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The current drawn by a 120 V dc motor with back emf of 110 V and armature resistance of 0.4 ohm is

The current drawn by a 120 V dc motor with back emf of 110 V and armature resistance of 0.4 ohm is

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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**Question 3:** The current drawn by a 120 V DC motor with a back EMF of 110 V and an armature resistance of 0.4 ohm is **Explanation:** To find the current drawn by the DC motor, we can use Ohm's Law and the formula for back electromotive force (EMF). Ohm's Law is given by: \[ I = \frac{V}{R} \] Where: - \( I \) is the current, - \( V \) is the voltage, - \( R \) is the resistance. The voltage across the armature is the difference between the supply voltage and the back EMF: \[ V = V_{\text{supply}} - V_{\text{back EMF}} \] Given: - Supply voltage (\( V_{\text{dc}} \)) = 120 V, - Back EMF (\( V_{\text{back EMF}} \)) = 110 V, - Armature resistance (\( R_{\text{arm}} \)) = 0.4 ohms. First, calculate the voltage across the armature: \[ V_{\text{armature}} = 120\text{ V} - 110\text{ V} = 10\text{ V} \] Now, use Ohm's Law to find the current: \[ I = \frac{V_{\text{armature}}}{R_{\text{arm}}} = \frac{10\text{ V}}{0.4\text{ ohms}} = 25\text{ A} \] Therefore, the current drawn by the motor is **25 A**. --- This text can be placed on an educational website to explain how to calculate the current drawn by a DC motor given certain parameters.
Transcribed Image Text:**Question 3:** The current drawn by a 120 V DC motor with a back EMF of 110 V and an armature resistance of 0.4 ohm is **Explanation:** To find the current drawn by the DC motor, we can use Ohm's Law and the formula for back electromotive force (EMF). Ohm's Law is given by: \[ I = \frac{V}{R} \] Where: - \( I \) is the current, - \( V \) is the voltage, - \( R \) is the resistance. The voltage across the armature is the difference between the supply voltage and the back EMF: \[ V = V_{\text{supply}} - V_{\text{back EMF}} \] Given: - Supply voltage (\( V_{\text{dc}} \)) = 120 V, - Back EMF (\( V_{\text{back EMF}} \)) = 110 V, - Armature resistance (\( R_{\text{arm}} \)) = 0.4 ohms. First, calculate the voltage across the armature: \[ V_{\text{armature}} = 120\text{ V} - 110\text{ V} = 10\text{ V} \] Now, use Ohm's Law to find the current: \[ I = \frac{V_{\text{armature}}}{R_{\text{arm}}} = \frac{10\text{ V}}{0.4\text{ ohms}} = 25\text{ A} \] Therefore, the current drawn by the motor is **25 A**. --- This text can be placed on an educational website to explain how to calculate the current drawn by a DC motor given certain parameters.
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