DC Machine Fundamentals
EMF of DC Generator
E = \frac{P \phi N Z}{60 A}
| Symbol | Description | Unit |
|---|---|---|
| E | Generated EMF | V |
| P | Number of poles | |
| \phi | Flux per pole | Wb |
| N | Speed | rpm |
| Z | Total armature conductors | |
| A | Number of parallel paths |
Worked example
4-pole DC generator, lap wound (A=4), φ=0.02 Wb, Z=400, N=1500 rpm. Find EMF.
Given: P=4, φ=0.02 Wb, Z=400, N=1500, A=4
- E = (P × φ × N × Z) / (60 × A)
- E = (4 × 0.02 × 1500 × 400) / (60 × 4)
- Numerator = 4 × 0.02 × 1500 × 400 = 48000
- Denominator = 240
- E = 48000 / 240 = 200 V
Answer: E = 200 V
DC Motor Speed Equation
N = K \frac{V - I_a R_a}{\phi}
| Symbol | Description | Unit |
|---|---|---|
| N | Motor speed | rpm |
| V | Terminal voltage | V |
| I_a | Armature current | A |
| R_a | Armature resistance | Ω |
| \phi | Flux per pole | Wb |
Worked example
DC shunt motor: V=220 V, Ra=0.5 Ω, Ia=30 A, K=1, φ=0.05 Wb. Find speed.
Given: V=220 V, Ra=0.5 Ω, Ia=30 A, φ=0.05 Wb
- Back-EMF Eb = V − Ia×Ra = 220 − 30×0.5 = 220 − 15 = 205 V
- N = Eb/(K×φ) = 205/(1×0.05) = 4100 rpm
Answer: N = 4100 rpm
DC Motor Torque
T = \frac{P \phi Z I_a}{2\pi A}
| Symbol | Description | Unit |
|---|---|---|
| T | Electromagnetic torque | N·m |
| I_a | Armature current | A |
Worked example
4-pole lap wound DC motor: φ=0.02 Wb, Z=400, Ia=50 A, A=4. Find torque.
Given: P=4, φ=0.02 Wb, Z=400, Ia=50 A, A=4
- T = (P × φ × Z × Ia) / (2π × A)
- T = (4 × 0.02 × 400 × 50) / (2π × 4)
- Numerator = 1600
- Denominator = 8π = 25.13
- T = 1600/25.13 = 63.67 N·m
Answer: T = 63.67 N·m
Efficiency of DC Machine
\eta = \frac{P_{out}}{P_{in}} = \frac{P_{in} - P_{losses}}{P_{in}} \times 100\%
| Symbol | Description | Unit |
|---|---|---|
| \eta | Efficiency | % |
| P_{out} | Output power | W |
| P_{in} | Input power | W |
Worked example
DC motor: V=200 V, IL=25 A, total losses=1000 W. Find efficiency.
Given: V=200 V, IL=25 A, P_losses=1000 W
- P_in = V × IL = 200 × 25 = 5000 W
- P_out = P_in − P_losses = 5000 − 1000 = 4000 W
- η = (4000/5000) × 100 = 80%
Answer: η = 80%
Transformer Equivalent Circuit and Testing
Turns Ratio
a = \frac{N_1}{N_2} = \frac{V_1}{V_2} = \frac{I_2}{I_1}
| Symbol | Description | Unit |
|---|---|---|
| a | Turns ratio | |
| N_1, N_2 | Primary and secondary turns | |
| V_1, V_2 | Primary and secondary voltages | V |
Worked example
Transformer: N1=500, N2=100, V1=2200 V. Find V2 and turns ratio.
Given: N1=500, N2=100, V1=2200 V
- a = N1/N2 = 500/100 = 5
- V2 = V1/a = 2200/5 = 440 V
Answer: a = 5, V2 = 440 V
Transformer Efficiency
\eta = \frac{x \cdot S \cdot pf}{x \cdot S \cdot pf + P_{core} + x^2 P_{cu}} \times 100\%
| Symbol | Description | Unit |
|---|---|---|
| x | Fraction of full load | |
| S | Full-load VA rating | VA |
| pf | Power factor of load | |
| P_{core} | Core (iron) loss | W |
| P_{cu} | Full-load copper loss | W |
Worked example
50 kVA transformer: P_core=500 W, P_cu=1000 W. Find η at full load, pf=0.8.
