Question

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 $15V$ 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. $7.5\times {10}^{-2}A$

Answer 1

Answer: (D)

$7.5\times {10}^{-2}A$

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 $15V$ is applied across $200\Omega$ then current through it
$I=\frac{15}{200}=7.5\times {10}^{-2}A$