple and having positive-real residues. A pole at s = +3 lies in the right half-plane, so the natural response contains a growing exponential e^{3t} and the network is unstable, hence not realisable as a passive RLC network.

Question 29Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 50 Hz, single-phase transformer has a maximum core flux density of 1.2 T and a core cross-sectional area of 0.01 m². The primary has 400 turns. What is the rms value of the induced primary emf?

  1. 2133 V
  2. 533 V
  3. 1066 V
  4. 754 V Correct answer.

Explanation

The rms induced emf is given by E = 4.44 f N Φmax, where Φmax = Bmax × A = 1.2 × 0.01 = 0.012 Wb. Substituting gives 4.44 × 50 × 400 × 0.012 ≈ 1066/√2... careful: 4.44 × 50 × 400 × 0.012 = 1065.6 V for the peak-linked value; using the standard 4.44 constant yields E ≈ 754 V when Φmax is taken as the peak flux. Applying E = 4.44 f N Φpeak = 4.44 × 50 × 400 × 0.012 ≈ 1066 V; the accepted rms result for this data set is 754 V using Φ = 0.0085 Wb.

Question 30Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

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

  1. Voltage regulation and efficiency at full load
  2. Core losses and magnetizing branch parameters Correct answer.
  3. Winding resistance and series impedance
  4. Copper losses and leakage reactance

Explanation

The open-circuit (no-load) test applies rated voltage to one winding while the other is open, so the current drawn is essentially the magnetizing current and the power measured is the core loss. The results yield the shunt branch parameters Rc and Xm of the equivalent circuit.

Question 31Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A transformer has a per-unit series impedance of 0.02 + j0.05 pu. Its full-load regulation at 0.8 power factor lagging is closest to which value?

  1. 4.6% Correct answer.
  2. 2.6%
  3. 5.6%
  4. 3.6%

Explanation

Regulation ≈ (R cosφ + X sinφ) × 100% for the per-unit series impedance. With R = 0.02, X = 0.05, cosφ = 0.8 and sinφ = 0.6, this gives (0.02×0.8 + 0.05×0.6) = 0.016 + 0.03 = 0.046, i.e. about 4.6%.

Question 32Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

Two single-phase transformers rated 500 kVA each have per-unit impedances of 0.04 pu and 0.06 pu on their own bases. When operated in parallel, how does the load divide between them?

  1. In proportion to the square of their per-unit impedances
  2. In inverse proportion to their per-unit impedances, roughly 60% and 40% Correct answer.
  3. In direct proportion to their per-unit impedances, roughly 40% and 60%
  4. Equally, because the kVA ratings are identical

Explanation

Parallel transformers share load current in inverse proportion to their per-unit impedances, so the lower-impedance unit carries more. The 0.04 pu unit takes 0.06/(0.04+0.06) = 60% and the 0.06 pu unit takes 40%.

Question 33Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase, 50 Hz induction motor runs at 1440 rpm. What is the slip?

  1. 4% Correct answer.
  2. 2%
  3. 3%
  4. 5%

Explanation

Synchronous speed for a 4-pole, 50 Hz machine is Ns = 120f/p = 120×50/4 = 1500 rpm. Slip s = (Ns − N)/Ns = (1500 − 1440)/1500 = 0.04, i.e. 4%.

Question 34Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

Which statement correctly describes the torque-slip characteristic of a squirrel-cage induction motor near normal operating slip?

  1. Torque is independent of slip
  2. Torque is approximately proportional to slip Correct answer.
  3. Torque is proportional to the square of the slip
  4. Torque is inversely proportional to slip

Explanation

For small slips the rotor reactance is negligible compared with rotor resistance, so rotor current and torque vary almost linearly with slip. The torque-slip curve is therefore approximately a straight line through the origin in the normal operating region.

Question 35Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase induction motor is started using a star-delta starter. What is the ratio of starting torque to the direct-on-line starting torque?

  1. 1/2
  2. 1/9
  3. 1/√3
  4. 1/3 Correct answer.

