What would be the current drawn from a 16 V supply by the network of resistances shown in fig below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3 amp
The network is made of four identical sections joined end to end, so the sections are in series. Within each section the upper path (two \( 1\,\Omega \) resistors) and the lower path (two \( 2\,\Omega \) resistors) are in parallel between the same two junctions.
Upper path: \( 1 + 1 = 2\,\Omega \)
Lower path: \( 2 + 2 = 4\,\Omega \)
\( \displaystyle R_{\mathrm{section}} = \frac{2\times4}{2+4} = \frac{4}{3}\,\Omega \)
\( \displaystyle R_{\mathrm{total}} = 4\times\frac{4}{3} = \frac{16}{3}\,\Omega \)
\( \displaystyle I = \frac{V}{R_{\mathrm{total}}} = \frac{16}{16/3} = 3\,\mathrm{A} \)