Question

Five equal resistaces each of resistance $r$ are connected as shown in figure. A battery of $V$ volt is connected between A and B. The current flowing in AFCEB will be

• A. $V/2r$
• B. $V/r$
• C. $3V/r$
• D. $2V/r$

Answer 1

The correct option is A $V/2r$

A balanced wheatstone's bridge exist between A and B.
So resistance CD is no use because no current will flow.
Now as branch FD-DE and FC-CE are in series so these will become 2r and 2r.
As both of these(FDE-FCE) are in parallel, then the equivalent resistance will be $\frac{1}{R_{eq}}=\frac{1}{2r}+\frac{1}{2r}$
$R_{eq}=r$
current throuth circuit$V=IR$
$I=V/R_{eq}=V/r$
current through AFCEB=$V/2r$