Question

Resistances 1 $\Omega$,2 $\Omega$ and 3 $\Omega$ are connected to form a triangle.If a 1.5 V cell of negligible internal resistance is connected across the 3 $\Omega$ resistor, the current flowing through this resistor will be

• A. 0.25 A
• B. 0.5 A
• C. 1.0 A
• D. 1.5 A

Answer 1

Answer: (B)

0.5 A

Answer 2

Equivalent resistance between A and B = series combination of 1 $\Omega$ and 2 $\Omega$ in parallel with 3 $\Omega$resistor..
$\frac{1}{R}=\frac{1}{3}+\frac{1}{3}=\frac{2}{3} \, or \, R=1.5 \Omega.$
$\therefore\quad$ Current in the circuit is I = V/R = 1.5/1.5 =1A.
Since the resistance in arm ACB = resistance in arm AB = 3 $\Omega\quad$, the current divides equally in the two arms. Hence the current through the 3 $\Omega\quad$ resistor =I/2 = 0.5 A.