Thank you for reporting, we will resolve it shortly
Q.
Three resistors $1 \Omega , 2 \Omega,$ and $\,3 \Omega$ are connected to form a triangle. Across $3 \Omega$ resistor a $3 \,V$ battery is connected. The current through $3 \Omega$ resistor is
Required arrangement is shown in figure.
The equivalent circuit will look like
(since the two resistances of $1 \Omega$ and $2 \Omega$ are in series, which form $3 \Omega$ which is in parallel with $3 \Omega$ resistance).
Therefore, the effective resistance is
$\frac{(1+2) \times 3}{(1+2)+3}=\frac{3}{2} \Omega$
$\therefore $ Current in the circuit,
$I=\frac{3}{(3 / 2)}=2\, A$
$\therefore $ Current in $3 \,\Omega$ resistor $=\frac{I}{2}=1 \,A$
