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TSSPDCL Sub-Engineer practice questions

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Question 1Electrical Circuits and Network Theory

A 12 V source is connected across a 4 Ω resistor. What current flows through the resistor?

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

Explanation

Ohm's law states that current equals voltage divided by resistance, so I = V/R = 12/4 = 3 A. The relationship is linear, so doubling the voltage would double the current for the same resistance.

Question 2Electrical Circuits and Network Theory

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

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

Question 3Electrical Circuits and Network Theory

The unit of electrical conductance is the:

  1. henry
  2. ohm
  3. farad
  4. siemens Correct answer.

Explanation

Conductance is the reciprocal of resistance and is measured in siemens, where one siemens equals one reciprocal ohm. It expresses how readily a material permits current flow for a given applied voltage.

Question 4Electrical Circuits and Network Theory

In a purely capacitive AC circuit, the current:

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

Explanation

In a capacitor the current reaches its peak before the voltage does, giving a 90° leading phase relationship. This occurs because current depends on the rate of change of voltage across the capacitor.

Question 5Electrical Circuits and Network Theory

Kirchhoff's current law is a direct consequence of which physical principle?

  1. Conservation of energy
  2. Faraday's law of induction
  3. Conservation of charge Correct answer.
  4. Conservation of momentum

Explanation

Kirchhoff's current law states that the algebraic sum of currents at a node is zero, which reflects the fact that charge cannot accumulate at a junction. Charge entering a node must equal charge leaving it.

Question 6Electrical Circuits and Network Theory

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

  1. 100 V
  2. 70.7 V Correct answer.
  3. 141.4 V
  4. 63.7 V

Explanation

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

Question 7Electrical Circuits and Network Theory

The time constant of an RC series circuit with R = 10 kΩ and C = 100 μF is:

  1. 0.1 s
  2. 10 s
  3. 1 s Correct answer.
  4. 0.01 s

Explanation

The time constant is the product of resistance and capacitance: 10,000 Ω × 0.0001 F = 1 s. It is the time taken for the capacitor voltage to reach about 63.2% of its final value during charging.

Question 8Electrical Circuits and Network Theory

The superposition theorem is applicable to circuits that are:

  1. non-linear only
  2. time-varying
  3. linear and bilateral Correct answer.
  4. unilateral only

Explanation

Superposition requires linearity so that responses add proportionally, and bilateral elements so that direction of current does not change the element's behaviour. It is therefore valid for linear bilateral networks with multiple independent sources.

Question 9Electrical Circuits and Network Theory

A coil has an inductive reactance of 30 Ω and a resistance of 40 Ω at a given frequency. What is its impedance magnitude?

  1. 70 Ω
  2. 35 Ω
  3. 10 Ω
  4. 50 Ω Correct answer.

Explanation

Impedance magnitude is the phasor sum of resistance and reactance: Z = √(40² + 30²) = √(1600 + 900) = √2500 = 50 Ω. Resistance and reactance are 90° apart, so they combine by the Pythagorean relationship, not by simple addition.

Question 10Electrical Circuits and Network Theory

In a series RLC circuit at resonance, the current is:

  1. purely reactive
  2. minimum
  3. zero
  4. maximum and limited only by resistance Correct answer.

Explanation

At series resonance the inductive and capacitive reactances are equal and cancel, leaving only resistance in the impedance. The circuit impedance is therefore minimum and the current reaches its maximum value, limited by R.

Question 11Electrical Circuits and Network Theory

The quality factor Q of a series RLC circuit at resonance is given by:

  1. ωC/R
  2. 1/(ωRC)
  3. ωL/R Correct answer.
  4. R/ωL

Explanation

For a series resonant circuit the quality factor equals the reactance of the inductor at resonance divided by the resistance, Q = ω₀L/R. It also equals the resonant frequency divided by the bandwidth, indicating the sharpness of resonance.

Question 12Electrical Circuits and Network Theory

The Thevenin equivalent of a network consists of:

  1. a current source in parallel with a resistance
  2. two voltage sources in series
  3. a voltage source in series with a resistance Correct answer.
  4. a voltage source in parallel with a resistance

Explanation

Thevenin's theorem replaces any linear two-terminal network with a single voltage source equal to the open-circuit voltage in series with the Thevenin resistance. The resistance is found by deactivating independent sources and looking into the terminals.

Question 13Electrical Circuits and Network Theory

A 100 W lamp operating at 250 V draws a current of:

  1. 0.4 A Correct answer.
  2. 25 A
  3. 0.25 A
  4. 2.5 A

Explanation

Power equals voltage multiplied by current, so I = P/V = 100/250 = 0.4 A. The lamp's resistance at this operating point would be V/I = 625 Ω.

