Let's break down and **explain each part** of the problem step by step. --- ## Given Data - Magnetic field, \( B = .5 \) T - Maximum voltage, \( V_{e(max)} = 250 \) V - Length of conductor, \( l = 1 \) m - Armature resistance, \( R = 2.5 \) Ω - Radius, \( r = .5 \) m - Angular velocity, \( \omega = 500 \) rad/s --- ### a) **Starting the Machine with \( V_B(\text{max}) \)** **Question:** If the maximum instantaneous current the rotor can tolerate is 75 A, can we start the machine with \( V_B(\text{max}) \)? Why? **Explanation:** When starting, the machine is at rest (\(\omega = \)), so the back EMF (\(V_e\)) is zero. - The entire applied voltage (\(V_B\)) appears across the armature resistance. - Starting current: \( I_{start} = \frac{V_B}{R} \) - Using \( V_B = V_{e(max)} = 250 \) V and \( R = 2.5 \) Ω: \[ I_{start} = \frac{250}{2.5} = 100 \text{ A} \] - **Maximum allowed current = 75 A**, but starting current is **100 A**. **Conclusion:** **No, you cannot start the machine with \( V_B(\text{max}) \) because the starting current (100 A) exceeds the maximum allowed (75 A).** --- ### b) **Operating at \( \omega = 500 \) rad/s with \( V_B = V_{e(max)} \)** **Question:** Is the machine working as a motor or generator? Why? **Explanation:** - Back EMF at \(\omega = 500\) rad/s is \( V_{e(max)} = 250 \) V. - Applied voltage \( V_B = 250 \) V. The current in the armature: \[ I = \frac{V_B - V_e}{R} = \frac{250 - 250}{2.5} = \text{ A} \] - **No current flows:** No net force, and the machine neither delivers nor absorbs mechanical energy. **Conclusion:** **The machine is neither acting as a motor nor a generator; it is in a "no-load" condition.** --- ### c) **Operating at \( \omega = 550 \) rad/s** **Find Current and Direction.** - First, calculate new back EMF: \[ V_e = V_{e(max)} \times \frac{\omega}{500} = 250 \times \frac{550}{500} = 275 \text{ V} \] - Current: \[ I = \frac{V_B - V_e}{R} = \frac{250 - 275}{2.5} = \frac{-25}{2.5} = -10 \text{ A} \] - **Negative current**: Means current flows from the machine to the battery (generator action). **Conclusion:** **The machine is acting as a generator; current flows out of the armature to the battery.** --- ### d) **Operating at \( \omega = 450 \) rad/s** **Find Current and Direction.** - New back EMF: \[ V_e = 250 \times \frac{450}{500} = 225 \text{ V} \] - Current: \[ I = \frac{250 - 225}{2.5} = \frac{25}{2.5} = 10 \text{ A} \] - **Positive current**: Current flows from battery to machine (motor action). **Conclusion:** **The machine is acting as a motor; current flows into the armature from the battery.** --- ### e) **At \( \omega = 600 \) rad/s: Charging Current, Power Dissipated in R, Shaft Torque** #### **Step 1: Find Back EMF** \[ V_e = 250 \times \frac{600}{500} = 300 \text{ V} \] #### **Step 2: Charging Current** \[ I = \frac{V_B - V_e}{R} = \frac{250 - 300}{2.5} = \frac{-50}{2.5} = -20 \text{ A} \] - The negative sign means the current is charging the battery (generator mode). #### **Step 3: Power Dissipated in \( R \)** \[ P_R = I^2 R = (20)^2 \times 2.5 = 400 \times 2.5 = 100 \text{ W} \] #### **Step 4: Shaft Torque** - Power delivered by the shaft: \( P_{shaft} = V_e \times I \) - \( P_{shaft} = 300 \times 20 = 600 \) W (use absolute value of \( I \) for power calculation) - Torque: \[ P = \tau \omega \implies \tau = \frac{P}{\omega} = \frac{600}{600} = 10 \text{ Nm} \] --- ## **Summary Table** | Condition | \( V_e \) (V) | Current | Mode | Current Direction | \( P_R \) (W) | Torque (Nm) | |------------------|:------------:|:-------:|:----------|:-------------------|:-------------:|:-----------:| | Start | | 100 A | Motor | Into machine | 25,000 | N/A | | 500 rad/s (b) | 250 | A | None | None | | | | 550 rad/s (c) | 275 | -10 A | Generator | Out of machine | 250 | 5.5 | | 450 rad/s (d) | 225 | 10 A | Motor | Into machine | 250 | 4.5 | | 600 rad/s (e) | 300 | -20 A | Generator | Out of machine | 100 | 10 | --- ## **Key Points** - At rest, starting current is dangerously high—**never start with full voltage**. - If \( V_B = V_e \), no current flows. - If \( V_e > V_B \), the machine acts as a **generator**. - If \( V_B > V_e \), the machine acts as a **motor**. - The direction of current tells you the mode of operation. --- If you need further clarification on any step or formula, let me know!

