JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Current Electricity

Each pair of equal resistors in parallel reduces as $4\Vert 4=2\Omega$, $15\Vert 10=6\Omega$, $8\Vert 8=4\Omega$ and $12\Vert 12=6\Omega$. The upper arm is $2+2+6=10\Omega$ and the lower arm is $4+6=10\Omega$; in parallel these give $5\Omega$. Adding the series $1\Omega$ makes the total $6\Omega$, so the source current is $I=12/6=2$ A, splitting equally into $1$ A in each $10\Omega$ arm. The $15\Vert 10$ section of the upper arm carries this $1$ A, so the voltage across it, and hence across the $15\Omega$ resistor, is