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RRB JE Junior Engineer · Technical Abilities (CBT-2)

Electrical Engineering (CBT-2)

DC/AC circuits, electrical machines, power systems, instruments, utilization, and basic electronics at diploma depth for RRB JE CBT-2 Technical Abilities (Electrical).

Six concepts — one per Electrical diploma pillar on RRB JE CBT-2 Technical. A circuit is a closed path for charge; a machine converts that charge-flow into another form of power; the power system carries megawatts to a load; an instrument reads what happened; utilization is what the kilowatt-hour was for; electronics is the panel of diodes and amplifiers that sit beside the machines. Each card names the object, then works one diploma number.

  • RRB JE Junior Engineer
  • Medium level
  • 6 concepts
  • 55 practice questions

1DC and AC circuits

A circuit is a closed path that lets charge move. Voltage is the energy given to each coulomb; current is how many coulombs cross a cut each second. In a diploma network the first job is to name what is series — those parts share one current — and what is parallel — those parts share one voltage. Kirchhoff's current law says the currents into a node add to the currents out; Kirchhoff's voltage law says the rises and drops around any loop add to zero.

Resistance belongs to that particular piece of conductor. On AC the same piece also has reactance X_L = 2\pi f L or X_C = 1/(2\pi f C), so the opposition the source actually sees is the impedance Z = \sqrt{R^2 + X^2}. Power factor is R over Z, not X over Z: it is the fraction of the volt-amperes that does real work. Series resonance is the frequency where X_L = X_C, so Z collapses to R and the current peaks.

Figure. The three bars are the magnitudes from the worked coil: R = 8 Ω, XL = 6 Ω, Z = 10 Ω. Z is the hypotenuse, not R + XL. Power factor is the R bar over the Z bar.

How a network is read

  1. Name series and parallelSame current means series; same voltage means parallel. Write KCL at each node and KVL around each loop before combining.
  2. Build Z from R and XOn AC, Z = \sqrt{R^2 + X^2} with X = X_L - X_C. The current is I = V/Z, not V/R.
  3. Read power from RReal power is P = I^2 R = VI\cos\phi with \cos\phi = R/Z. Reactive power sits in X, not in R.

Series RL on 50 V

A coil is modelled as R = 8\,\Omega in series with X_L = 6\,\Omega across V = 50 V. Find Z, I, power factor and real power.

  • Z = \sqrt{8^2 + 6^2} = \sqrt{64 + 36}10\,\Omega
  • I = V/Z = 50/105 A
  • \cos\phi = R/Z = 8/100.8 lag
  • P = I^2 R = 25 \times 8200 W

Pro tip. Q = I²X_L = 150 VAR, and S = VI = 250 VA. Check: 200² + 150² = 250². Using X/Z as the power factor would have given 0.6 and a 150 W answer that is the reactive power, not the real power.

A series circuit has R = 8 Ω and X_L = 6 Ω. Someone quotes the power factor as 6/10 = 0.6. What did they compute, and what is the power factor?
  1. They computed R/Z; the power factor is 0.6 lead
  2. They computed X/Z, which is sin φ; the power factor is R/Z = 0.8 lag
  3. They computed Z/R; the power factor is 1.25

cos φ = R/Z = 8/10 = 0.8 lag. X/Z = 0.6 is sin φ, the reactive fraction. Z/R is the inverse of the power factor, not a power factor.

2Transformers and rotating machines

An electrical machine converts power from one form to another. A transformer stays electrical on both sides: two windings share one core, and V_1/V_2 = N_1/N_2 = I_2/I_1 for an ideal unit. A DC machine has a commutator so the current at the brushes is unidirectional. An induction motor has no electrical connection to the rotor — slip s = (N_s - N)/N_s is what induces the rotor currents. A synchronous machine locks to N_s = 120f/P and can run as motor or generator.

The paper's trap is mixing those jobs. A transformer does not produce mechanical torque. An induction motor does not run at synchronous speed on load — if it did, slip would be zero and there would be no rotor current. A DC series motor, not a shunt motor, is the usual traction choice because torque is large at start.

Figure. The two bars are the worked 4-pole, 50 Hz, 4% slip case. Ns is 1500 rpm; the shaft runs at 1440 rpm. The 60 rpm gap is the slip that induces the rotor currents. Values are not to a 1 rpm visual tolerance — the labels carry the numbers.

