--- title: "Commutator function and DC generator emf equation" subject: "Electrical Machines I" chapter: 3 key: "semester_4/electrical_machines_i:commutator-function-emf-equation" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/commutator-function-emf-equation site: BCT Engineering --- # Commutator function and DC generator emf equation Electrical Machines I · Chapter 3: DC Generator - Asked in: 2073 Magh, 4+4 marks - Video: [EMF Equation of DC Generator - DC Machines - Basic Electrical Engineering](https://www.youtube.com/watch?v=S8VIewhwqZs) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/commutator-function-emf-equation - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-3 ## Question Describe the function of commutator segments in DC generator. Derive the expression of voltage generated in DC generator. ## Short answer Commutator segments act as a mechanical rectifier, converting alternating conductor emf to unidirectional DC, collecting current via brushes, and connecting coils in parallel paths. Generated emf is E_g = P φ Z N / (60 A), where A is number of parallel paths; for lap A=P, for wave A=2. ## Answer #### Function of commutator segments The **commutator** is a cylinder made of wedge-shaped, hard-drawn copper segments insulated from each other by mica, mounted on the shaft. Each segment is connected to the end of an armature coil, and carbon brushes press on its surface. The emf induced in each armature conductor is **alternating**, because a conductor passes under N and S poles in turn. The commutator, together with the brushes, works as a **mechanical rectifier**: 1. **Rectification:** as a coil side moves from under one pole to the next, the coil connection to the brush is reversed at the same instant the emf reverses. So the external circuit always gets current in one direction (unidirectional DC). 2. **Collecting current:** it provides a sliding contact so current can be taken from the rotating armature to the stationary external circuit through the brushes. 3. **Connecting coils in series/parallel paths:** the segments join the coils so that their emfs add up between brushes (lap or wave paths). 4. **Commutation:** it reverses the current in a coil while the coil is short-circuited by the brush, keeping the output steady with many segments (low ripple). ```text Coil emf (AC) After commutator (DC) /\ /\ /\ /\ /\ / \ / \ / \/ \/ \ / \/ \ / \ (reverses) (always positive) ``` #### EMF equation of a DC generator Let - $P$ = number of poles, $\phi$ = flux per pole (Wb) - $Z$ = total number of armature conductors - $A$ = number of parallel paths ($A = P$ for lap, $A = 2$ for wave) - $N$ = speed (rpm) By Faraday's law, average emf in one conductor = flux cut per revolution ÷ time for one revolution. Flux cut by one conductor in one revolution: $$ d\phi = P\phi\ \text{Wb} $$ Time for one revolution: $$ dt = \frac{60}{N}\ \text{s} $$ So emf per conductor: $$ e = \frac{d\phi}{dt} = \frac{P\phi N}{60}\ \text{V} $$ The $Z$ conductors are divided into $A$ parallel paths, so each path has $Z/A$ conductors in series. The generated emf equals the emf of one path: $$ \begin{aligned} E_g &= \frac{P\phi N}{60} \times \frac{Z}{A} \\ &= \frac{P \phi Z N}{60 A}\ \text{V} \end{aligned} $$ - Lap winding ($A = P$): $E_g = \dfrac{\phi Z N}{60}$ - Wave winding ($A = 2$): $E_g = \dfrac{P \phi Z N}{120}$ For a given machine $P$, $Z$, $A$ are fixed, so $E_g = k \phi N$ (or $E_g = k\phi\omega$), i.e. emf is proportional to flux and speed. **Example:** $P = 4$, lap, $Z = 400$, $\phi = 0.02$ Wb, $N = 1500$ rpm gives $E_g = 0.02 \times 400 \times 1500 / 60 = 200$ V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the commutator called a mechanical rectifier? A. It converts AC to DC electronically using diodes B. It reverses the coil connection to the brushes when the coil emf reverses, giving unidirectional current C. It increases the voltage D. It removes ripple by filtering Answer: B. Conductor emf is alternating; reversing the connection at the same instant makes the external current unidirectional. (Key point: Commutator as rectifier) 2. In the derivation, what flux does one conductor cut per revolution? A. phi B. P phi C. phi/P D. 60 phi/N Answer: B. One conductor passes under P poles, each with flux phi, so it cuts P phi per revolution. (Key point: Flux cut per revolution) 3. What is the average emf of one conductor? A. P phi N/60 B. P phi Z N/60 C. phi N/60A D. P phi/N Answer: A. e = P phi/(60/N) = P phi N/60. (Key point: Emf per conductor) 4. Why is Z/A used in the final formula? A. Conductors are split into A parallel paths; the emf equals that of one path of Z/A conductors in series B. Z conductors are all in parallel C. A is the area of the pole D. Only half the conductors are active Answer: A. Terminal emf equals one path's emf, i.e. Z/A conductors times e. 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