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PE Electrical Power practice questions

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Question 1General Power Engineering (circuit analysis, measurement and instrumentation)

A 12 V ideal voltage source is connected across a series combination of a 4 Ω resistor and an 8 Ω resistor. What current flows through the circuit?

  1. 3.0 A
  2. 1.0 A Correct answer.
  3. 0.5 A
  4. 12 A

Explanation

The resistors are in series, so the total resistance is 4 Ω + 8 Ω = 12 Ω. Ohm's law then gives I = 12 V / 12 Ω = 1.0 A.

Question 2General Power Engineering (circuit analysis, measurement and instrumentation)

A sinusoidal voltage has an RMS value of 120 V. What is its peak value?

  1. 85 V
  2. 240 V
  3. 120 V
  4. 170 V Correct answer.

Explanation

For a sinusoid, the peak value is √2 times the RMS value. Thus Vpeak = 1.414 × 120 V ≈ 170 V.

Question 3General Power Engineering (circuit analysis, measurement and instrumentation)

A 100 μF capacitor is charged to 200 V. How much energy is stored?

  1. 0.5 J
  2. 4.0 J
  3. 2.0 J Correct answer.
  4. 1.0 J

Explanation

The energy stored in a capacitor is E = ½CV². Substituting C = 100 μF and V = 200 V gives E = ½ × 100×10⁻⁶ × 200² = 2.0 J.

Question 4General Power Engineering (circuit analysis, measurement and instrumentation)

An energy meter registers a constant 5 kW load for 4 hours. What energy is consumed?

  1. 80 kWh
  2. 20 kWh Correct answer.
  3. 1.25 kWh
  4. 9 kWh

Explanation

Energy equals power multiplied by time, so 5 kW × 4 h = 20 kWh.

Question 5General Power Engineering (circuit analysis, measurement and instrumentation)

An oscilloscope displays a waveform that spans 5 vertical divisions. The vertical scale is 2 V/division. What is the peak-to-peak voltage?

  1. 2.5 V
  2. 5 V
  3. 10 V Correct answer.
  4. 20 V

Explanation

Peak-to-peak voltage is the number of divisions multiplied by the volts-per-division setting. Thus 5 divisions × 2 V/division = 10 V.

Question 6General Power Engineering (circuit analysis, measurement and instrumentation)

A 2 kΩ resistor and a 50 μF capacitor form a series RC circuit. What is the time constant?

  1. 100 s
  2. 0.1 s Correct answer.
  3. 10 s
  4. 0.01 s

Explanation

The time constant of a series RC circuit is τ = RC. Converting 50 μF to 50×10⁻⁶ F gives τ = 2000 Ω × 50×10⁻⁶ F = 0.1 s.

Question 7General Power Engineering (circuit analysis, measurement and instrumentation)

A moving-coil meter has a full-scale current of 100 μA and a coil resistance of 50 Ω. What series resistance is needed to make a 10 V full-scale voltmeter?

  1. 100.00 kΩ
  2. 50 Ω
  3. 99.95 kΩ Correct answer.
  4. 10 kΩ

Explanation

The multiplier resistance is R = VFS / IFS − Rm. Here R = 10 V / 100 μA − 50 Ω = 100,000 Ω − 50 Ω = 99,950 Ω, or 99.95 kΩ.

Question 8General Power Engineering (circuit analysis, measurement and instrumentation)

A load draws 8 kW of real power and 10 kVA of apparent power. What is the power factor?

  1. 0.80 Correct answer.
  2. 0.60
  3. 1.25
  4. 0.75

Explanation

Power factor is real power divided by apparent power. Thus pf = 8 kW / 10 kVA = 0.80.

Question 9General Power Engineering (circuit analysis, measurement and instrumentation)

A 12 V independent source feeds a 6 Ω resistor in series with a 3 Ω resistor, and the load is connected across the 3 Ω resistor. What is the Thevenin resistance seen by the load?

  1. 9 Ω
  2. 3 Ω
  3. 2 Ω Correct answer.
  4. 6 Ω

Explanation

With the independent voltage source replaced by a short circuit, the 6 Ω and 3 Ω resistors are in parallel as seen from the load terminals. Thus Rth = 6 Ω || 3 Ω = 2 Ω.

