Question

Consider the circuit shown in the figure below:

The diodes $D_1$ and $D_2$ have forward cut-in voltage of O.7 V, then the value of current flowing through the diode $D_2$ is equal to _____ mA.

• A. 100

Answer 1

The correct option is A 100

Assuming $D_1$ and $D_2$ to be in on state. Thus redrawing the circuit , we get,

Applying KCL at node ‘V’, we get,

$\frac{V-6+0.7}{10}+\frac{V}{20}+\frac{V-5.7+0.7}{8}=0$

$V[\frac{1}{10}+\frac{1}{20}+\frac{1}{8}]=0.53+\frac{5}{8}$

$V\times \frac{11}{40}=1.155$

$V=\frac{1.155\times 40}{11}=4.2V$

$I_{D2}=\frac{5.7-0.7-4.2}{8}=\frac{0.8}{8}=100mA$