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

Answer 1

Answer: (B)

Equivalent resistance between $A$ and $B$ $=$ series combination of $1\text{\Omega }$ and $2 \text{\Omega }$ in parallel with $3 \text{\Omega }$ resistor.

So net resistance across the battery is given by
$\frac{1}{R} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$
or $R = 1.5 \text{\Omega }$ .

$\therefore $ total current supplied by the battery,
$I=\frac{V}{R}=\frac{1 . 5 }{1 . 5}=1A$
Since, the resistance in arm $ACB$ $=$ resistance in arm $AB=3\Omega$ , the current is divided equally in the two arms.
Hence, the current through the $3 \text{\Omega }$ resistor = $\frac{I}{2}=0.5A$ .