Question 10General Power Engineering (circuit analysis, measurement and instrumentation)

When applying superposition to a linear circuit, how is an independent current source treated while finding the contribution of the other sources?

  1. Replaced by a short circuit
  2. Replaced by a voltage source equal to its current
  3. Replaced by an open circuit Correct answer.
  4. Left at its original value

Explanation

Superposition deactivates each independent source in turn: voltage sources become short circuits and current sources become open circuits. Dependent sources are not deactivated because their values depend on circuit variables.

Question 11General Power Engineering (circuit analysis, measurement and instrumentation)

A linear circuit has a Thevenin equivalent of 10 V in series with 2 Ω. What is the Norton equivalent current?

  1. 20 A
  2. 5 A Correct answer.
  3. 0.2 A
  4. 12 A

Explanation

The Norton current is the short-circuit current, equal to the Thevenin voltage divided by the Thevenin resistance. Thus IN = 10 V / 2 Ω = 5 A.

Question 12General Power Engineering (circuit analysis, measurement and instrumentation)

A balanced three-phase load operates at 480 V line-to-line and 10 A line current with a power factor of 0.90. What total real power does the load draw?

  1. 8.31 kW
  2. 2.49 kW
  3. 7.48 kW Correct answer.
  4. 4.32 kW

Explanation

For a balanced three-phase load, total real power is P = √3 VL IL cos θ. Substituting gives P = √3 × 480 V × 10 A × 0.90 ≈ 7.48 kW.

Question 13General Power Engineering (circuit analysis, measurement and instrumentation)

A 200:5 current transformer carries 160 A in its primary. What current appears in the secondary?

  1. 8 A
  2. 2.5 A
  3. 5 A
  4. 4 A Correct answer.

Explanation

The current transformer ratio is 200:5, or 40:1. Secondary current is therefore 160 A / 40 = 4 A.

Question 14General Power Engineering (circuit analysis, measurement and instrumentation)

A Wheatstone bridge has arms R1 = 100 Ω, R2 = 200 Ω, and R3 = 300 Ω, with Rx forming the fourth arm. At balance, what is Rx?

  1. 150 Ω
  2. 900 Ω
  3. 600 Ω Correct answer.
  4. 300 Ω

Explanation

At bridge balance, the ratio of adjacent arms is equal: R1/R2 = R3/Rx. Solving gives Rx = R2R3/R1 = (200 × 300)/100 = 600 Ω.

Question 15General Power Engineering (circuit analysis, measurement and instrumentation)

A 1 mA full-scale meter with a 50 Ω internal resistance is to be used as a 10 mA ammeter. What shunt resistance is required?

  1. 50 Ω
  2. 0.50 Ω
  3. 5.56 Ω Correct answer.
  4. 5 Ω

Explanation

The shunt carries the excess current, and the required value is Rsh = Rm/(n − 1), where n is the current multiplication factor. Here n = 10 mA / 1 mA = 10, so Rsh = 50/9 ≈ 5.56 Ω.

Question 16General Power Engineering (circuit analysis, measurement and instrumentation)

A 12 V source is suddenly applied to a series RL circuit with R = 6 Ω and L = 2 H. What current flows at one time constant after the switch closes?

  1. 1.00 A
  2. 1.26 A Correct answer.
  3. 0.74 A
  4. 2.00 A

Explanation

The final current is 12 V / 6 Ω = 2 A, and the time constant is L/R = 2/6 = 0.333 s. At t = τ, i = 2(1 − e⁻¹) ≈ 1.26 A.

Question 17General Power Engineering (circuit analysis, measurement and instrumentation)

A series RLC circuit has L = 10 mH and C = 100 μF. At what frequency does it resonate?

  1. 318 Hz
  2. 100 Hz
  3. 159 Hz Correct answer.
  4. 50 Hz

Explanation

Resonance occurs at f = 1/(2π√(LC)). With L = 10 mH and C = 100 μF, LC = 1×10⁻⁶, so f = 1/(2π × 0.001) ≈ 159 Hz.

