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 $I _{ A }=4 A$
In circuit $B$, lower $p-n$-junction diode is reverse biased.
Hence, no current will flow but upper diode is forward biased so current, can flow through it
$V = I _{ B } R$
or $8= I _{ B } \times 4$
or $I_{B}=2 A$