IN STEP 1 GIVE THE INTRODUCTION OF THE CONCEPT AND GIVE ANSWER FOR EACH PART OF THE QUESTION IN EACH DIFFERENT STEP WITH CLEAR EXPLANATION AND IN THE FINAL STEP GIVE THE WHOLE FINAL ANSWER IN JUST VER

Question

IN STEP 1 GIVE THE INTRODUCTION OF THE CONCEPT AND GIVE ANSWER FOR EACH PART OF THE QUESTION IN EACH DIFFERENT STEP WITH CLEAR EXPLANATION AND IN THE FINAL STEP GIVE THE WHOLE FINAL ANSWER IN JUST VERY FEW SENTENCES AND MOREOVER I NEED COMPLETE AND CLEAR ANSWER at last explain what we did in each step in just few sentences8.

A certain separately excited DC motor is rated
as follows: 160VDC, 500W, 1.5A, 1200rpm.
If the armature winding resistance is 5Q and
the field winding resistance is 50Q, please
determine:

(a) The torque constant

(b) The rated torque

(c) The speed at 2 times the rated torque
(d) The speed at 0.8 times the rated torque
(Each part 5%)

VIPSolutions · Accepted Answer

**Step 1: Introduction to the Concept**

A separately excited DC motor uses an external supply for the field winding, making the field flux independent of the armature current. The key equations for such a DC motor are:
- **Back emf, $ E = V - I_a R_a $**
- **Mechanical power output, $ P_{mech} = T \cdot \omega $**
- **Torque, $ T = K_T I_a $**
- **Back emf, $ E = K_E \omega $**
- For SI units, the torque constant $ K_T $ and the back emf constant $ K_E $ are numerically equal.

Given Data:
- Rated Voltage ($ V $): 160 V
- Output Power ($ P $): 500 W
- Rated Current ($ I $): 1.5 A
- Rated Speed ($ N $): 1200 rpm
- Armature Resistance ($ R_a $): 5 Ω
- Field Resistance ($ R_f $): 50 Ω

---

**Step 2: (a) The Torque Constant**

$$
P_{mech} = T \cdot \omega
$$
$$
\omega = 2\pi N/60 = 2\pi \times 1200/60 = 125.66~\text{rad/s}
$$
$$
T = \frac{P_{mech}}{\omega} = \frac{500}{125.66} = 3.98~\text{Nm}
$$

Since $ T = K_T I_a $ (for SI units):

$$
K_T = \frac{T}{I_a} = \frac{3.98}{1.5} = 2.65~\text{Nm/A}
$$

---

**Step 3: (b) The Rated Torque**

Already found above:

$$
T_{rated} = 3.98~\text{Nm}
$$

---

**Step 4: (c) The Speed at 2 Times the Rated Torque**

If torque doubles, armature current also doubles (since $ T = K_T I_a $):
$$
I_a' = 2 \times 1.5 = 3~\text{A}
$$

Back emf at rated conditions:
$$
E = V - I_a R_a = 160 - (1.5 \times 5) = 152.5~\text{V}
$$

At doubled torque:
$$
E' = V - I_a' R_a = 160 - (3 \times 5) = 145~\text{V}
$$

$$
\frac{E'}{E} = \frac{\omega'}{\omega}
\implies \omega' = \omega \times \frac{E'}{E} = 125.66 \times \frac{145}{152.5} = 119.59~\text{rad/s}
$$

Convert to rpm:
$$
N' = \frac{\omega' \times 60}{2\pi} = \frac{119.59 \times 60}{2\pi} = 1143~\text{rpm}
$$

---

**Step 5: (d) The Speed at 0.8 Times the Rated Torque**

$$
I_a'' = 0.8 \times 1.5 = 1.2~\text{A}
$$
$$
E'' = 160 - (1.2 \times 5) = 154~\text{V}
$$
$$
\omega'' = \omega \times \frac{E''}{E} = 125.66 \times \frac{154}{152.5} = 127.08~\text{rad/s}
$$
$$
N'' = \frac{127.08 \times 60}{2\pi} = 1214~\text{rpm}
$$

---

**Final Step: Summary of Answers in Few Sentences**

- **(a)** Torque constant ($ K_T $) is **2.65 Nm/A**.
- **(b)** Rated torque is **3.98 Nm**.
- **(c)** Speed at twice rated torque is **1143 rpm**.
- **(d)** Speed at 0.8 times rated torque is **1214 rpm**.

---

**What We Did in Each Step:**
1. Introduced the concept and relevant DC motor equations.
2. Calculated the torque constant using rated output and current.
3. Found the rated torque from output power and speed.
4. Determined speed at twice the rated torque by considering increased armature current and reduced back emf.
5. Found speed at 0.8 times rated torque by using decreased armature current and increased back emf.
6. Summarized all answers concisely.

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Study Help, Fast Answers

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Related Answers

Study Help, Fast Answers