Which machine, which number

  1. Name the conversionTransformer: AC voltage in, AC voltage out. DC machine: electrical ↔ mechanical with a commutator. Induction: electrical in, mechanical out, rotor induced. Synchronous: locked to Ns.
  2. Write Ns firstN_s = 120f/P in rpm. Every induction-motor speed is a slip below that. A 4-pole, 50 Hz machine has Ns = 1500 rpm, never 3000.
  3. Keep the transformer ratiosVoltage follows turns; current inverts. Ampere-turns balance: N_1 I_1 = N_2 I_2 on an ideal unit.
Four machines, four jobs
MachineConvertsExam cue
TransformerAC voltage ↔ AC voltageV1/V2 = N1/N2 = I2/I1
DC machineElectrical ↔ mechanicalCommutator; series = high start T
Induction motorAC in → shaft outSlip; no rotor supply leads
SynchronousMotor or generator at NsNs = 120f/P; can supply VARs

Slip on a 4-pole motor

A 4-pole induction motor is fed at 50 Hz and runs at 4% slip. Find synchronous speed, rotor speed and rotor frequency.

  • N_s = 120f/P = 120 \times 50 / 41500 rpm
  • N = N_s(1 - s) = 1500 \times 0.961440 rpm
  • f_r = sf = 0.04 \times 502 Hz

Pro tip. Rotor frequency is sf, not f and not (1 − s)f. At start s = 1 so fr = f; at sync s = 0 so fr = 0 and torque vanishes. Quoting 1500 rpm as the running speed is the no-load-sync trap.

A 4-pole, 50 Hz induction motor is said to be 'running at 1500 rpm on load'. Why is that impossible, and what must be true instead?
  1. 1500 rpm is Ns; on load there must be slip, so N < 1500 rpm and rotor currents exist
  2. 1500 rpm is too slow; a 4-pole machine always runs at 3000 rpm
  3. On load the motor must run above Ns so that slip is negative

Ns = 120 × 50 / 4 = 1500 rpm. Load torque needs rotor current, which needs slip, so N is below Ns. Super-synchronous speed is a generator / plugging story, not motoring on load. 3000 rpm is the 2-pole Ns.

3Generation, transmission, distribution

A power system is the chain that takes generated megawatts to a load: generator, step-up transformer, transmission line, step-down transformer, then distribution. Voltage is raised for the long haul because current on a three-phase line is I = P/(\sqrt{3}\,V\cos\phi), so I^2 R loss falls as voltage rises. Distribution then drops the voltage to what a motor or a house can use.

Protection is a decision plus an interruption: the relay decides that a fault is present; the circuit breaker opens the circuit. A CT or PT only scales the line quantity down to the relay. Raising voltage does not change the power being sent — it changes the current that carries that power.

Figure. Topology only — left to right, not a scale map. Generation, step-up, the long line (the I²R cost), step-down, load. Voltage is high on the line node so current there is small.

Reading a feeder

  1. Follow the chainGenerate at 11 kV class, step up to 132/220/400 kV for the line, step down to 33/11 kV, then 415 V at the board.
  2. Current from powerThree-phase line current is I = P/(\sqrt{3} V \cos\phi). Forgetting \sqrt{3} inflates I by 1.73 and the I^2 R loss with it.
  3. Relay then breakerThe relay compares the CT/PT secondary with a setting. The breaker interrupts. Neither device is the other.

Current on an 11 kV feeder

A 3-phase, 11 kV feeder delivers 1.5 MW at 0.8 lag. Find the line current. Take \sqrt{3} = 1.732.

  • \sqrt{3}\,V\cos\phi = 1.732 \times 11000 \times 0.815241.6
  • I = P / (\sqrt{3}\,V\cos\phi) = 1.5 \times 10^6 / 15241.698.4 A

Pro tip. Drop √3 and the same feeder looks like 170 A (1.5e6 / (11000 × 0.8)). That inflated current would size the conductor and the CT wrong. Power stayed 1.5 MW either way; only I changed.

A 3-phase feeder delivers a fixed 1.5 MW. The sending voltage is doubled and the power factor is unchanged. The line current
  1. Doubles, because more voltage means more current
  2. Halves, because I = P / (√3 V cos φ) at fixed P and pf
  3. Stays the same, because power did not change

P and pf are fixed, so I is inversely proportional to V. Doubling V halves I and cuts I²R loss to one quarter. Power is not I, and more voltage is not more current on a constant-power feeder.