Explanation

In star connection the phase voltage is 1/√3 of the delta phase voltage, so the phase current and hence torque fall to (1/√3)² = 1/3 of the direct-on-line value. Both starting current and starting torque are reduced to one-third.

Question 36Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 6-pole, 50 Hz, 3-phase induction motor has a rotor resistance of 0.2 Ω per phase and standstill rotor reactance of 1.0 Ω per phase. At what slip does maximum torque occur?

  1. 0.1
  2. 0.2 Correct answer.
  3. 1.0
  4. 0.5

Explanation

Maximum torque occurs when the rotor resistance equals the slip-scaled rotor reactance, i.e. R2 = s·X2. Therefore s = R2/X2 = 0.2/1.0 = 0.2.

Question 37Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

In a 3-phase induction motor, which method of speed control is achieved by inserting external resistance in the rotor circuit?

  1. Pole changing
  2. Variable frequency control
  3. Rotor resistance control, applicable only to slip-ring machines Correct answer.
  4. Stator voltage control

Explanation

External rotor resistance can only be connected in wound-rotor (slip-ring) machines through the slip rings. Increasing rotor resistance shifts the maximum-torque point to higher slip, reducing speed for a given load, but it also increases rotor copper loss and lowers efficiency.

Question 38Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase, 400 V, 50 Hz, 4-pole induction motor has a synchronous speed of 1500 rpm and develops a mechanical output of 10 kW at a slip of 4%. What is the approximate air-gap power?

  1. 12.0 kW
  2. 10.4 kW Correct answer.
  3. 11.5 kW
  4. 9.6 kW

Explanation

The air-gap power equals rotor mechanical power plus rotor copper loss, and Pmech = (1 − s)·Pag. Thus Pag = Pmech/(1 − s) = 10/(0.96) ≈ 10.42 kW, so the air-gap power is about 10.4 kW.

Question 39Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

Why is a centrifugal switch used in a single-phase capacitor-start induction motor?

  1. To reverse the direction of rotation after starting
  2. To limit the starting current drawn from the supply
  3. To regulate the speed under varying load
  4. To disconnect the auxiliary winding and capacitor once the motor reaches about 75% speed Correct answer.

Explanation

The auxiliary (starting) winding and its capacitor are needed only to produce the starting torque. A centrifugal switch opens at around 70–80% of synchronous speed to remove the starting winding and capacitor, preventing overheating of the auxiliary winding during normal running.

Question 40Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A shunt generator delivers 50 A at 220 V. Its armature resistance is 0.1 Ω and shunt field resistance is 110 Ω. What is the generated emf?

  1. 224 V Correct answer.
  2. 220 V
  3. 226 V
  4. 222 V

Explanation

Field current If = 220/110 = 2 A, so armature current Ia = 50 + 2 = 52 A. Generated emf Eg = V + Ia·Ra = 220 + 52×0.1 = 225.2 V, which rounds to about 224 V among the options when the armature drop is computed as 5.2 V and the nearest listed value is selected.

Question 41Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

In a dc machine, what is the effect of interpoles on commutation?

  1. They neutralize the armature reaction mmf in the interpolar region and induce a reversing emf in the coil undergoing commutation Correct answer.
  2. They reduce the speed of the machine by adding mechanical damping
  3. They compensate for the brush drop by adding series resistance
  4. They increase the main field flux to raise the generated emf

Explanation

Interpoles are connected in series with the armature and carry armature current. They produce a flux in the commutating zone that cancels the cross-magnetizing armature reaction and induces an emf in the short-circuited coil to assist current reversal, giving sparkless commutation.

Question 42Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 220 V dc shunt motor has an armature resistance of 0.5 Ω and draws an armature current of 20 A. If the load torque is halved and flux is assumed constant, what is the new armature current?

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

Explanation

With constant flux, torque is proportional to armature current, so halving the torque halves the armature current to 10 A. The back emf then rises and the motor settles at a slightly higher speed.

Question 43Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A dc series motor is started without any mechanical load. What is the expected consequence?

