Physics - Semiconductors Question with Solution | TestHub

Here current flows in circuit Aas both $(p-n)$ junction diode act as forward biasing. Total resistnace $R$. $$\frac{1}{R}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}\text{ or }R=2\Omega$$ According to Ohm's law $$V=I_{A}R$$ or $8=I_{\mathrm{A}}\times 2$ or $I_{\mathrm{A}}=4\mathrm{A}$ In circuit $B$, lower $p$ - $n$-junction diode is reverse biased. Hence, no current will flow but upper diode is forward biased so current cna flow through it $$V=I_{B}R$$ or $8=I_B\times 4$ or $I_B=2\mathrm{A}$