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If each diode in figure has a forward bias resistance of 25Omega and infinite resistance in reverse bias, what will be the values of the currentI_(1), I_(2), I_(3) and I_(4)? |
Answer» SOLUTION :In above figure, diode in branch CD is reverse biased. Hence its resistance is `oo` and so current passing through it is `I_(3)=0` and so it can be removed from given network.  In figure (2) `I_(1)=I_(2)+I_(4)""(1)` Here resistance in branches ABand EF are same and so: `I_(2)=I_(4)=(I_(1))/(2)""....(2)` Applying Kirchhoff.s second law along CLOSED PATH GEFHG, `-150I_(2)-25I_(1)= -5` `therefore 150I_(2)+25I_(1)=5` `therefore 150 I_(2)+25(2I_(2))=5` `therefore 200I_(2)=5` `therefore I_(2)=0.025A` `therefore I_(4)=0.025A ""(because I_(2)=I_(4))` From equation (1), `I_(1)=I_(2)+I_(4)` `therefore I_(1)=0.025+0.025=0.05A` |
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