Question

Figure shows a network of eight resistors, each equal to $2\, \Omega,$ connected to a $3\, V$ 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, $B E$ and $C F$ are ineffective.
$\therefore A B, B C, C D$ are in series. So, their equivalent resistance is $R_{1}=6\, \Omega$
$A E, E F, F D$ 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, $I=\frac{3\, V }{3\, \Omega}=1.0\, A$