Question 18General Power Engineering (circuit analysis, measurement and instrumentation)

A series circuit has R = 3 Ω and inductive reactance XL = 4 Ω. What is the phase angle of the impedance?

  1. 45.0°
  2. 53.1° Correct answer.
  3. 30.0°
  4. 36.9°

Explanation

The impedance phase angle is θ = arctan(XL/R). Thus θ = arctan(4/3) ≈ 53.1°, with current lagging the voltage.

Question 19General Power Engineering (circuit analysis, measurement and instrumentation)

At a node, a 10 A independent current source injects current. Two resistors, 5 Ω and 20 Ω, connect the node to ground. What is the node voltage?

  1. 50 V
  2. 40 V Correct answer.
  3. 10 V
  4. 8 V

Explanation

Applying KCL at the node gives V/5 + V/20 = 10. The equivalent conductance is 0.25 S, so V = 10/0.25 = 40 V.

Question 20General Power Engineering (circuit analysis, measurement and instrumentation)

A 10 V source feeds a divider made of two 100 kΩ resistors. An ideal voltmeter would read 5 V across the lower resistor. If the voltmeter has a 100 kΩ input resistance, what does it read?

  1. 3.33 V Correct answer.
  2. 2.50 V
  3. 10.0 V
  4. 5.00 V

Explanation

The voltmeter's 100 kΩ input appears in parallel with the lower 100 kΩ resistor, reducing that branch to 50 kΩ. The divider then gives 10 V × 50/(100 + 50) ≈ 3.33 V.

Question 21General Power Engineering (circuit analysis, measurement and instrumentation)

Two wattmeters connected to a balanced three-phase load read 5 kW and 2 kW. What is the load power factor?

  1. 0.87
  2. 0.71
  3. 0.50
  4. 0.80 Correct answer.

Explanation

For the two-wattmeter method, tan θ = √3(W1 − W2)/(W1 + W2). Here tan θ = √3(3)/(7) ≈ 0.742, so θ ≈ 36.6° and pf = cos θ ≈ 0.80.

Question 22General Power Engineering (circuit analysis, measurement and instrumentation)

When determining the Thevenin resistance of a circuit that contains a dependent source, what is the correct procedure?

  1. Apply a test source, deactivate independent sources, and keep dependent sources active Correct answer.
  2. Turn off all sources, including the dependent source, and calculate the resistance
  3. Open all dependent sources and short all independent sources before calculating
  4. Divide the open-circuit voltage by the short-circuit current while ignoring dependent sources

Explanation

A dependent source cannot simply be turned off because its value is controlled by another circuit variable. The standard method applies a test voltage or current, deactivates only independent sources, and keeps dependent sources active while measuring the terminal response.

Question 23General Power Engineering (circuit analysis, measurement and instrumentation)

A balanced delta-connected load with 30 Ω per phase is supplied from a 480 V line-to-line source. What is the line current?

  1. 16.0 A
  2. 9.24 A
  3. 27.7 A Correct answer.
  4. 48.0 A

Explanation

In a delta connection, the phase voltage equals the line-to-line voltage, so the phase current is 480/30 = 16 A. The line current is √3 times the phase current: 27.7 A.

Question 24General Power Engineering (circuit analysis, measurement and instrumentation)

A voltage waveform consists of a 100 V RMS fundamental and a 30 V RMS third harmonic. What is the true RMS value?

  1. 104 V Correct answer.
  2. 130 V
  3. 110 V
  4. 100 V

Explanation

For a waveform with multiple harmonics, the true RMS value is the square root of the sum of the squares of the individual RMS components. Thus Vrms = √(100² + 30²) ≈ 104 V.

Question 25General Power Engineering (circuit analysis, measurement and instrumentation)

An instrumentation amplifier has a differential gain of 1000 and a common-mode gain of 0.5. What is its common-mode rejection ratio in decibels?

  1. 2000 dB
  2. 34 dB
  3. 66 dB Correct answer.
  4. 60 dB

Explanation

CMRR is the differential gain divided by the common-mode gain, expressed in decibels as 20 log10(Ad/Acm). Here CMRR = 1000/0.5 = 2000, and 20 log10(2000) ≈ 66 dB.