# Explanation of Permanent Magnet DC Machine ## Given Data - **Magnetic field (B)**: 0.5 T - **Maximum voltage (V_{e(max)})**: 250 V - **Length of conductor (l)**: 1 m - **Armature resistance (R)**: 2.5 Ω - **Radius (r)**: 0.5 m - **Angular velocity (ω)**: 500 rad/s --- ## Part a: Starting the Machine with \( V_B(\text{max}) \) ### Question If the maximum instantaneous current the rotor can tolerate is 75 A, can we start the machine with \( V_B(\text{max}) \)? Why? ### Explanation 1. At start, the machine is at rest (\(\omega = 0\)), so back EMF (\(V_e\)) is zero. 2. The entire applied voltage \(V_B\) appears across the armature resistance. Calculating starting current: \[ I_{start} = \frac{V_B}{R} = \frac{250}{2.5} = 100 \text{ A} \] ### Conclusion **No**, you cannot start the machine with \( V_B(\text{max}) \) because the starting current (100 A) exceeds the maximum allowed (75 A). --- ## Part b: Operating at \( \omega = 500 \) rad/s ### Question Is the machine working as a motor or generator? Why? ### Explanation - Back EMF at \(\omega = 500\) rad/s is \( V_e = 250 \) V. - Applied voltage \( V_B = 250 \) V. Calculating current: \[ I = \frac{V_B - V_e}{R} = \frac{250 - 250}{2.5} = 0 \text{ A} \] ### Conclusion **The machine is in a "no-load" condition; it is neither acting as a motor nor a generator.** --- ## Part c: Operating at \( \omega = 550 \) rad/s ### Calculation of Current and Direction 1. New back EMF: \[ V_e = 250 \times \frac{550}{500} = 275 \text{ V} \] 2. Current: \[ I = \frac{V_B - V_e}{R} = \frac{250 - 275}{2.5} = -10 \text{ A} \] ### Conclusion **The machine is acting as a generator; current flows out of the armature to the battery.** --- ## Part d: Operating at \( \omega = 450 \) rad/s ### Calculation of Current and Direction 1. New back EMF: \[ V_e = 250 \times \frac{450}{500} = 225 \text{ V} \] 2. Current: \[ I = \frac{V_B - V_e}{R} = \frac{250 - 225}{2.5} = 10 \text{ A} \] ### Conclusion **The machine is acting as a motor; current flows into the armature from the battery.** --- ## Part e: At \( \omega = 600 \) rad/s ### Step 1: Find Back EMF \[ V_e = 250 \times \frac{600}{500} = 300 \text{ V} \] ### Step 2: Charging Current \[ I = \frac{V_B - V_e}{R} = \frac{250 - 300}{2.5} = -20 \text{ A} \] - The negative sign indicates the current is charging the battery. ### Step 3: Power Dissipated in \( R \) \[ P_R = I^2 R = (20)^2 \times 2.5 = 400 \times 2.5 = 100 \text{ W} \] ### Step 4: Shaft Torque - Power delivered by the shaft: \[ P_{shaft} = V_e \times I = 300 \times 20 = 600 \text{ W} \] - Torque: \[ \tau = \frac{P}{\omega} = \frac{600}{600} = 10 \text{ Nm} \] --- ## Summary of Findings 1. **Starting Condition:** Cannot start with maximum voltage (exceeds current limit). 2. **Operating at 500 rad/s:** No current (no-load condition). 3. **Operating at 550 rad/s:** Acts as a generator with current flowing out. 4. **Operating at 450 rad/s:** Acts as a motor with current flowing in. 5. **At 600 rad/s:** Charges the battery with a current of -20 A, dissipates 100 W, and produces 10 Nm of torque. --- If you have more questions or need further clarification, feel free to ask!