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 current `I_(1), I_(2), I_(3) and I_(4)`?

Explanation

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 AB and 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`