JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Current Electricity

Reducing the resistor network in the circuit gives a net resistance

$R_{net}=6\Omega$

With a 6 V source, the main-line (ammeter) current is

$I=\frac{V}{R_{net}}=\frac{6}{6}=1\text{ A}$

This 1 A splits equally between the two parallel $4\Omega$ branches, giving $0.5\text{ A}$ in each.

Potential across AB: $V_{AB}=0.5\times 4=2\text{ V}$

Potential across CD: $V_{CD}=1\times 4=4\text{ V}$

So the net resistance is $6\Omega$, the ammeter reads 1 A, and $V_{CD}=4\text{ V}$, while $V_{AB}$ is only 2 V. Statements A, B and D hold.