Given: x=1, S=50000 VA, pf=0.8, P_core=500, P_cu=1000
- Output = 1 × 50000 × 0.8 = 40000 W
- Total loss = 500 + 1²×1000 = 1500 W
- η = 40000/(40000+1500) × 100 = 40000/41500 × 100 = 96.39%
Answer: η = 96.39%
Voltage Regulation
VR = \frac{V_{nl} - V_{fl}}{V_{fl}} \times 100\%
| Symbol | Description | Unit |
|---|---|---|
| V_{nl} | No-load secondary voltage | V |
| V_{fl} | Full-load secondary voltage | V |
Worked example
Transformer no-load voltage = 440 V, full-load voltage = 420 V. Find VR.
Given: V_nl=440 V, V_fl=420 V
- VR = (440 − 420) / 420 × 100
- VR = 20/420 × 100 = 4.76%
Answer: VR = 4.76%
Maximum Efficiency Condition
x_{\eta max} = \sqrt{\frac{P_{core}}{P_{cu}}}
| Symbol | Description | Unit |
|---|---|---|
| x_{\eta max} | Load fraction at maximum efficiency | |
| P_{core} | Iron (core) loss | W |
| P_{cu} | Full-load copper loss | W |
Worked example
Transformer: P_core=400 W, P_cu=1600 W. At what fraction of load does η peak?
Given: P_core=400 W, P_cu=1600 W
- x = √(P_core/P_cu) = √(400/1600) = √0.25 = 0.5
Answer: Maximum efficiency at 50% of full load
Three-Phase Induction Motor
Synchronous Speed
N_s = \frac{120 f}{P}
| Symbol | Description | Unit |
|---|---|---|
| N_s | Synchronous speed | rpm |
| f | Supply frequency | Hz |
| P | Number of poles |
Worked example
4-pole induction motor, 50 Hz supply. Find synchronous speed.
Given: P=4, f=50 Hz
- Ns = 120 × 50 / 4 = 6000/4 = 1500 rpm
Answer: Ns = 1500 rpm
Slip
s = \frac{N_s - N_r}{N_s}
| Symbol | Description | Unit |
|---|---|---|
| s | Slip (per unit) | |
| N_s | Synchronous speed | rpm |
| N_r | Rotor speed | rpm |
Worked example
4-pole, 50 Hz induction motor runs at 1440 rpm. Find slip.
Given: Ns=1500 rpm, Nr=1440 rpm
- s = (1500 − 1440)/1500 = 60/1500 = 0.04
Answer: s = 0.04 (4%)
Rotor Copper Loss and Air Gap Power
P_{rotor Cu} = s \cdot P_{ag}, \quad P_{mech} = (1-s) P_{ag}
| Symbol | Description | Unit |
|---|---|---|
| P_{ag} | Air gap power (power transferred to rotor) | W |
| P_{rotor Cu} | Rotor copper loss | W |
| P_{mech} | Mechanical power developed | W |
Worked example
Induction motor: Pag = 20 kW, s = 0.04. Find rotor copper loss and mechanical power.
Given: P_ag=20000 W, s=0.04
- P_rotor_Cu = 0.04 × 20000 = 800 W
- P_mech = (1 − 0.04) × 20000 = 0.96 × 20000 = 19200 W
Answer: P_rotor_Cu = 800 W, P_mech = 19.2 kW
Slip at Maximum Torque
s_{mT} = \frac{R_2}{X_2}
| Symbol | Description | Unit |
|---|---|---|
| s_{mT} | Slip at maximum torque | |
| R_2 | Rotor resistance (referred to stator) | Ω |
| X_2 | Rotor leakage reactance at standstill | Ω |
Worked example
Induction motor: R2=0.3 Ω, X2=1.5 Ω. Find slip at max torque.
Given: R2=0.3 Ω, X2=1.5 Ω
- s_mT = R2/X2 = 0.3/1.5 = 0.2
Answer: s_mT = 0.2 (20%)
Torque Ratio (Running to Starting)
\frac{T}{T_{max}} = \frac{2}{\frac{s}{s_{mT}} + \frac{s_{mT}}{s}}
| Symbol | Description | Unit |
|---|---|---|
| T | Torque at slip s | N·m |
| T_{max} | Maximum torque | N·m |
| s_{mT} | Slip at maximum torque |
Worked example
s_mT = 0.2, operating slip s = 0.04. Find T/T_max.
