Set 7 — DC Machine Fundamentals and EMF Equation
A dc machine generates a voltage the moment its armature turns in a field, and one equation fixes how much. This set works that equation in both directions — emf from flux, speed and winding on one side, and speed or winding from a required emf on the other — then attaches it to the circuit that surrounds the armature. Compound generators supply the circuit work: long shunt and short shunt, brush contact drops, and a divertor across the series field.
The drill is to identify the connection first, compute the armature current second, and only then sum the drops. The last two problems return to the winding itself and show that lap and wave differ by a single symbol.
The emf equation is the whole of this set. Every conductor cuts \(P\phi\) webers per revolution, the machine turns \(N/60\) revolutions per second, and the \(Z\) conductors are shared between \(A\) parallel paths:
\[ E_a = \frac{\phi P N}{60}\times\frac{Z}{A} \]The winding fixes only \(A\). A lap winding gives \(A = P\), a simplex wave winding gives \(A = 2\). Nothing else in the emf equation changes, so a wave machine generates \(P/2\) times the emf of the same machine lap-connected.
Count the conductors, not the slots. With \(S\) slots and \(n\) conductors per slot, \(Z = Sn\).
A generator's emf is above its terminal voltage by every drop in the armature circuit — armature resistance, series field, and brush contact:
\[ E_a = V_t + I_a\left(R_a + R_{se}\right) + V_{\text{brush}} \]Long shunt or short shunt decides one number: the shunt-field voltage. In a long-shunt machine the shunt field is across the terminals, so \(I_f = V_t/R_{sh}\). In a short-shunt machine the series field is in the load line, so the shunt field sees \(V_t + I_L R_{se}\). In both, \(I_a = I_L + I_f\).
A divertor is a resistance in parallel with the series field, used to weaken the compounding. Replace the pair by \(R_{se}R_d/(R_{se}+R_d)\) and carry on.
The power balance closes the problem. The armature develops \(E_aI_a\); the load, the shunt field and the copper losses share it out:
\[ E_aI_a = V_tI_L + V_tI_f + I_a^2\left(R_a+R_{se}\right) \]
A long-shunt compound generator delivers a load current of 50 A at 500 V. Its armature, series field and shunt field resistances are 0.05 Ω, 0.03 Ω and 250 Ω respectively. Allow 1 V per brush for contact drop. Calculate
- the armature current
- the generated emf
Read the connection before touching a number. In a long-shunt machine the series field is in series with the armature, and that whole branch is bridged by the shunt field. The shunt field therefore sits directly across the output terminals and sees the full 500 V:
The armature supplies both the load and its own field. Current leaving the armature splits at the terminals:
This is the current that flows through the series field as well, because in long shunt the two windings carry the same current.
Total the drops around the armature circuit. Three of them stand between the generated emf and the terminals:
The brush drop is quoted per brush, and the current crosses two brushes — one positive, one negative — so the figure is doubled.
A generator must generate more than it delivers:
A short-shunt compound generator delivers a load current of 30 A at 220 V. Its armature, series field and shunt field resistances are 0.05 Ω, 0.30 Ω and 200 Ω respectively. Allow 1 V per brush for contact drop. Calculate
- the armature current
- the induced emf
In short shunt the series field is in the load line, not in the armature branch, so it carries the load current and nothing else. Its drop can be found immediately:
The shunt field is connected across the armature, which sits on the supply side of the series field. It therefore sees the terminal voltage plus the series drop:
This is the step that distinguishes short shunt from long shunt. Using 220 V here would give 1.100 A and quietly corrupt everything downstream.
Armature current and its drop:
Sum the drops. The emf must cover the terminal voltage, the series field, the brushes and the armature resistance:
A long-shunt compound generator has a terminal voltage of 230 V while delivering 150 A. The shunt field, series field, divertor and armature resistances are 92 Ω, 0.015 Ω, 0.03 Ω and 0.032 Ω respectively. Brush contact drop may be neglected. Determine
- the induced emf
- the total power generated in the armature
- the distribution of that power
Shunt field and armature currents. Long shunt again, so the shunt field is across the 230 V terminals:
Absorb the divertor into the series field. A divertor is a low resistance bolted across the series winding so that only part of the armature current passes through the turns, weakening the compounding without rewinding the machine. Electrically the pair is one resistance:
The divertor here is twice the series-field resistance, so one-third of the armature current is diverted and the effective series mmf falls to two-thirds of its plain value.
