Q. The value of $i_{1}$ in the circuit diagram will be
A) $1A$
B) $\frac{1}{2}A$
C) $\frac{3}{4}A$
D) $\frac{3}{2}A$
Explanation
(a) $ABCD$ forms a balanced Wheatstone bridge. Hence the resistances of arm $BD$ will be ineffective. Now we have resistance $(2 + 2)\ \Omega,\ \ (4 + 4)\Omega$ and $(2 + 2)\ \Omega$ in parallel between $A$ and $B$.
Potential difference across $A$ and $C$ =EMF of battery$= \ 8V$
$\therefore$ Current$\ i_{1} = \frac{8}{(4 + 4)} = 1A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.