Question 14Electrical Circuits and Network Theory

The maximum power transfer theorem states that maximum power is delivered to a load when the load resistance is:

  1. much larger than the source resistance
  2. much smaller than the source resistance
  3. zero
  4. equal to the source resistance Correct answer.

Explanation

Maximum power is transferred when the load resistance equals the internal or Thevenin resistance of the source. At this condition the efficiency is 50%, which is acceptable for signal circuits but not for power transmission.

Question 15Electrical Circuits and Network Theory

A sinusoidal current of RMS value 5 A flows through a 10 Ω resistor. The average power dissipated is:

  1. 50 W
  2. 500 W
  3. 250 W Correct answer.
  4. 25 W

Explanation

Average power in a resistor is I²R using the RMS current: 5² × 10 = 250 W. Using the RMS value automatically accounts for the time-averaged heating effect of the sinusoid.

Question 16Electrical Circuits and Network Theory

The capacitance of a parallel-plate capacitor is directly proportional to:

  1. the area of the plates Correct answer.
  2. the charge stored
  3. the separation between the plates
  4. the square of the plate separation

Explanation

Capacitance is given by C = εA/d, so it increases with plate area and decreases with separation. It depends on geometry and dielectric permittivity, not on the charge stored at a given instant.

Question 17Electrical Circuits and Network Theory

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

  1. 200 V
  2. 400 V
  3. 231 V Correct answer.
  4. 693 V

Explanation

In a star connection the phase voltage equals the line voltage divided by the square root of three: 400/1.732 ≈ 231 V. The line current equals the phase current in this configuration.

Question 18Electrical Circuits and Network Theory

The initial current through an inductor in a circuit, just after a switch closes, is determined by:

  1. the rate of change of voltage
  2. the inductor's resistance
  3. the current that was flowing through it just before switching Correct answer.
  4. the applied voltage divided by the inductance

Explanation

Inductor current cannot change instantaneously because a sudden change would require infinite voltage. Therefore the current immediately after switching equals the current immediately before switching, and the inductor behaves as a current source at that instant.

Question 19Electrical Circuits and Network Theory

Norton's equivalent circuit of a network is a current source in parallel with a resistance. The Norton current is equal to:

  1. the current through the load
  2. the short-circuit current between the terminals Correct answer.
  3. the open-circuit voltage divided by the resistance
  4. the open-circuit voltage

Explanation

The Norton current is the current that flows through a short circuit placed across the output terminals. The Norton resistance is found the same way as the Thevenin resistance, and the two equivalents are duals related by source transformation.

Question 20Electrical Circuits and Network Theory

In a two-port network, the h-parameter h₁₁ represents:

  1. the reverse voltage ratio
  2. the forward current gain
  3. the open-circuit output admittance
  4. the short-circuit input impedance Correct answer.

Explanation

The hybrid parameter h₁₁ is defined as input voltage divided by input current with the output port short-circuited, so it is the short-circuit input impedance. It is widely used for modelling transistors such as the common-emitter configuration.

Question 21Electrical Circuits and Network Theory

A circuit has two independent voltage sources. To apply superposition, the procedure is to:

  1. retain all sources and solve the equations simultaneously
  2. replace each voltage source by a short circuit one at a time and add the responses Correct answer.
  3. replace each voltage source by an open circuit one at a time and add the responses
  4. replace each voltage source by a current source

Explanation

When suppressing a voltage source for superposition, its voltage is set to zero, which is equivalent to a short circuit. Each source is considered alone and the individual responses are added algebraically.

Question 22Electrical Circuits and Network Theory

In a series circuit containing a resistor and a capacitor, the voltage across the resistor:

  1. leads the current by 90°
  2. lags the total applied voltage
  3. is exactly opposite to the applied voltage
  4. is in phase with the current through it Correct answer.

Explanation

Resistor voltage and current are always in phase, regardless of what other elements share the series circuit. In an RC circuit the capacitor voltage lags the current by 90°, which is what produces the overall phase shift.

Question 23Electrical Circuits and Network Theory

A 10 Ω resistor carries a current of 2 A. The power dissipated is 40 W. If the current is doubled while the resistance stays constant, the new power is:

  1. 160 W Correct answer.
  2. 320 W
  3. 120 W
  4. 80 W

Explanation

Power varies with the square of current when resistance is fixed, since P = I²R. Doubling the current multiplies power by four, so 40 × 4 = 160 W.

Question 24Electrical Circuits and Network Theory

The resonant frequency of a series circuit with L = 10 mH and C = 100 μF is approximately:

  1. 503 Hz
  2. 1.59 kHz
  3. 15.9 Hz
  4. 159 Hz Correct answer.

