5 hours · 5 marks
A DC machine (motor or generator) has the following main parts:
| Part | Function / notes |
|---|---|
| Yoke (frame) | Outer cast-iron/steel casing; mechanical support; carries magnetic flux of the field system. |
| Poles & pole shoes | Field poles bolted to yoke; laminated shoes spread flux and support field winding. |
| Field winding | Copper coils on poles; produce main magnetic field when excited. |
| Armature core | Laminated cylindrical iron; slots hold armature conductors; rotates in magnetic field. |
| Armature winding | Lap or wave winding in slots; where EMF is induced / torque produced. |
| Commutator | Cylinder of copper segments insulated by mica; mechanical rectifier/inverter — converts AC in armature to DC at brushes (generator) or DC supply to alternating currents in coils (motor). |
| Brushes | Carbon/graphite; press on commutator; connect rotating armature circuit to external stationary circuit. |
| Shaft, bearings, interpoles | Mechanical support; interpoles (compoles) improve commutation under load. |
The armature winding carries alternating current relative to the rotating conductors. The commutator + brushes form a switching arrangement that:
Good commutation requires brush at magnetic neutral axis (or slightly shifted / with interpoles). Sparking indicates poor commutation (high reactance voltage, worn brushes, wrong brush position).
Principle: Faraday’s law — when armature conductors cut magnetic flux (driven by prime mover), an EMF is induced. Commutator delivers DC at the brushes.
Average EMF per conductor $= B\ell v$. For $Z$ conductors, $P$ poles, $A$ parallel paths, flux/pole $\Phi$, speed $N$ rpm:
$$E_g = \frac{\Phi Z N}{60}\times\frac{P}{A}$$Also $E_g = V + I_a R_a$ (+ brush drop if considered) for a separately/shunt generator supplying terminal voltage $V$.
Principle: When a current-carrying conductor lies in a magnetic field, it experiences a force $F=B I \ell$ (Fleming’s left-hand rule). Forces on armature conductors produce a torque about the shaft.
Under a N-pole, conductors carrying current into the page experience force in one tangential direction; under S-pole, reverse current (via commutator) experiences force in the same rotational sense → continuous unidirectional torque.
Gross (developed) torque: electrical power converted $= E_b I_a = T\omega$, with $\omega=2\pi N/60$:
$$T = \frac{E_b I_a}{\omega} = 0.159\,\Phi Z I_a\frac{P}{A}\quad(\mathrm{N\cdot m})$$In proportional form: $T \propto \Phi I_a$. For constant field (shunt), $T \propto I_a$.
| Type | Field connection | Notes |
|---|---|---|
| Separately excited | Field from independent DC source | Flexible control; lab / Ward–Leonard. |
| Shunt | Field parallel with armature | Nearly constant speed (motor); flat voltage (generator). |
| Series | Field in series with armature | High starting torque; never run without load (motor). |
| Compound (long/short shunt) | Both series + shunt fields | Cumulative: good regulation / torque; differential: special uses. |
When a DC motor armature rotates in the field, an EMF is induced that opposes the applied voltage — the back EMF $E_b$:
$$E_b = \frac{\Phi Z N}{60}\frac{P}{A} = V - I_a R_a$$Roles of back EMF:
At start $E_b=0$, so a variable resistance (starter) is inserted in series with the armature to limit $I_{\mathrm{st}}$ to a safe value (≈1.5–2.5× FL).
Three-point starter (shunt motor) terminals: L (line), A (armature), F (field).
“Three-point” = L, A, F connection points. (Four-point starter puts NVR across supply for better field-control compatibility.)
| Method | How | Effect |
|---|---|---|
| Armature voltage / rheostat | Series resistance with armature, or variable $V_a$ | Speed below rated (rheostat); Ward–Leonard / chopper for wide range. Rheostat: inefficient ($I_a^2 R$ loss). |
| Field control | Rheostat in shunt-field circuit; weaken $\Phi$ | Speed above rated; limited by commutation and mechanical strength. |
Iron + friction + windage often grouped as constant losses.
$$\eta_{\mathrm{commercial}} = \frac{\text{output}}{\text{input}} = \frac{\text{output}}{\text{output}+\text{total losses}}$$For generator: output $= V I_L$; input = mechanical power from prime mover ≈ $E_g I_a$ (gross electrical generated).