Question 26General Power Engineering (circuit analysis, measurement and instrumentation)

In a purely capacitive AC circuit, how does the current relate to the voltage?

  1. It lags the voltage by 90°.
  2. It leads the voltage by 90°. Correct answer.
  3. It is in phase with the voltage.
  4. It is 180° out of phase with the voltage.

Explanation

For a capacitor, current is proportional to the rate of change of voltage. With a sinusoidal voltage, this makes the current lead the voltage by 90°.

Question 27General Power Engineering (circuit analysis, measurement and instrumentation)

A 100 Ω resistor carries 0.5 A. What power is dissipated in the resistor?

  1. 100 W
  2. 12.5 W
  3. 25 W Correct answer.
  4. 50 W

Explanation

Use P = I²R. Substituting gives P = (0.5 A)² × 100 Ω = 25 W.

Question 28General Power Engineering (circuit analysis, measurement and instrumentation)

What is the internal resistance of an ideal ammeter?

  1. Infinite
  2. Zero Correct answer.
  3. Equal to the circuit resistance
  4. 1 Ω per ampere of range

Explanation

An ammeter is inserted in series and should not alter the current. An ideal ammeter therefore has zero internal resistance and zero voltage drop.

Question 29General Power Engineering (circuit analysis, measurement and instrumentation)

Two 10 μF capacitors are connected in series. What is the equivalent capacitance?

  1. 2.5 μF
  2. 20 μF
  3. 5 μF Correct answer.
  4. 10 μF

Explanation

For two capacitors in series, 1/Ceq = 1/C1 + 1/C2. Thus 1/Ceq = 1/10 + 1/10 = 1/5 μF, so Ceq = 5 μF.

Question 30General Power Engineering (circuit analysis, measurement and instrumentation)

The RMS value of a sinusoidal voltage is 120 V. What is its peak value?

  1. 120 V
  2. 240 V
  3. 340 V
  4. 170 V Correct answer.

Explanation

For a sine wave, Vpeak = √2 × Vrms. Therefore Vpeak ≈ 1.414 × 120 V ≈ 170 V.

Question 31General Power Engineering (circuit analysis, measurement and instrumentation)

Which instrument is used to measure the insulation resistance of a de-energized cable?

  1. A megohmmeter Correct answer.
  2. A clamp-on ammeter
  3. A power factor meter
  4. A watt-hour meter

Explanation

Insulation resistance is very high, so it is measured with a megohmmeter that applies a high DC test voltage. A clamp-on ammeter, watt-hour meter, and power factor meter measure other electrical quantities.

Question 32General Power Engineering (circuit analysis, measurement and instrumentation)

Why cannot the current through an inductor change instantaneously?

  1. The magnetic-field energy cannot change instantly. Correct answer.
  2. The DC resistance of an ideal inductor is infinite.
  3. An ideal inductor conducts only steady direct current.
  4. The voltage across an ideal inductor is always zero.

Explanation

Inductor current is linked to magnetic-field energy, W = ½LI². A sudden current change would require a sudden energy change, which an ideal inductor cannot produce, so current is continuous.

Question 33General Power Engineering (circuit analysis, measurement and instrumentation)

A 12 V battery delivers 2 A for 3 hours. How much energy is delivered?

  1. 6 Wh
  2. 36 Wh
  3. 72 Wh Correct answer.
  4. 24 Wh

Explanation

Energy in watt-hours is voltage × current × time. Thus 12 V × 2 A × 3 h = 72 Wh.

Question 34General Power Engineering (circuit analysis, measurement and instrumentation)

A 6 Ω resistor and a 3 Ω resistor are connected in parallel, and that combination is in series with a 4 Ω resistor. What is the total resistance?

  1. 4 Ω
  2. 9 Ω
  3. 13 Ω
  4. 6 Ω Correct answer.

Explanation

The parallel pair is (6 × 3)/(6 + 3) = 2 Ω. Adding the series 4 Ω resistor gives 6 Ω total.