4Electrical measurements

A measuring instrument turns a quantity into a readable deflection. Permanent-magnet moving-coil (PMMC) instruments read the average of a DC current and read zero on AC — the coil tries to follow both halves and the average is nothing. Moving-iron instruments read RMS on AC or DC. A dynamometer wattmeter multiplies voltage and current to read power. An induction energy meter integrates that power over time: disc revolutions are watt-hours.

CTs and PTs are not decorative: they scale a dangerous line quantity to 5 A or 110 V so the instrument and the relay can live. A megger injects a high DC voltage to read insulation resistance, not ordinary continuity.

Figure. Three instrument classes, not a circuit. PMMC is the AC trap (terracotta): it is a DC-average device. Moving iron reads RMS. The dynamometer wattmeter reads power. Classification only — boxes are equal on purpose, not to scale as 'importance'.

Pick the instrument

  1. Name the quantityCurrent or voltage → ammeter/voltmeter. Power → dynamometer wattmeter. Energy → induction meter. Insulation → megger.
  2. Name AC or DCPMMC is DC only. Moving iron is RMS on both. A PMMC on 50 Hz AC reads zero, not the RMS.
  3. Scale with CT/PTLine current / CT ratio = secondary current. A 400/5 CT at 240 A primary reads 3 A secondary, not 5 A.
Which instrument, which reading
InstrumentReadsAC / DC
PMMCAverage current or voltageDC only — AC reads 0
Moving ironRMS current or voltageAC or DC
DynamometerPower (and some VA)AC or DC
Induction meterEnergy (kWh)AC

Energy-meter disc to kilowatts

A meter is marked 1800 rev/kWh. The disc makes 45 revolutions in 90 s. Find the load in kW.

  • Energy in 90 s = 45/18000.025 kWh
  • Hours = 90/36000.025 h
  • Power = 0.025/0.0251.0 kW

Pro tip. Power = (revolutions / meter constant) / hours. Using 90 s as 90 h, or forgetting to divide by 1800, are the two ways this stem goes wrong. 45/90 = 0.5 is a revolution rate, not a kilowatt.

A PMMC ammeter is connected in series with a 5 A, 50 Hz AC load. The pointer reads
  1. 5 A, because 5 A is the RMS
  2. Zero, because a PMMC reads the average of AC, which is zero
  3. 7.07 A, the peak of a 5 A RMS sine

PMMC torque follows the instantaneous current; over a cycle the average of a sine is zero, so the pointer sits at zero. RMS is a moving-iron reading. Peak is not what any of these instruments is scaled to show on AC.

5Utilization of electrical energy

Utilization is what the electrical energy is for: light, heat, a weld, or a train. Illumination on a surface is E = I \cos\theta / d^2 lux when I is the candela intensity of the source, d is the distance in metres, and \theta is the angle from the normal. A lumen is the light flux; a lux is that flux per square metre. Electric heating is I^2 R in a resistance, or induction or dielectric heating when the job needs heat inside the work.

Traction wants a large starting torque, which is why the classical locomotive used a DC series motor. Arc welding wants a high current at a low voltage so the arc stays, not a high-voltage low-current supply. Those are different end-uses of the same kilowatt-hour.

Figure. Static geometry of the worked lamp: 800 cd source, 4 m normal distance, 50 lux on the bench. The dashed line is the normal, not a light ray being animated. Inverse-square is a relation in one still frame.

Match the end-use

  1. Light: inverse squareE = I \cos\theta / d^2. Double the mounting height and lux falls to one quarter, not one half. Cosine is 1 when the ray is normal to the plane.
  2. Heat: I²R or inducedResistance ovens and irons are I^2 R. Induction heating puts the current in the work. Dielectric heating puts it in an insulator.
  3. Traction and weldDC series (or a modern drive that copies its torque curve) for trains. High current, low voltage for an arc weld.

Lux under a lamp

A lamp of 800 candela hangs 4 m above a bench. The ray is normal to the bench (\theta = 0). Find the illumination.

  • \cos\theta = \cos 01
  • d^2 = 4^216\,\mathrm{m}^2
  • E = I / d^2 = 800 / 1650 lux

Pro tip. 800/4 = 200 lux is the missing-square trap. If the bench is offset so cos θ = 0.8, E falls to 40 lux — the cosine is a real factor, not decoration. Candela is the source; lux is the bench.

The same 800 cd lamp is raised from 4 m to 8 m, still on the normal. Illumination on the bench
  1. Halves to 25 lux, because height doubled
  2. Falls to one quarter, 12.5 lux, because E goes as 1/d²
  3. Stays 50 lux, because candela did not change

E = I/d². Height 2× means d² is 4×, so 50/4 = 12.5 lux. Halving is the linear-distance trap. Candela is the source intensity; lux on the bench must change with d.