Induced emf. The armature current flows through \(R_a\) and through the parallel combination in series with it:
Total power generated is the emf times the current that produces it — not the terminal voltage times the load current:
Where the 36.05 kW goes. Four destinations, and they must add up:
| Destination | Expression | Power |
|---|---|---|
| Delivered to the load | \(V_tI_L = 230\times150\) | 34 500 W |
| Shunt field copper loss | \(V_tI_f = 230\times2.5\) | 575 W |
| Armature copper loss | \(I_a^2R_a = 152.5^2\times0.032\) | 744 W |
| Series field and divertor loss | \(I_a^2(R_{se}\parallel R_d) = 152.5^2\times0.01\) | 233 W |
| Total | \(E_aI_a\) | 36 052 W |
The balance closes to the last watt, which is the only real check available on a problem of this kind.
A 300 kW, 600 V long-shunt compound generator has a shunt field resistance of 75 Ω, an armature resistance including brush resistance of 0.03 Ω, a commutating field winding resistance of 0.011 Ω, a series field resistance of 0.012 Ω and a divertor resistance of 0.036 Ω. With the machine delivering full load, calculate the voltage generated by the armature and the power generated in it.
Full load fixes the terminal current. The rating is the output the machine delivers at rated voltage:
Add the field current. Long shunt, so the 75 Ω shunt winding is across 600 V:
Assemble the armature-circuit resistance. Three windings lie in the path of \(I_a\): the armature itself (brush resistance already included), the commutating field, and the series field shunted by its divertor.
The commutating (interpole) winding always carries the armature current and always belongs in this sum; only the series field can be divided by a divertor.
Generated voltage and generated power:
Check by difference. The armature-circuit copper loss is
and \(317\,703 - 12\,903 = 304\,800\ \text{W} = 600 \times 508\), exactly the power arriving at the terminals. Of that, 4.8 kW feeds the shunt field and 300 kW reaches the load, as the rating requires.
A four-pole generator with a lap-wound armature has 51 slots, each slot containing 20 conductors. What voltage is generated when the machine is driven at 1500 rpm, the flux per pole being 7 mWb?
Convert slots to conductors. The emf equation counts conductors, and the slot count is only a route to them:
A lap winding has as many parallel paths as poles. Each coil group is brought back to a brush under the adjacent pole, so
Hence 255 conductors in series per path, the other three paths simply adding current capacity.
Substitute into the emf equation:
The first factor, 0.7 V, is the emf of a single conductor; the second is the number of them in series between the brushes.
An 8-pole dc generator has 500 armature conductors and a useful flux of 0.05 Wb per pole. Find
- the emf generated when the armature is lap-connected and driven at 1200 rpm
- the speed at which the same machine must be driven, wave-wound, to produce the same emf
Lap winding: eight parallel paths. With \(A = P = 8\) the ratio \(P/A\) is unity, so the emf equation reduces to flux times conductors per path times speed:
Wave winding: two parallel paths, whatever the pole count. Setting \(A = 2\) puts all 250 conductors of each path in series:
Equate to 500 V and solve for the speed:
A quarter of the lap speed, which is exactly the ratio \(A_{\text{wave}}/A_{\text{lap}} = 2/8\) — the emf is inversely proportional to the number of parallel paths at fixed speed.
