What would be the current drawn from a 16 V supply by the network of resistances shown in fig below:

3 amp

The correct answer is Option (2) → 3 amp

as,

$\sum\limits_{i=1}^4=\frac{4}{3}$ [R_eq of wheatstone bridge]

Now,

$R_{eq}=R_1+R_2+R_3+R_4$ [In series]

$=4R_1=\frac{4×4}{3}=\frac{16}{3}Ω$

$V=16V$ [given]

$∴I=\frac{V}{R}$ [By ohm's law]

$I=\frac{16×3}{16}=3$

$I=3A$