Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.5 A
An ideal diode has zero resistance when forward biased (its arrow points from the higher to the lower potential). It passes no current when reverse biased.
From the figure, the positive terminal of the 5 V cell is on the E side. So A and C are at the higher potential, and B and D are at the lower potential.
D1 points from A to B, so it is forward biased and the \( 10\,\Omega \) branch conducts.
D2 points from D to C, so it is reverse biased and the \( 20\,\Omega \) branch carries no current.
The whole battery current flows through the \( 10\,\Omega \) resistor:
\( \begin{aligned} I &= \frac{V}{R} \\ &= \frac{5\,\mathrm{V}}{10\,\Omega} \\ &= 0.5\,\mathrm{A} \end{aligned} \)