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An automobile starter motor draws a current of 1.7 A from a 17.2 V battery when operating at normal speed. A broken pulley locks the armature in position, and the current increases to 12.3 A. What was the back emf of the motor when operating normally? Answer in units of V.

Respuesta :

Answer:

14.82 Volts

Explanation:

[tex]V_{back}[/tex] = Back emf of the motor

[tex]V_{battery}[/tex] = battery voltage = 17.2 Volts

[tex]i_{locked}[/tex] = Current in locked condition = 12.3 A

R = resistance

In locked condition, using ohm's law

[tex]R = \frac{V_{battery}}{i_{locked}}[/tex]

[tex]R = \frac{17.2}{12.3}[/tex]

R = 1.4 Ω

[tex]i_{normal}[/tex] = Current in normal condition = 1.7 A

Back emf of the motor is given as

[tex]V_{back}[/tex] = [tex]V_{battery}[/tex] - [tex]i_{normal} R[/tex]

[tex]V_{back}[/tex] = 17.2 - (1.7 x 1.4)

[tex]V_{back}[/tex] = 14.82 Volts

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