Physics - Semiconductors Question with Solution | TestHub

The given circuit can be redrawn as,



This means,${\mathrm{D}}_2$is reversed based. Hence, this arm of the circuit will act as open arm i.e., the close circuit will have two resistors$2 \mathrm{\Omega }$and$4 \mathrm{\Omega }$which will be in series connection.

$\Rightarrow R_{eq}=2 \mathrm{\Omega }+4 \mathrm{\Omega }$

$=6 \mathrm{\Omega }$

So, current through$4 \mathrm{\Omega }$resistor will be,

$I=\frac{V}{R}$

$=\frac{12}{6}$

$=2\mathrm{ A}$