Q. A shunt generator delivers 96 A at 200 V. Armature resistance $R_a=0.1\,\Omega$, shunt-field resistance $R_{\mathrm{sh}}=50\,\Omega$. Iron and friction losses are 2500 W. Find the generated EMF and commercial efficiency.
Shunt current:
$$I_{\mathrm{sh}} = \frac{V}{R_{\mathrm{sh}}} = \frac{200}{50} = 4\,\mathrm{A}$$Armature current (generator: armature supplies load + shunt):
$$I_a = I_L + I_{\mathrm{sh}} = 96 + 4 = 100\,\mathrm{A}$$Generated EMF:
$$E_g = V + I_a R_a = 200 + 100\times 0.1 = \mathbf{210\,\mathrm{V}}$$Output $= V I_L = 200\times 96 = 19200\,\mathrm{W}$
Copper losses:
$$I_a^2 R_a + I_{\mathrm{sh}}^2 R_{\mathrm{sh}} = 100^2\times 0.1 + 4^2\times 50 = 1000 + 800 = 1800\,\mathrm{W}$$Iron + friction $= 2500\,\mathrm{W}$
Total losses $= 1800 + 2500 = 4300\,\mathrm{W}$
Input $= \text{output} + \text{losses} = 19200 + 4300 = 23500\,\mathrm{W}$
$$\eta_{\mathrm{comm}} = \frac{19200}{23500}\times 100\% = \mathbf{81.70\%}$$(Check: generated power $E_g I_a = 210\times 100 = 21000\,\mathrm{W}$; after Cu losses in armature+field path accounting, commercial $\eta$ uses terminal output / mechanical input.)
Q. A 240 V shunt motor has $R_a=0.4\,\Omega$ and $R_f=120\,\Omega$. It draws 30 A at half load and runs at 1400 rpm. A resistance of $1.2\,\Omega$ is connected in series with the armature and the torque is reduced by 20%. Find the new speed. (Field unchanged.)
Field current: $I_f = V/R_f = 240/120 = 2\,\mathrm{A}$
Half-load armature current: $I_{a1} = 30 - 2 = 28\,\mathrm{A}$
$$E_{b1} = V - I_{a1} R_a = 240 - 28\times 0.4 = 240 - 11.2 = 228.8\,\mathrm{V}$$Since $T\propto\Phi I_a$ and $\Phi$ constant (shunt field unchanged):
$$T_2 = 0.8\,T_1 \;\Rightarrow\; I_{a2} = 0.8\times 28 = 22.4\,\mathrm{A}$$With $1.2\,\Omega$ in series with armature:
$$E_{b2} = V - I_{a2}(R_a+1.2) = 240 - 22.4\times(0.4+1.2) = 240 - 22.4\times 1.6 = 240 - 35.84 = 204.16\,\mathrm{V}$$$N\propto E_b/\Phi$; $\Phi$ constant:
$$N_2 = N_1\times\frac{E_{b2}}{E_{b1}} = 1400\times\frac{204.16}{228.8} = \mathbf{1249\,\mathrm{rpm}}\text{ (approx.)}$$Commutator: segmented copper cylinder; converts generated AC in coils to DC at brushes (generator), and ensures torque remains unidirectional (motor) by reversing coil current under brush.
Brushes: carbon blocks sliding on commutator; connect external circuit; placed near GNA for sparkless commutation.
Torque: $F=BI\ell$ on each conductor; commutator maintains correct current direction under N and S poles so all forces contribute to the same rotation; $T=0.159\,\Phi Z I_a (P/A)$.