In the given circuit, D1 and D2 are normal diodes. The current drawn from the battery by the diode network is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.0\,\mathrm{A} \)
A p-n junction diode conducts when it is forward biased and blocks current when it is reverse biased. The diodes are taken as ideal: zero resistance when forward biased and infinite resistance when reverse biased. The positive terminal of the cell is its longer plate, on the left.
Branch with D1 and \( 5\,\Omega \): the current leaves the positive terminal of the cell and flows in the direction of the arrow of D1, so D1 is forward biased and conducts.
Branch with D2 and \( 10\,\Omega \): the arrow of D2 points against the direction of the current, so D2 is reverse biased and no current flows in this branch.
Only the \( 5\,\Omega \) branch carries current:
\( \displaystyle I = \frac{V}{R} = \frac{5\,\mathrm{V}}{5\,\Omega} = 1.0\,\mathrm{A} \)