Electro Magnetic Induction And Alternating Currents Question 502
Question: A simple electric motor has an armature resistance of $ 1\Omega ~ $ 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:
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Let initial e.m.f. induced =e
$ \therefore $ Initial current $ i=\frac{E-e}{R} $ i.e., $ 2=\frac{12-e}{1} $This gives $ e=12-2\text{=}10 $ volt. As $ e\propto \omega $ .
when speed is halved, the value of induced e.m.f. becomes $ \frac{e}{2}=\frac{10}{2}=5,volt $
$ \therefore $ New value of current $ i’=\frac{E-e}{R}=\frac{12-5}{1}=7A $