What is bought and what is paid for. The wave machine reaches 500 V at a quarter of the speed, but each of its two paths must carry the whole armature current in half the number of paths, so its current rating falls in the same proportion. The product — the power — is unchanged, as it must be for the same iron and copper.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Generated emf | \(E_a = \dfrac{\phi P N}{60}\times\dfrac{Z}{A}\) | \(N\) in rpm — Problems 5, 6 |
| Conductors | \(Z = S \times n\) | Slots × conductors per slot — Problem 5 |
| Parallel paths, lap | \(A = P\) | Problems 5, 6 |
| Parallel paths, wave | \(A = 2\) | Any pole count — Problem 6 |
| EMF constant form | \(E_a = k_a\phi\omega_m,\ k_a = \dfrac{ZP}{2\pi A}\) | \(\omega_m = 2\pi N/60\) |
| Generator emf balance | \(E_a = V_t + I_a(R_a+R_{se}) + V_{\text{brush}}\) | Problems 1–4 |
| Brush contact drop | \(V_{\text{brush}} = 2 \times (\text{drop per brush})\) | Two brushes in the path — Problems 1, 2 |
| Long shunt | \(I_f = V_t/R_{sh},\ I_a = I_L + I_f\) | Problems 1, 3, 4 |
| Short shunt | \(I_f = (V_t + I_LR_{se})/R_{sh}\) | Problem 2 |
| Divertor | \(R_{se}\parallel R_d = \dfrac{R_{se}R_d}{R_{se}+R_d}\) | Problems 3, 4 |
| Power developed | \(P_{\text{gen}} = E_aI_a\) | Not \(V_tI_L\) — Problems 3, 4 |
| Armature copper loss | \(I_a^2R_a\) | Problem 3 |
| Shunt field loss | \(V_tI_f\) | Problem 3 |
| Power balance | \(E_aI_a = V_tI_L + V_tI_f + I_a^2(R_a+R_{se})\) | Closes the loss table — Problems 3, 4 |
| Output rating | \(I_L = P_{\text{rated}}/V_t\) | Full-load line current — Problem 4 |
Common Mistakes
Using the load current as the armature current. The armature must also supply the shunt field, so \(I_a = I_L + I_f\) in a generator. Missing the 2 A here costs 0.16 V of emf — small, but it also corrupts every loss in the balance sheet — Problems 1 and 3.
Putting the terminal voltage across a short-shunt field winding. That field is across the armature, which is one series-field drop above the terminals: 229 V, not 220 V — Problem 2.
Counting one brush instead of two. The quoted drop is per brush and the current crosses a positive and a negative brush, so the figure doubles — Problems 1 and 2.
Treating the divertor as being in series with the series field. It is deliberately in parallel, to bypass part of \(I_a\) and weaken the compounding; in series it would strengthen the drop instead — Problems 3 and 4.
Leaving the commutating field out of the armature circuit. Interpole and compensating windings carry \(I_a\) and must be added to \(R_a\) — Problem 4.
Quoting \(V_tI_L\) as the power generated. That is the power delivered. The power generated is \(E_aI_a\), and the difference is the entire loss table — Problems 3 and 4.
Subtracting the drops in a generator. A generator's emf is above its terminal voltage; only a motor's back emf is below it — Problems 1–4.
Using \(A = 2\) for a lap winding. Lap gives \(A = P\); the confusion changes the answer by a factor of \(P/2\) — Problems 5 and 6.
Using the slot count as \(Z\). Each slot holds several conductors, and 51 in place of 1020 is a twentyfold error — Problem 5.
Feeding the speed into the emf equation in rev/s. The 60 in the denominator already converts rpm; using rev/s inflates the emf sixtyfold — Problems 5 and 6.
Every problem on this page rested on two things: the emf equation, and a correct reading of how the field windings are connected. The emf equation itself never varied — the same six symbols served a 178 V laboratory machine and a 300 kW industrial generator — and the only quantity the winding designer controls in it is the number of parallel paths.
That quantity has so far been handed over as a rule: \(A = P\) for lap, \(A = 2\) for wave. It deserves better than assertion. Where the paths come from, why a wave winding closes on itself after \(P/2\) coils, how the coil pitch and commutator pitch are chosen, and what equalisers are for, all follow from the geometry of the winding rather than from any electrical principle.
Next: Set 8 — Armature Windings, Lap and Wave, where the parallel paths are counted from the winding table instead of quoted, and the pitches that make a winding closed and symmetrical are worked out.