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

Answer 1

Answer: (D)

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

Answer 2

Here, the resistance for $10\, \Omega,\, 60\, \Omega$ 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 is
$1=\frac{15}{200}=7.5 \times 10^{-2} A$