Question

Figure shows a network of eight resistors, each equal to $2\Omega ,$ connected to a $3V$ battery of negligible internal resistance. The current $I$ in the circuit is

• A. 0.25 A
• B. 0.50 A
• C. 0.75 A
• D. 1.0 A

Answer 1

Answer: (D)

Because of symmetry, $BE$ and $CF$ are ineffective.
$\therefore AB,BC,CD$ are in series. So, their equivalent resistance is $R_1=6\Omega$
$AE,EF,FD$ are in series. So, their equivalent resistance is $R_2=6\Omega$
$R_1$ and $R_2$ are in parallel. So the equivalent resistance of the circuit is
$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{6\Omega }+\frac{1}{6\Omega }$ or $R=3\Omega$
$\therefore$ Current, $I=\frac{3V}{3\Omega }=1.0A$