Question 35General Power Engineering (circuit analysis, measurement and instrumentation)

A source has a Thevenin voltage of 20 V and a Thevenin resistance of 5 Ω. What load resistance absorbs maximum power?

  1. 0 Ω
  2. 10 Ω
  3. 20 Ω
  4. 5 Ω Correct answer.

Explanation

Maximum power transfer to a resistive load occurs when the load resistance equals the Thevenin resistance. Here that is 5 Ω.

Question 36General Power Engineering (circuit analysis, measurement and instrumentation)

A planar network has 6 branches and 4 nodes. How many independent mesh equations are needed for mesh analysis?

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

Explanation

For a planar network, the number of independent mesh equations is b − n + 1. Substituting gives 6 − 4 + 1 = 3.

Question 37General Power Engineering (circuit analysis, measurement and instrumentation)

A full-wave rectified sine wave has a peak value of 10 V. What is its average value?

  1. 10 V
  2. 5 V
  3. 7.07 V
  4. 6.37 V Correct answer.

Explanation

The average value of a full-wave rectified sine is 2Vpeak/π. Thus (2 × 10)/π ≈ 6.37 V.

Question 38General Power Engineering (circuit analysis, measurement and instrumentation)

A series RLC circuit has a resistance of 10 Ω and an inductive reactance of 100 Ω at resonance. What is its quality factor?

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

Explanation

At resonance, the quality factor of a series RLC circuit is Q = XL/R. Substituting gives 100/10 = 10.

Question 39General Power Engineering (circuit analysis, measurement and instrumentation)

A platinum RTD has a resistance of 100 Ω at 0°C and a temperature coefficient of 0.00385/°C. What is its approximate resistance at 100°C using the linear approximation?

  1. 385 Ω
  2. 61.5 Ω
  3. 138.5 Ω Correct answer.
  4. 100.0 Ω

Explanation

The linear RTD approximation is R = R0(1 + αΔT). Thus R = 100[1 + 0.00385(100)] = 138.5 Ω.

Question 40General Power Engineering (circuit analysis, measurement and instrumentation)

A strain gauge has a gauge factor of 2.0 and a nominal resistance of 120 Ω. It experiences a strain of 500 microstrain. What is the change in resistance?

  1. 0.24 Ω
  2. 0.12 Ω Correct answer.
  3. 0.06 Ω
  4. 1.20 Ω

Explanation

Gauge factor is defined by GF = (ΔR/R)/ε. Thus ΔR = GF × ε × R = 2.0 × 500×10⁻⁶ × 120 Ω = 0.12 Ω.

Question 41General Power Engineering (circuit analysis, measurement and instrumentation)

A watt-hour meter has a constant of 600 revolutions per kWh. If its disk makes 10 revolutions in 1 minute, what is the power?

  1. 1 kW Correct answer.
  2. 100 W
  3. 600 W
  4. 3.6 kW

Explanation

At 600 rev/kWh, 10 revolutions represent 10/600 kWh = 1/60 kWh. That energy is used in 1/60 hour, so power is (1/60 kWh)/(1/60 h) = 1 kW.

Question 42General Power Engineering (circuit analysis, measurement and instrumentation)

A current transformer has a rated secondary current of 5 A and a total burden impedance of 0.8 Ω. What is the burden VA?

  1. 40 VA
  2. 6.25 VA
  3. 20 VA Correct answer.
  4. 4 VA

Explanation

The burden in volt-amperes is I²Z for the secondary circuit. Thus (5 A)² × 0.8 Ω = 20 VA.

Question 43General Power Engineering (circuit analysis, measurement and instrumentation)

A 10:1 oscilloscope probe is used with an oscilloscope having a 1 MΩ input resistance. What input resistance does the probe present to the circuit under test?

  1. 9 MΩ
  2. 10 MΩ Correct answer.
  3. 1 MΩ
  4. 100 kΩ

Explanation

A 10:1 probe adds a 9 MΩ series resistor to the 1 MΩ scope input. The circuit sees 9 MΩ + 1 MΩ = 10 MΩ, and the signal is attenuated by 10.

