Question

Figure shows a network of resistances connected to a 2 V battery. If the internal resistance of the battery is negligible current I in the circuit is

• A. 0.25 A
• B. 0.5 A
• C. 0.75 A
• D. 1 A

Answer 1

The correct option is B 0.5 A

Because $\frac{P}{Q}=\frac{R}{S}$ the circuit is balanced Wheatstone's bride.Therefore potential at B = potential at D. Hence no current flows through the 5 ohm resistor. It is therefore not effective The equivalent resistance between A and C is the resistance of a parallel combination of P + R = 2 + 4 = 6 ohm and Q + S = 4 + 8 = 12 ohm which is given by
$\frac{1}{R}+\frac{1}{6}+\frac{1}{12}$
R = 4 ohm
Therefore current l is
$l=\frac{2}{4}=0.5$ A