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?
A.
$$7\,A$$
B.
$$3\,A$$
C.
$$5\,A$$
D.
$$4\,A$$
Answer :
$$7\,A$$
Solution :
Let initial e.m.f. induced $$= e.$$
$$\therefore $$ Initial current $$i = \frac{{E - e}}{R}$$
$${\text{i}}{\text{.e}}{\text{.,}}\,2 = \frac{{12 - e}}{1}$$
This gives $$e = 12 - 2 = 10\,{\text{volt}}{\text{.}}$$ As $$e \propto \omega .$$ when speed is halved, the value of induced e.m.f. becomes
$$\frac{e}{2} = \frac{{10}}{2} = 5\,{\text{volt}}$$
$$\therefore $$ New value of current
$$i' = \frac{{E - e}}{R} = \frac{{12 - 5}}{1} = 7\,A$$
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A.
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