Question 44General Power Engineering (circuit analysis, measurement and instrumentation)

A digital multimeter reads 50.0 V on the 100 V range. Its accuracy is specified as ±(0.5% of reading + 2 counts), with a resolution of 0.1 V. What is the maximum error?

  1. 0.20 V
  2. 0.45 V Correct answer.
  3. 0.25 V
  4. 0.50 V

Explanation

The percentage term is 0.5% of 50.0 V = 0.25 V. Two counts at 0.1 V per count add 0.20 V, so the total maximum error is 0.45 V.

Question 45General Power Engineering (circuit analysis, measurement and instrumentation)

A four-wire Kelvin connection is used in resistance measurement mainly to do what?

  1. Cancel thermoelectric EMF by reversing polarity
  2. Remove lead and contact resistance effects Correct answer.
  3. Measure insulation resistance at very high values
  4. Increase the test voltage to a higher level

Explanation

In a four-wire measurement, current is forced through the unknown resistor with one pair of leads, while voltage is sensed with a separate high-impedance pair. Because the sense leads carry almost no current, lead and contact resistance drops do not appreciably affect the reading.

Question 46General Power Engineering (circuit analysis, measurement and instrumentation)

Two voltage sources, 10 V in series with 2 Ω and 5 V in series with 3 Ω, are connected to a common node. A 6 Ω load connects that node to ground. What is the voltage at the common node?

  1. 6.67 V Correct answer.
  2. 7.50 V
  3. 5.00 V
  4. 8.00 V

Explanation

Millman's theorem gives V = (10/2 + 5/3 + 0/6)/(1/2 + 1/3 + 1/6). The numerator is 6.667 A and the denominator is 1 S, so V = 6.67 V.

Question 47General Power Engineering (circuit analysis, measurement and instrumentation)

A source has a Thevenin impedance of 8 + j6 Ω. What load impedance absorbs maximum average power?

  1. 6 + j8 Ω
  2. 6 − j8 Ω
  3. 8 + j6 Ω
  4. 8 − j6 Ω Correct answer.

Explanation

For maximum average power transfer with a complex source impedance, the load should be the complex conjugate of the Thevenin impedance. The conjugate of 8 + j6 Ω is 8 − j6 Ω.

Question 48General Power Engineering (circuit analysis, measurement and instrumentation)

A 10 μF capacitor has an initial voltage of 4 V. At t = 0, it is connected through a 2 kΩ resistor to a 12 V DC source. What is the capacitor voltage at t = 10 ms?

  1. 8.00 V
  2. 10.00 V
  3. 12.00 V
  4. 7.15 V Correct answer.

Explanation

The time constant is τ = RC = (2 kΩ)(10 μF) = 20 ms. With initial voltage 4 V and final voltage 12 V, v(t) = 12 + (4 − 12)e^(−t/τ). At 10 ms, v = 12 − 8e^(−0.5) ≈ 7.15 V.

Question 49General Power Engineering (circuit analysis, measurement and instrumentation)

A Wheatstone bridge has three arms of 120 Ω and a variable fourth arm initially balanced at 120 Ω. It is excited by 10 V. If the variable arm increases by 0.24 Ω, what is the approximate bridge output voltage?

  1. 5.0 mV Correct answer.
  2. 10 mV
  3. 20 mV
  4. 2.5 mV

Explanation

For a small change in one arm, the bridge output is approximately (Vs/4)(ΔR/R). Here (10 V/4)(0.24/120) = 0.005 V = 5.0 mV.

Question 50General Power Engineering (circuit analysis, measurement and instrumentation)

A 100 Ω resistor is measured using an ammeter in series with the parallel combination of the resistor and a voltmeter. The ammeter reads 0.5 A and the voltmeter reads 50 V. The voltmeter resistance is 10 kΩ. What is the actual resistance?

  1. 101.0 Ω Correct answer.
  2. 99.0 Ω
  3. 100.0 Ω
  4. 105.0 Ω

Explanation

The voltmeter draws 50 V / 10 kΩ = 5 mA. The resistor current is therefore 0.5 A − 0.005 A = 0.495 A, so R = 50 V / 0.495 A ≈ 101.0 Ω.