Let's break down the problem: ### Given: - **4-pole** DC machine - **2-layer lap winding** - **64 coils** - **Each coil has 5 turns** Let's answer each part step by step: --- ## a) How many conductors are connected in series within each parallel path? ### Step 1: Calculate the total number of armature conductors (Z) Each coil has 5 turns, and each turn is made of 2 conductors (one going and one returning). So, **each coil = 5 × 2 = 10 conductors**. - Total number of coils = **64** - Total conductors (**Z**) = **64 coils × 10 conductors/coil = 640 conductors** --- ### Step 2: Number of parallel paths (A) in lap winding For a lap winding, Number of parallel paths (**A**) = Number of poles (**P**) So, **A = 4** --- ### Step 3: Conductors in series per parallel path Number of conductors in series per parallel path = \[ \text{Conductors per path} = \frac{Z}{A} = \frac{640}{4} = 160 \] --- ## **Answer to (a):** \[ \boxed{160} \] conductors are connected in series within each parallel path. --- ## b) At what rotational speed does each conductor generate 1 V of EMF? What is the total EMF observed at the external terminals? Let's proceed step by step. --- ### Step 1: EMF generated by one conductor The EMF generated per conductor per revolution is: \[ \text{EMF per conductor} = \frac{P \cdot \Phi \cdot N}{60} \] But actually, the full armature EMF equation is: \[ E = \frac{P \cdot \Phi \cdot Z \cdot N}{60 \cdot A} \] Where: - \( E \) = total EMF (V) - \( P \) = number of poles = 4 - \( \Phi \) = flux per pole (in Weber) - \( Z \) = total number of conductors = 640 - \( N \) = speed in rpm - \( A \) = number of parallel paths = 4 But the EMF generated by a single conductor in one revolution is: \[ e_{cond} = \text{Flux cut per second} = \frac{P \cdot \Phi \cdot N}{60} \] But per turn, it's \(\Phi \cdot N\) per second. However, **per conductor**, the EMF generated is proportional to the rate at which it cuts the flux. #### Alternatively, the EMF per conductor per revolution is: Each conductor cuts the total flux per revolution: - In one revolution, a conductor under each pole cuts a flux \(\Phi\), so in one revolution (for a P pole machine), a conductor cuts \(P \cdot \Phi\). - Number of revolutions per second: \( n = \frac{N}{60} \) So, EMF generated per conductor: \[ e_{cond} = P \cdot \Phi \cdot n = P \cdot \Phi \cdot \frac{N}{60} \] Set \( e_{cond} = 1 \) V: \[ 1 = P \cdot \Phi \cdot \frac{N}{60} \] \[ N = \frac{60}{P \cdot \Phi} \] But we do **not** know \(\Phi\). **Normally, with only the number of conductors and not the flux, we cannot numerically determine the speed.** But, if you want the speed at which each conductor generates 1V of EMF, the answer is: \[ N = \frac{60}{P \cdot \Phi} \] --- ### Step 2: Find total EMF at terminals The total EMF of the machine is: \[ E = \frac{P \cdot \Phi \cdot Z \cdot N}{60 \cdot A} \] But since each conductor generates 1V, and there are 160 conductors in series per path (from part a), and 4 parallel paths, then: - **EMF per path = 1 V × 160 = 160 V** - Since paths are in parallel, total EMF at terminals = EMF per path = **160 V** --- ## **Final Answers:** ### a) **Number of conductors in series within each parallel path =** \[ \boxed{160} \] --- ### b) **(i) Rotational speed at which each conductor generates 1 V of EMF:** \[ N = \frac{60}{P \cdot \Phi} \] (where \(\Phi\) is the flux per pole in Weber) **(ii) Total EMF observed at the external terminals:** \[ \boxed{160\ \text{V}} \] --- **Summary Table:** | Part | Answer | |------|--------| | a) | 160 conductors in series per path | | b) | (i) \(N = \dfrac{60}{4\Phi}\) rpm, (ii) 160 V at the terminals |