Q. For the network shown in the figure the value of the current $i$ is
A) $\frac{9V}{35}$
B) $\frac{5V}{18}$
C) $\frac{5V}{9}$
D) $\frac{18V}{5}$
Explanation
(b) The given network is a balanced Wheatstone bridge. It’s equivalent resistance will be $R = \frac{18}{5}\Omega$
So current from the battery $i = \frac{V}{R} = \frac{V}{18/5} = \frac{5V}{18}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.