Single Choice

Four resistance of $$10\Omega$$, $$60\Omega$$, $$100\Omega$$ and $$ 200\Omega$$ respectively taken in order are used to form a Wheatstone's bridge. A $$15 V$$ battery is connected to the ends of a $$200\Omega $$ resistance, the current through it will be

A$$7.5\times { 10 }^{ -5 }A$$
B$$7.5\times { 10 }^{ -4 }A$$
C$$7.5\times { 10 }^{ -3 }A$$
D$$7.5\times { 10 }^{ -2 }A$$
Correct Answer

Solution

Resistances $$10\Omega$$, $$60\Omega$$ and $$100\Omega$$ are in series and they together are in parallel to $$200\Omega$$ resistance. When a potential difference of $$15 V$$ is applied across $$200\Omega$$ then current through it
$$I=\dfrac { 15 }{ 200 } =7.5\times { 10 }^{ -2 }A$$


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