Explanation

The resonant frequency is f = 1/(2π√(LC)). Here LC = 1e-3 × 1e-4 = 1e-7, whose square root is about 3.16e-4, giving f ≈ 1/(2π × 3.16e-4) ≈ 159 Hz.

Question 25Electrical Circuits and Network Theory

In a series RLC circuit the bandwidth is 200 rad/s and the resonant angular frequency is 2000 rad/s. If the inductance is 0.5 H, the resistance is:

  1. 1000 Ω
  2. 10 Ω
  3. 100 Ω Correct answer.
  4. 50 Ω

Explanation

For a series circuit the bandwidth equals R/L, so R = bandwidth × L = 200 × 0.5 = 100 Ω. The quality factor is ω₀ divided by bandwidth, giving Q = 10 in this case.

Question 26Electrical Circuits and Network Theory

A 230 V, 50 Hz supply is applied to a pure inductor of 0.2 H. What is the magnitude of the current drawn from the supply?

  1. 0.46 A
  2. 1.83 A
  3. 7.32 A
  4. 3.66 A Correct answer.

Explanation

The inductive reactance is XL = 2πfL = 2π × 50 × 0.2 ≈ 62.8 Ω, and in a purely inductive circuit the current magnitude is V/XL = 230/62.8 ≈ 3.66 A. The current lags the applied voltage by 90°, but its magnitude depends only on reactance.

Question 27Electrical Circuits and Network Theory

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

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

Explanation

In a balanced delta connection the line current equals √3 times the phase current, so IL = 1.732 × 10 ≈ 17.32 A. The line current also lags or leads the corresponding phase current by 30°, but its magnitude follows the √3 ratio.

Question 28Electrical Circuits and Network Theory

A coil of resistance 8 Ω and inductive reactance 6 Ω is connected across a 100 V AC supply. What is the power dissipated in the coil?

  1. 600 W
  2. 1200 W
  3. 1000 W
  4. 800 W Correct answer.

Explanation

The impedance magnitude is Z = √(R² + XL²) = √(64 + 36) = 10 Ω, so the current is 100/10 = 10 A. Only the resistance dissipates power, giving P = I²R = 100 × 8 = 800 W, which also equals VI cos φ with cos φ = 0.8.

Question 29Electrical Circuits and Network Theory

A capacitor of 100 μF charged to 200 V is connected across a 50 Ω resistor. What is the total energy dissipated in the resistor as the capacitor discharges?

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

Explanation

The stored energy in a capacitor is W = ½CV² = 0.5 × 100 × 10⁻⁶ × 200² = 2 J. During discharge all of this stored energy is converted to heat in the resistor, since the final capacitor voltage is zero. The resistance value affects only how fast the energy is released, not the total.

Question 30Electrical Circuits and Network Theory

A series RLC circuit has R = 20 Ω, L = 0.2 H and C = 50 μF. What is the bandwidth of the circuit in rad/s?

  1. 100 rad/s Correct answer.
  2. 50 rad/s
  3. 200 rad/s
  4. 400 rad/s

Explanation

For a series RLC circuit the bandwidth is Δω = R/L = 20/0.2 = 100 rad/s. This is independent of C, and the resonant frequency ω₀ = 1/√(LC) ≈ 316 rad/s then gives Q = ω₀/Δω ≈ 3.16.

Question 31Electrical Machines (Transformers, DC and AC Machines)

A 100 kVA, 6600/415 V single-phase transformer has a full-load secondary current closest to which value?

  1. 144 A
  2. 60 A
  3. 15.2 A
  4. 241 A Correct answer.

Explanation

Full-load secondary current equals rated kVA divided by rated secondary voltage: 100000/415 ≈ 241 A. The primary current would be 100000/6600 ≈ 15.2 A, so the value must be matched to the winding asked about.

Question 32Electrical Machines (Transformers, DC and AC Machines)

In a transformer on no-load, the core loss is measured by an open-circuit test because at rated voltage the no-load current is small and the copper loss is negligible. Which quantity does the no-load power input essentially represent?

  1. Mechanical friction and windage loss
  2. Full-load copper loss of both windings
  3. Hysteresis and eddy-current loss in the core Correct answer.
  4. Leakage reactance drop in the secondary

Explanation

With the secondary open, only the small magnetising current flows, so I²R loss is negligible and the wattmeter reads almost pure core loss. Core loss comprises hysteresis and eddy-current components set by the flux and frequency.

Question 33Electrical Machines (Transformers, DC and AC Machines)

The condition for maximum efficiency of a transformer is that which two quantities are equal?