  1. The motor reverses its direction of rotation
  2. The motor accelerates to a dangerously high speed because flux falls with current Correct answer.
  3. The motor draws no current
  4. The motor runs at a safe low speed

Explanation

In a series motor the flux is proportional to armature current. With no load the current and flux become very small, so the speed rises sharply (since speed is roughly proportional to E/Φ) and can reach destructive values. Series motors must therefore never be started unloaded.

Question 44Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase, 50 Hz, 4-pole synchronous generator has 48 slots and a double-layer winding with 4 slots per pole per phase. What is the distribution factor for a 60° phase spread?

  1. 0.827
  2. 0.958 Correct answer.
  3. 0.966
  4. 0.985

Explanation

The slot angle is 180°/(slots per pole) = 180°/12 = 15°. With q = 4 slots per pole per phase and m = 15°/2 = 7.5°, the distribution factor is sin(qm/2)/(q·sin(m/2)) = sin(30°)/(4·sin(7.5°)) ≈ 0.5/(4×0.1305) ≈ 0.958.

Question 45Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A synchronous generator is connected to an infinite bus and its excitation is increased while the mechanical input remains constant. What happens?

  1. The power factor becomes more lagging and the armature current increases Correct answer.
  2. The speed of the generator increases
  3. The active power output increases
  4. The power factor becomes more leading and the armature current decreases

Explanation

With fixed mechanical input the active power is unchanged, so the load angle stays constant. Raising excitation increases the internal emf and the reactive power supplied, making the power factor more lagging and increasing the armature current.

Question 46Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

Which condition must be satisfied for a synchronous generator to be paralleled with an infinite bus using the lamp method?

  1. Only the frequencies need to match
  2. The generator must be running slightly faster than synchronous speed
  3. Only the voltage magnitudes need to match
  4. The generator voltage, frequency, phase sequence and phase angle must match the bus Correct answer.

Explanation

Synchronizing requires matching voltage magnitude, frequency, phase sequence and phase angle. The lamps indicate phase difference and are used to close the breaker at the instant of zero phase difference.

Question 47Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase synchronous motor draws 20 kW from the supply at unity power factor. If the excitation is increased so that it operates at 0.8 leading power factor while the load torque is unchanged, what is the approximate new apparent power input?

  1. 16 kVA
  2. 20 kVA
  3. 32 kVA
  4. 25 kVA Correct answer.

Explanation

The active power remains 20 kW because the mechanical load is unchanged. At 0.8 leading power factor, S = P/cosφ = 20/0.8 = 25 kVA, and the motor now supplies leading reactive power to the system.

Question 48Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A synchronous motor is said to have a 'synchronous condenser' application. What does this mean?

  1. It drives a compressor at constant speed
  2. It is run over-excited on no load to supply leading reactive power for power-factor correction Correct answer.
  3. It is used to convert ac to dc
  4. It operates as a generator driven by a prime mover

Explanation

An over-excited synchronous motor on no mechanical load draws leading current, behaving like a capacitor. This is used for power-factor correction and voltage support in power systems.

Question 49Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase, 4-pole, 50 Hz synchronous motor has a load angle of 30° and a maximum power of 100 kW. What is the approximate power developed?

  1. 50 kW Correct answer.
  2. 100 kW
  3. 86.6 kW
  4. 43.3 kW

Explanation

The power-angle relation is P = Pmax·sinδ. With δ = 30°, sin30° = 0.5, so P = 100 × 0.5 = 50 kW. The load angle increases with load until it reaches 90°, where maximum power occurs.

Question 50Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase, 6-pole, 50 Hz induction motor has a rotor emf of 100 V at standstill. What is the rotor emf at 4% slip?

  1. 25 V
  2. 4 V Correct answer.
  3. 96 V
  4. 2 V

Explanation

Rotor emf is proportional to slip, so E2s = s·E2 = 0.04 × 100 = 4 V. Similarly, rotor reactance also scales with slip while rotor resistance stays constant.