The current drawn from the battery in the given network is (Internal resistance of the battery is negligible)
- 1.2 A
- 4 A
- 2.4 A
- 4.8 A
Solution
From the figure, we can see that this is a balanced Wheatstone bridge. $\begin{array}{ll} \therefore & \frac{1}{R}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5} \\ \therefore & R=5 \Omega \\ \therefore & I=\frac{12}{5}=2.4 \mathrm{~A} \end{array}$
Asked in: MHT CET 2024 (11 May Shift 1)