  1. Core flux and leakage flux
  2. Primary current and secondary current
  3. Copper loss and iron loss Correct answer.
  4. Leakage reactance and magnetising reactance

Explanation

Efficiency peaks when the variable I²R copper loss equals the essentially constant iron loss. At that loading the derivative of efficiency with respect to load current becomes zero.

Question 34Electrical Machines (Transformers, DC and AC Machines)

A transformer has a voltage regulation of 4 percent at full load at unity power factor. For the same load current at a leading power factor, the regulation will be

  1. Exactly 4 percent
  2. Greater than 4 percent
  3. Less than 4 percent and may become negative Correct answer.
  4. Independent of power factor

Explanation

Regulation depends on the phasor combination of resistance and leakage-reactance drops with the load current. A leading current partly cancels the reactive drop, reducing regulation and possibly making the secondary voltage rise above no-load value.

Question 35Electrical Machines (Transformers, DC and AC Machines)

A 500 kVA transformer has iron loss of 3 kW and full-load copper loss of 5 kW. The load at which efficiency is maximum is approximately

  1. 500 kVA
  2. 387 kVA Correct answer.
  3. 250 kVA
  4. 645 kVA

Explanation

Maximum efficiency occurs at the load fraction given by the square root of iron loss divided by full-load copper loss. Here that is sqrt(3/5) ≈ 0.775, giving about 387 kVA.

Question 36Electrical Machines (Transformers, DC and AC Machines)

In a parallel operation of two single-phase transformers, which condition is essential to avoid a large circulating current when the secondaries are connected together?

  1. Equal kVA ratings only
  2. Equal per-unit impedances only
  3. Equal no-load currents only
  4. Equal voltage ratios and identical polarity at the terminals Correct answer.

Explanation

Any mismatch in voltage ratio or reversed polarity produces a resultant EMF around the closed secondary loop, driving a circulating current limited only by the small leakage impedances. Equal ratings and impedances improve load sharing but are not prerequisites for safe paralleling.

Question 37Electrical Machines (Transformers, DC and AC Machines)

The secondary of a current transformer must never be open-circuited while the primary carries current mainly because

  1. The burden resistor would overheat from excess current
  2. The secondary insulation is rated only for zero voltage
  3. The flux and secondary voltage rise to dangerous values with core saturation Correct answer.
  4. The primary current would fall to zero

Explanation

With the secondary open, the demagnetising ampere-turns vanish and the whole primary current becomes magnetising current. Core flux and the induced secondary voltage then rise steeply, endangering insulation and the operator.

Question 38Electrical Machines (Transformers, DC and AC Machines)

A transformer core is built from thin laminations insulated from each other primarily to reduce

  1. Eddy-current loss Correct answer.
  2. Copper loss
  3. Leakage flux
  4. Hysteresis loss

Explanation

Laminations interrupt the paths of induced eddy currents, raising the effective resistance and cutting eddy-current loss. Hysteresis loss depends on the core material and flux density, not on lamination thickness.

Question 39Electrical Machines (Transformers, DC and AC Machines)

In a DC generator, the function of the commutator and brushes is to

  1. Reduce the armature reaction
  2. Convert the alternating EMF in the armature coils into a unidirectional terminal voltage Correct answer.
  3. Provide the main magnetic field
  4. Increase the magnitude of the generated EMF

Explanation

The EMF induced in rotating armature coils is alternating, and the commutator with brushes acts as a mechanical rectifier. It reverses coil connections at the right instants so the external terminal voltage stays unidirectional.

Question 40Electrical Machines (Transformers, DC and AC Machines)

A separately excited DC motor runs at 1000 rpm with a certain armature voltage and no additional armature resistance. If the flux is reduced by 20 percent while the armature voltage and torque are held constant, the speed becomes approximately

  1. 800 rpm
  2. 1600 rpm
  3. 1000 rpm
  4. 1250 rpm Correct answer.

Explanation

For a DC motor, torque is proportional to flux times armature current, so constant torque with lower flux demands higher current. Speed is proportional to back EMF divided by flux, and with flux reduced to 0.8 of its value the speed rises by 1/0.8, that is to 1250 rpm.

Question 41Electrical Machines (Transformers, DC and AC Machines)

In a DC shunt motor, the field winding is connected in parallel with the armature. What is the main consequence of a break in the field circuit while the motor is running on load?

  1. The motor speed rises dangerously because flux collapses to residual value Correct answer.
  2. The armature current falls to zero
  3. The motor reverses direction smoothly
  4. The motor stops immediately and safely

Explanation

Loss of field leaves only residual flux, so the back EMF falls and armature current rises sharply while speed tends to run away. This is why shunt motors must never be started with an open field circuit.

