The current drawn from the battery in the given network is (Internal resistance of battery is neglected)
The current drawn from the battery in the given network is (Internal resistance of battery is neglected)
- $2 \cdot 4 \mathrm{~A}$
- $0.6 \mathrm{~A}$
- $3.6 \mathrm{~A}$
- $1 \cdot 2 \mathrm{~A}$
Solution
Bridge is in balanced condition
$\therefore \frac{1}{R}=\frac{1}{5}+\frac{1}{5} \quad \therefore R=\frac{5}{2} \Omega$
$\therefore \mathrm{i}=\frac{6 \times 2}{5}=\frac{12}{5}=2.4 \mathrm{~A}$
Asked in: MHT CET 2020 (20 Oct Shift 2)
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