Class 12 NEET Physics Semiconductor Electronics Part 2

Concept:An ideal diode behaves as a closed switch when forward biased and as an open switch when reverse biased.A diode conducts only if its anode is at a higher potential than its cathode.Formula:Ohm's law: $I = \frac{V}{R}$Parallel resistors: $R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$Solution:The $+$ terminal of the $10\ \mathrm{V}$ battery is on the left, so the left rail is at higher potential.Branch with $4\ \Omega$: diode points right, anode on positive left rail, so it is forward biased and conducts.Branch with $3\ \Omega$: diode points left, cathode on positive left rail, so it is reverse biased and carries no current.Branch with $2\ \Omega$: diode points right, anode on positive left rail, so it is forward biased and conducts.Branch with $5\ \Omega$: diode points left, cathode on positive left rail, so it is reverse biased and carries no current.So only the $4\ \Omega$ and $2\ \Omega$ branches are active, connected in parallel across $10\ \mathrm{V}$.$I_{4\Omega} = \frac{10}{4} = \frac{5}{2}\ \mathrm{A}$$I_{2\Omega} = \frac{10}{2} = 5\ \mathrm{A}$$I = \frac{5}{2} + 5 = \frac{15}{2}\ \mathrm{A}$Shortcut:$R_{\text{eq}} = \frac{4 \times 2}{4 + 2} = \frac{4}{3}\ \Omega$$I = \dfrac{V}{R_{\text{eq}}} = \frac{10}{4/3} = \frac{15}{2}\ \mathrm{A}$Answer:$I = \frac{15}{2}\ \mathrm{A}$, so the correct option is D.