Given: s=0.04, s_mT=0.2
- T/Tmax = 2 / (s/s_mT + s_mT/s)
- = 2 / (0.04/0.2 + 0.2/0.04)
- = 2 / (0.2 + 5) = 2/5.2 = 0.385
Answer: T/T_max = 0.385
Synchronous Machine
Synchronous Generator EMF
E_f = V_t + I_a(R_a + jX_s)
| Symbol | Description | Unit |
|---|---|---|
| E_f | Excitation voltage (no-load EMF) | V |
| V_t | Terminal voltage | V |
| I_a | Armature current | A |
| X_s | Synchronous reactance | Ω |
Worked example
Sync generator: Vt=11 kV (phase), Ia=200 A at 0.8 pf lag, Ra=0, Xs=10 Ω. Find Ef.
Given: Vt=11000∠0° V, Ia=200∠−36.87° A, Xs=10 Ω
- Ia(jXs) = 200∠−36.87° × 10∠90° = 2000∠53.13° V
- = 2000(0.6+j0.8) = 1200+j1600 V
- Ef = 11000 + 1200 + j1600 = 12200 + j1600
- |Ef| = √(12200² + 1600²) = √(148840000+2560000) = √151400000 = 12305 V
Answer: |Ef| = 12.305 kV
Power Angle (Salient Pole — Simplified)
P = \frac{E_f V_t}{X_s} \sin\delta
| Symbol | Description | Unit |
|---|---|---|
| P | Active power per phase | W |
| \delta | Power angle (torque angle) | ° |
| X_s | Synchronous reactance | Ω |
Worked example
Synchronous generator: Ef=1.3 pu, Vt=1.0 pu, Xs=1.2 pu, δ=30°. Find P.
Given: Ef=1.3, Vt=1.0, Xs=1.2, δ=30°
- P = (1.3 × 1.0 / 1.2) × sin(30°)
- P = (1.3/1.2) × 0.5
- P = 1.0833 × 0.5 = 0.5417 pu
Answer: P = 0.5417 pu
Synchronous Speed (rad/s)
\omega_s = \frac{2\pi N_s}{60}
| Symbol | Description | Unit |
|---|---|---|
| \omega_s | Synchronous speed in rad/s | rad/s |
| N_s | Synchronous speed in rpm | rpm |
Worked example
2-pole synchronous machine at 50 Hz. Find Ns and ωs.
Given: P=2, f=50 Hz
- Ns = 120×50/2 = 3000 rpm
- ωs = 2π×3000/60 = 2π×50 = 314.16 rad/s
Answer: Ns = 3000 rpm, ωs = 314.16 rad/s
DC Motor Speed Control
Speed Ratio (Armature Resistance Control)
\frac{N_2}{N_1} = \frac{V - I_a(R_a + R_{ext})}{V - I_a R_a}
| Symbol | Description | Unit |
|---|---|---|
| N_2 | Speed with external resistance | rpm |
| N_1 | Original speed | rpm |
| R_{ext} | External series resistance | Ω |
Worked example
DC shunt motor: V=220 V, Ra=1 Ω, Ia=20 A, N1=1000 rpm. Rext=4 Ω inserted. Find N2.
Given: V=220, Ra=1, Ia=20, N1=1000, Rext=4
- Eb1 = V − Ia×Ra = 220 − 20×1 = 200 V
- Eb2 = V − Ia×(Ra+Rext) = 220 − 20×(1+4) = 220 − 100 = 120 V
- N2/N1 = Eb2/Eb1 = 120/200 = 0.6
- N2 = 0.6 × 1000 = 600 rpm
Answer: N2 = 600 rpm
Ward-Leonard Speed Ratio
N \propto \frac{V_{gen}}{\phi_{motor}}
| Symbol | Description | Unit |
|---|---|---|
| V_{gen} | Variable generator voltage applied to motor | V |
| \phi_{motor} | Motor field flux (constant in armature control) | Wb |
Worked example
Ward-Leonard drive: at V_gen=200 V motor runs at 900 rpm. Find speed at V_gen=150 V (constant flux).
