SATHEE: Electro Magnetic Induction And Alternating Currents Question 502

Electro Magnetic Induction And Alternating Currents Question 502

Question: A simple electric motor has an armature resistance of $1 Ω$ and runs from a dc source of 12 volt. When running unloaded it draws a current of 2 amp. When a certain load is connected, its speed becomes one-half of its unloaded value. What is the new value of current drawn?

Options:

A) 7 A

B) 3 A

C) 5 A

D) 4 A

Show Answer

Answer:

Correct Answer: A

Solution:

Let initial e.m.f. induced =e
$∴$ Initial current $i = \frac{E - e}{R}$ i.e., $2 = \frac{12 - e}{1}$

This gives $e = 12 - 2 = 10$ volt. As $e \propto ω$ .

when speed is halved, the value of induced e.m.f. becomes $\frac{e}{2} = \frac{10}{2} = 5, v o l t$
$∴$ New value of current $i^{'} = \frac{E - e}{R} = \frac{12 - 5}{1} = 7 A$

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Electro Magnetic Induction And Alternating Currents Question 502

Question: A simple electric motor has an armature resistance of 1Ω and runs from a dc source of 12 volt. When running unloaded it draws a current of 2 amp. When a certain load is connected, its speed becomes one-half of its unloaded value. What is the new value of current drawn?

Options:

Answer:

Solution:

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Question: A simple electric motor has an armature resistance of $1 Ω$ and runs from a dc source of 12 volt. When running unloaded it draws a current of 2 amp. When a certain load is connected, its speed becomes one-half of its unloaded value. What is the new value of current drawn?