Class 12 NEET Physics Semiconductor Electronics Part 2

Concept:Ideal diodes act as a short circuit (zero resistance) when forward biased and as an open circuit (infinite resistance) when reverse biased.Formula:Ohm's law: $I = V/R$. For resistors in parallel, total current is the sum of individual branch currents.Solution:1. Identify the bias of each diode. In the given circuit, both diodes are forward biased (anode at higher potential than cathode).2. Replace each forward-biased diode with a short circuit (wire of zero resistance).3. Redraw the circuit: The $10\ \mathrm{V}$ battery now directly connects across a $2\ \Omega$ resistor and a $4\ \Omega$ resistor in parallel.4. Calculate current in each branch: $I_{2\Omega} = \frac{10}{2} = 5\ \mathrm{A}$; $I_{4\Omega} = \frac{10}{4} = 2.5\ \mathrm{A}$.5. Total current supplied by the battery: $I = I_{2\Omega} + I_{4\Omega} = 5 + 2.5 = 7.5\ \mathrm{A} = \frac{15}{2}\ \mathrm{A}$.Shortcut:When both diodes are forward biased, the equivalent resistance is the parallel combination: $R_{\text{eq}} = \frac{2 \times 4}{2+4} = \frac{8}{6} = \frac{4}{3}\ \Omega$. Then total current: $I = \frac{V}{R_{\text{eq}}} = \frac{10}{4/3} = 10 \times \frac{3}{4} = \frac{30}{4} = \frac{15}{2}\ \mathrm{A}$.Answer:$\frac{15}{2}\ \mathrm{A}$ (Option D).