Find the total current drawn from the battery as shown in the circuit diagram :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.33 A
The network is symmetric: the top row and the bottom row each have three \( 10\,\Omega \) resistors in series, and the two \( 50\,\Omega \) resistors join matching points of the two rows. By symmetry these matching points are at the same potential, so no current flows through the \( 50\,\Omega \) resistors and they can be removed.
Each row is then \( 10 + 10 + 10 = 30\,\Omega \), and the two rows are in parallel:
\( \displaystyle R_{\text{network}} = \frac{30 \times 30}{30 + 30} = 15\,\Omega \)
The \( 3\,\Omega \) internal resistance of the battery is in series with the network:
\( R_{\text{total}} = 15 + 3 = 18\,\Omega \)
\( \displaystyle I = \frac{E}{R_{\text{total}}} = \frac{6}{18} \approx 0.33\,\mathrm{A} \)