Question

For the network shown in the figure the value of the current i is :-

• A. $\frac{18V}{5}$
• B. $\frac{5V}{9}$
• C. $\frac{9V}{35}$
• D. $\frac{5V}{18}$

Answer 1

The correct option is D $\frac{5V}{18}$

This is a Wheatstone bridge circuit. here four arms resistances , $P=4\Omega ,Q=2\Omega ,R=6\Omega ,S=3\Omega$
since, $\frac{P}{Q}=\frac{R}{S}$ so it is in balanced condition and no current flow the middle resistance $4\Omega$. Thus, it is removed from circuit.
Now equivalent resistance of the circuit is $R_{eq}=(4+2)||(6+3)=6||9=\frac{6\times 9}{6+9}=54/15=18/5\Omega$
Current, $i=V/R_{eq}=5V/18$