The current drawn from the battery in the given network is (Internal resistance of the battery is negligible)

• A

  1.2 A
• B

  4 A
• Correct answer
  C

  2.4 A
• D

  4.8 A

Solution

The given circuit can be drawn as:
From the figure, we can see that this is a balanced Wheatstone bridge. $\begin{array}{ll} \therefore & \frac{1}{R}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5} \\ \therefore & R=5 \Omega \\ \therefore & I=\frac{12}{5}=2.4 \mathrm{~A} \end{array}$

Hence, the correct answer is option: 'C'.