Given: V1=200 V, N1=900 rpm, V2=150 V
- N ∝ V_gen (constant flux)
- N2/N1 = V2/V1 = 150/200 = 0.75
- N2 = 0.75 × 900 = 675 rpm
Answer: N2 = 675 rpm
Single-Phase Transformer — OC and SC Tests
Core Loss from OC Test
P_{core} = V_0 I_0 \cos\phi_0
| Symbol | Description | Unit |
|---|---|---|
| P_{core} | Core (iron) loss | W |
| V_0 | Applied voltage (OC test) | V |
| I_0 | No-load current | A |
| \phi_0 | No-load power factor angle | ° |
Worked example
OC test: V0=220 V, I0=1.5 A, W0=100 W. Find core loss and no-load PF.
Given: V0=220 V, I0=1.5 A, W0=100 W
- P_core = W0 = 100 W
- cos(φ0) = W0/(V0×I0) = 100/(220×1.5) = 100/330 = 0.303
Answer: P_core = 100 W, pf0 = 0.303 lagging
Equivalent Resistance and Reactance from SC Test
R_{eq} = \frac{P_{sc}}{I_{sc}^2}, \quad Z_{eq} = \frac{V_{sc}}{I_{sc}}, \quad X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2}
| Symbol | Description | Unit |
|---|---|---|
| R_{eq} | Total equivalent resistance referred to HV side | Ω |
| Z_{eq} | Total equivalent impedance | Ω |
| P_{sc} | SC test power (copper loss at full load) | W |
Worked example
SC test on 10 kVA, 2200/220 V transformer (HV side): Vsc=120 V, Isc=4.55 A, Wsc=215 W.
Given: Vsc=120 V, Isc=4.55 A, Wsc=215 W
- Zeq = Vsc/Isc = 120/4.55 = 26.37 Ω
- Req = Wsc/Isc² = 215/(4.55)² = 215/20.7 = 10.39 Ω
- Xeq = √(26.37² − 10.39²) = √(695.4 − 107.9) = √587.5 = 24.24 Ω
Answer: Req=10.39 Ω, Zeq=26.37 Ω, Xeq=24.24 Ω (HV side)
Quick reference
| Formula | Expression |
|---|---|
| DC Generator EMF | E = P\phi NZ/(60A) |
| DC Motor Speed | N = K(V - I_a R_a)/\phi |
| DC Motor Torque | T = P\phi Z I_a/(2\pi A) |
| Turns Ratio | a = N_1/N_2 = V_1/V_2 |
| Transformer Efficiency | \eta = xS\cdot pf/(xS\cdot pf + P_{core}+x^2P_{cu}) |
| Max Efficiency Load | x = \sqrt{P_{core}/P_{cu}} |
| Voltage Regulation | VR = (V_{nl}-V_{fl})/V_{fl} |
| Synchronous Speed | Ns = 120f/P |
| Slip | s = (Ns-Nr)/Ns |
| Rotor Copper Loss | P_{Cu} = s \cdot P_{ag} |
| Slip at Max Torque | s_{mT} = R_2/X_2 |
| Generator Power | P = (E_f V_t/X_s)\sin\delta |
| Core Loss (OC test) | P_{core} = V_0 I_0 \cos\phi_0 |
| SC Test Req | R_{eq} = P_{sc}/I_{sc}^2 |
| Mechanical Power | P_{mech} = (1-s)P_{ag} |
Exam tips
- GATE consistently asks the condition for maximum efficiency in transformers — always equate copper loss to iron loss (x²Pcu = Pcore) and solve for x.
- Induction motor questions on slip at maximum torque require s_mT = R2/X2 referred to the stator side — never use un-referred rotor values directly.
- For synchronous generator phasor diagrams, a lagging power factor means Ef leads Vt in the phasor diagram; drawing this incorrectly loses marks in ESE.
- DC motor torque and speed formulae share the same back-EMF: always derive Eb first from Eb = V − IaRa before computing speed or efficiency.
- The three-test method (no-load, blocked-rotor, stator resistance) for induction motors maps directly onto OC/SC/resistance tests for transformers — the GATE examiner exploits this parallel frequently.
- Voltage regulation sign convention: positive VR means voltage drops from no-load to full-load (lagging load); negative VR (leading load) is a common trick question in university papers.