6Basic electronics for electrical JE

Electronics for an electrical JE is the handful of devices on a panel or in a drive. A diode conducts one way: anode more positive than cathode, past about 0.7 V for silicon. A rectifier uses diodes to turn AC into pulsating DC. A half-wave circuit keeps one half-cycle, so V_{dc} = V_m/\pi; a full-wave bridge keeps both, so V_{dc} = 2V_m/\pi. A shunt capacitor afterwards is there to cut the ripple, not to raise the DC by a new formula.

A bipolar transistor is used as a switch (cutoff or saturation) or as an amplifier (active region). An inverting op-amp has closed-loop gain -R_f/R_{in}. A Zener diode is used in reverse breakdown as a voltage reference, not as a rectifier.

Figure. The two bars are the worked averages at Vm = 31.4 V: half-wave 10 V, full-wave 20 V. Full-wave is twice, not √2 times. Static comparison of two numbers — not a waveform being drawn.

Device, then the number

  1. Diode directionForward: anode positive, silicon drop ≈ 0.7 V. Reverse: blocks, except a Zener which is deliberately broken down as a reference.
  2. Rectifier averageHalf-wave V_{dc} = V_m/\pi. Full-wave V_{dc} = 2V_m/\pi. The factor of two is the second half-cycle, not a different Vm.
  3. Gain or switchInverting op-amp A_v = -R_f/R_{in}. Transistor switch: cutoff = off, saturation = on. Active region is the amplifier, not the contactor.

Half-wave and full-wave from one Vm

A sine has V_m = 31.4 V. Take \pi = 3.14. Find V_{dc} for a half-wave rectifier and for a full-wave bridge, each into a resistive load.

  • Half-wave V_{dc} = V_m/\pi = 31.4/3.1410 V
  • Full-wave V_{dc} = 2V_m/\pi = 2 \times 1020 V
  • Half-wave I_{dc} into 100\,\Omega = 10/1000.10 A

Pro tip. Vm/2 = 15.7 V is not a rectifier average. RMS = Vm/√2 = 22.2 V is the AC meter reading, not Vdc. The capacitor filter reduces ripple around the average; it does not replace Vm/π with Vm.

A full-wave bridge and a half-wave rectifier see the same Vm. The full-wave DC average is
  1. The same as half-wave, because Vm did not change
  2. Twice the half-wave average, 2Vm/π versus Vm/π
  3. √2 times the half-wave average, because two diodes conduct

Full-wave keeps both half-cycles, so the average doubles: 2Vm/π. Vm is the same peak; the extra factor is time, not a new peak. √2 is the RMS-to-peak ratio of a sine, not a rectifier identity.

Notes

  • Series shares current, parallel shares voltage: Kirchhoff’s current law balances a node and his voltage law closes a loop. On AC the same element also has reactance, X_{L} = 2\pi f L or X_{C} = 1/(2\pi f C), so the source sees an impedance Z = \sqrt{R^2 + X^2}; power factor is R/Z and never X/Z, and series resonance is the frequency where X_{L} = X_{C} and the current peaks.
  • Four machines, four jobs: A transformer stays electrical on both sides, V_{1}/V_{2} = N_{1}/N_{2} = I_{2}/I_{1}, and makes no shaft torque. A DC machine uses a commutator, and the series motor’s large starting torque is why it was the traction choice. An induction motor’s rotor has no supply leads and must run below N_{s} = 120f/P, because zero slip would mean no rotor current.
  • The power system raises voltage to cut current: Three-phase line current is I = P/(\sqrt{3} V \cos\phi), so raising the transmission voltage lowers the current and the I^2 R loss without changing the power being sent. Protection is a decision plus an interruption - the relay decides that a fault is present and the breaker opens the circuit, while a CT or PT only scales the line quantity down.
  • Which instrument reads what: A permanent-magnet moving-coil meter reads a DC average and reads zero on AC, while a moving-iron meter reads RMS on either. A dynamometer wattmeter reads power, an induction meter integrates that power into kilowatt-hours, and a megger injects a high DC voltage to read insulation resistance rather than continuity.
  • Utilization is the end use: Illumination on a surface follows the inverse square, E = I \cos\theta / d^2 lux, so doubling the mounting height leaves a quarter of the lux. Resistance heating is I^2 R, while induction and dielectric heating put the energy inside the work; arc welding wants a high current at low voltage, and traction wants a large starting torque.
  • Panel electronics: A silicon diode conducts one way past about 0.7 V. A half-wave rectifier averages V_{m}/\pi and a full-wave bridge 2V_{m}/\pi, and a shunt capacitor after it cuts ripple rather than changing that average. A transistor is a switch at cut-off or saturation and an amplifier in the active region, an inverting op-amp has gain -R_{f}/R_{in}, and a Zener works in reverse breakdown as a reference.

