Question

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

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

Answer 1

Answer: (B)

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 $R=1.5\Omega$ .
$\therefore$ Current in the circuits , $I=\frac{V}{R}=\frac{1.5}{1.5}=1A$
Since the resistance in arm ACB = resistance in arm $AB=3\Omega$ , the current divided equally in the two arms. Hence the current through the $3\Omega$ resistor = $\frac{I}{2}=0.5A$ .