Question 42Electrical Machines (Transformers, DC and AC Machines)

A 4-pole DC generator has a lap-wound armature with 600 conductors and runs at 750 rpm. The flux per pole is 0.02 Wb. The generated EMF is

  1. 300 V
  2. 600 V
  3. 37.5 V
  4. 150 V Correct answer.

Explanation

For a lap winding the number of parallel paths equals the number of poles. Using E = phi Z N P / (60 A) with A = P gives E = 0.02 × 600 × 750 / 60 = 150 V.

Question 43Electrical Machines (Transformers, DC and AC Machines)

Interpoles in a DC machine are connected in series with the armature winding so that they

  1. Supply the main field excitation
  2. Produce a flux opposing the cross-magnetising effect of armature reaction in the commutating zone Correct answer.
  3. Limit the starting current
  4. Improve the insulation of the commutator

Explanation

Interpoles carry armature current and are placed in the interpolar axis, so their flux opposes the armature cross-flux and induces a reversing EMF in the commutating coils. This improves commutation and reduces sparking.

Question 44Electrical Machines (Transformers, DC and AC Machines)

A 3-phase, 4-pole, 50 Hz induction motor runs at 1440 rpm. The slip is

  1. 3 percent
  2. 6 percent
  3. 2.8 percent
  4. 4 percent Correct answer.

Explanation

Synchronous speed is 120 f / P = 120 × 50 / 4 = 1500 rpm. Slip equals (1500 − 1440)/1500 = 0.04, or 4 percent.

Question 45Electrical Machines (Transformers, DC and AC Machines)

At standstill, a squirrel-cage induction motor behaves like a transformer with a short-circuited secondary. The rotor frequency at standstill is

  1. Equal to the supply frequency Correct answer.
  2. Twice the supply frequency
  3. Zero
  4. Half the supply frequency

Explanation

Rotor frequency equals slip times supply frequency, and slip is unity at standstill. Hence the rotor quantities alternate at the same frequency as the stator supply.

Question 46Electrical Machines (Transformers, DC and AC Machines)

A 3-phase induction motor develops maximum torque when the rotor resistance per phase equals

  1. The magnetising reactance per phase
  2. The rotor standstill leakage reactance per phase Correct answer.
  3. The stator resistance per phase
  4. Zero

Explanation

The torque-slip characteristic peaks when the rotor circuit impedance is such that rotor resistance equals rotor standstill reactance. Adding external rotor resistance in a slip-ring motor therefore shifts the maximum-torque point to a higher slip.

Question 47Electrical Machines (Transformers, DC and AC Machines)

Which starting method is used for a slip-ring induction motor to obtain high starting torque with low starting current?

  1. Capacitor starting of the stator
  2. Star-delta starting of the stator
  3. Adding external resistance in the rotor circuit Correct answer.
  4. Direct-on-line starting

Explanation

External rotor resistance raises the rotor power factor at start and shifts the maximum-torque point towards standstill. This gives high starting torque while limiting the stator current drawn.

Question 48Electrical Machines (Transformers, DC and AC Machines)

A 3-phase induction motor is running at a slip of 2 percent. The ratio of rotor copper loss to the air-gap power transferred to the rotor is

  1. 0.02 Correct answer.
  2. 0.98
  3. 50
  4. 1.02

Explanation

Rotor copper loss equals slip times the air-gap power, since the mechanical power developed is (1 − s) times the air-gap power. With s equal to 0.02 the loss fraction is 0.02.

Question 49Electrical Machines (Transformers, DC and AC Machines)

In a single-phase capacitor-start induction motor, the auxiliary winding is placed in space quadrature with the main winding and the capacitor causes the current in it to

  1. Lead the main winding current by nearly 90 electrical degrees Correct answer.
  2. Be zero at starting
  3. Lag the main winding current by nearly 90 electrical degrees
  4. Be exactly in phase with the main winding current

Explanation

A rotating field requires a phase displacement between the two winding currents. The capacitor makes the auxiliary current lead, and with a suitable value the displacement approaches 90 degrees, producing a strong starting torque.

Question 50Electrical Machines (Transformers, DC and AC Machines)

The synchronous speed of a 6-pole alternator supplying a 50 Hz system is

  1. 1000 rpm Correct answer.
  2. 750 rpm
  3. 3000 rpm
  4. 1500 rpm

Explanation

Synchronous speed equals 120 f / P, giving 120 × 50 / 6 = 1000 rpm. The number of poles and the supply frequency together fix the rotational speed.