Formulas

  • AC single element: X_{L} = 2\pi f L, X_{C} = 1/(2\pi f C), Z = \sqrt{R^2 + X^2}, power factor = R/Z. Series resonance is where X_{L} = X_{C} and current peaks.
  • Transformer: V_{1}/V_{2} = N_{1}/N_{2} = I_{2}/I_{1}, electrical on both sides, no shaft torque.
  • Induction motor synchronous speed N_{s} = 120f/P; slip must be non-zero or there is no rotor current. Slip s = (N_{s} - N)/N_{s}.
  • Three-phase line current I = P/(\sqrt{3} V \cos\phi), so raising transmission voltage lowers current and the I^2 R loss for the same power.
  • Instrument to quantity: PMMC reads a DC average (zero on AC), moving iron reads RMS on either, dynamometer reads power, induction meter integrates kWh, megger reads insulation resistance.
  • Utilization: illumination E = I \cos\theta / d^2 lux, resistance heating = I^2 R, half-wave rectifier average V_{m}/\pi, full-wave bridge 2V_{m}/\pi, inverting op-amp gain -R_{f}/R_{in}.

Exam traps & shortcuts

  • Power factor is R/Z, never X/Z. Sketch the impedance triangle if the option set offers both.
  • A transformer changes voltage and current but produces no shaft torque — it is not a machine that can drive a load.
  • An induction motor cannot reach synchronous speed: zero slip means zero rotor current and zero torque.
  • Raising transmission voltage cuts the current and the I^2 R loss without changing the power delivered.
  • A PMMC meter reads zero on AC. A relay decides; only the breaker interrupts, and a CT or PT merely scales.
  • Doubling the mounting height leaves a quarter of the lux — illumination follows the inverse square, not the inverse.

Reference tables

The six numbers the Electrical paper actually asks. Each row is the identity from one concept, not a new method.

Diploma formula sheet
QuantityIdentityConcept
Impedance / pfZ=\sqrt{R^2+X^2}, \cos\phi=R/ZDC and AC circuits
Synchronous speedN_s=120f/P, N=N_s(1-s)Machines
3-ph line currentI=P/(\sqrt{3} V \cos\phi)Power systems
Energy-meter loadP=(\mathrm{rev}/K)/\mathrm{hours}Measurements
IlluminationE=I\cos\theta/d^2 luxUtilization
Rectifier averageHW V_m/\pi; FW 2V_m/\piElectronics

Classification pairs the paper uses as distractors. Name the object; do not locate a row by position.

Instrument and machine cues
If the stem saysReach for
PMMC on 50 Hz ACReading is zero — average of a sine
Moving iron on ACRMS
Induction motor at Ns on loadImpossible — slip must be non-zero
Traction starting torqueDC series motor (or a drive that copies it)
Zener in a power supplyReverse breakdown as a reference, not a rectifier
Relay versus breakerRelay decides; breaker interrupts

Recap

If you keep only pegs, keep these.

Series / parallel
Series shares current; parallel shares voltage. KCL at the node, KVL around the loop.
Z and pf
Z = √(R² + X²). Power factor is R/Z, never X/Z. Worked coil: 8 Ω, 6 Ω → Z = 10 Ω, pf = 0.8, P = 200 W.
Ns and slip
Ns = 120f/P. 4-pole 50 Hz → 1500 rpm. On-load induction speed is below Ns; fr = sf.
Transformer ratios
V1/V2 = N1/N2 = I2/I1. A transformer does not make shaft torque.
Feeder current
I = P / (√3 V cos φ). Raise V, I falls, I²R falls. 1.5 MW at 11 kV, 0.8 pf → 98.4 A.
PMMC vs MI
PMMC = DC average, zero on AC. Moving iron = RMS. Wattmeter = power. Energy meter = kWh.
Lux and traction
E = I cosθ / d². 800 cd at 4 m → 50 lux. Traction wants series-motor starting torque.
Rectifier and Zener
Half-wave Vm/π; full-wave 2Vm/π. Zener is a reverse reference, not a rectifier.

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