40. A 120.0-V motor draws a current of 7.00 A when running at normal speed. The resistance of the armature wire is 0.720 . (a) Determine the back emf generated by the motor. (b) What is the current at the instant when the motor is just turned on and has not begun to rotate? (c) What series resistance must be added to limit the starting current to 15.0 A?
40. A 120.0-V motor draws a current of 7.00 A when running at normal speed. The resistance of the armature wire is 0.720 . (a) Determine the back emf generated by the motor. (b) What is the current at the instant when the motor is just turned on and has not begun to rotate? (c) What series resistance must be added to limit the starting current to 15.0 A?
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![### Problem 40 - Electric Motor Analysis
#### Given Data:
- Voltage (V): 120.0 V
- Current (I): 7.00 A (when running at normal speed)
- Resistance of the armature wire (R): 0.720 Ω
#### Questions:
(a) **Determine the back emf generated by the motor.**
(b) **What is the current at the instant when the motor is just turned on and has not begun to rotate?**
(c) **What series resistance must be added to limit the starting current to 15.0 A?**
#### Solutions:
**(a) Back emf (E) Calculation:**
When the motor is running at normal speed:
- The back emf can be expressed using the formula:
\[
E = V - I \cdot R
\]
Where:
- \( V \) is the supply voltage,
- \( I \) is the current,
- \( R \) is the resistance.
Substituting the given values:
\[
E = 120.0 \text{ V} - (7.00 \text{ A} \times 0.720 \text{ Ω})
\]
\[
E = 120.0 \text{ V} - 5.04 \text{ V}
\]
\[
E = 114.96 \text{ V}
\]
**(b) Current when the Motor is just Turned On:**
When the motor is just turned on and has not yet started to rotate, the back emf \( E \) is zero.
- The current \( I_0 \) at this instant can be determined using Ohm's Law:
\[
I_0 = \frac{V}{R}
\]
Substituting the given values:
\[
I_0 = \frac{120.0 \text{ V}}{0.720 \text{ Ω}}
\]
\[
I_0 = 166.67 \text{ A}
\]
**(c) Additional Series Resistance to Limit the Starting Current:**
To limit the starting current to 15.0 A:
- Let \( R_s \) be the additional series resistance required.
\[
I = \frac{V}{R + R_s}
\]
Rearranging](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc02bfde4-a78d-4551-93f6-03d589a5c8dc%2Fe5386761-cd7e-4a32-a6c4-e8a843f3055f%2Fozui3h9_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 40 - Electric Motor Analysis
#### Given Data:
- Voltage (V): 120.0 V
- Current (I): 7.00 A (when running at normal speed)
- Resistance of the armature wire (R): 0.720 Ω
#### Questions:
(a) **Determine the back emf generated by the motor.**
(b) **What is the current at the instant when the motor is just turned on and has not begun to rotate?**
(c) **What series resistance must be added to limit the starting current to 15.0 A?**
#### Solutions:
**(a) Back emf (E) Calculation:**
When the motor is running at normal speed:
- The back emf can be expressed using the formula:
\[
E = V - I \cdot R
\]
Where:
- \( V \) is the supply voltage,
- \( I \) is the current,
- \( R \) is the resistance.
Substituting the given values:
\[
E = 120.0 \text{ V} - (7.00 \text{ A} \times 0.720 \text{ Ω})
\]
\[
E = 120.0 \text{ V} - 5.04 \text{ V}
\]
\[
E = 114.96 \text{ V}
\]
**(b) Current when the Motor is just Turned On:**
When the motor is just turned on and has not yet started to rotate, the back emf \( E \) is zero.
- The current \( I_0 \) at this instant can be determined using Ohm's Law:
\[
I_0 = \frac{V}{R}
\]
Substituting the given values:
\[
I_0 = \frac{120.0 \text{ V}}{0.720 \text{ Ω}}
\]
\[
I_0 = 166.67 \text{ A}
\]
**(c) Additional Series Resistance to Limit the Starting Current:**
To limit the starting current to 15.0 A:
- Let \( R_s \) be the additional series resistance required.
\[
I = \frac{V}{R + R_s}
\]
Rearranging
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