Discussion Forum



1)

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:

Option D

Explanation:

Resistance $10 \Omega$,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 

$I=\frac{15}{200}=7.5 \times 10^{-2} A$