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  • The cut-off voltage of the diodes (shown in the figure) in forward bias is <img alt="\mathrm{0.6 \mathrm{~V}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B0.6%20%5Cmathrm%7B%7E

The cut-off voltage of the diodes (shown in the figure) in forward bias is \mathrm{0.6 \mathrm{~V}}. The current through the resister of \mathrm{40 \, \Omega} is_________ \mathrm{\mathrm{mA}}.

 

Option: 1

4


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer



Since \mathrm{D_{2}} in reverse
therefore, there will be no current in \mathrm{D_{2}}.
Voltage in \mathrm{D_{1}= 0\cdot 6\, V}.
Remaining \mathrm{1-0.6= 0.4\, V} will cause current in \mathrm{\left ( 60+40 \right )\Omega :}

\mathrm{\therefore\; i= \frac{0.4}{100}= 4\, mA}
Ans : (4)
 

Posted by

seema garhwal

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