Question

Currents flowing in each of the following circuits $A$ and $B$ respectively are

• A. 1 A, 2 A
• B. 2 A, 1 A
• C. 4 A, 2 A
• D. 2 A, 4 A

Answer 1

Answer: (C)

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