The current drawn from the battery in the give network is (Internal resistance of the battery is negligible)
- 2.4 A
- 1.6 A
- 2.0 A
- 3.0 A
Solution
It is a balanced Wheatstone bridge. No current will flow through $5 \Omega$ resistance, and hence it can be removed from the circuit. $3 \Omega$ and $2 \Omega$ resistance are in series. Hence we have two branches with $5 \Omega$ resistances each, connected in parallel. Their equivalent resistance $2.5 \Omega$
$\therefore \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6}{2.5}=2.4 \mathrm{~A}$Asked in: MHT CET 2021 (24 Sep Shift 2)