If each diode in figure has a forward has a forward bias resistance of 25 Omega and infinite resistance in reverse bias, what will be the values of the current ` I_1 , I_2 , I_3 and I_4? `

Explanation

Given, forward biased reistance `= 25 Omega`
Reverse biased resistance `= oo`
As the diode in branch CD is in reverse biased which having resistance infinite,
so `I_(3) = 0`
Resistance in branch `AB = 25 + 125 = 150 Omega " say " R_(1)`
Resistance in branch `EF = 25 + 125 = 150 Omega " say " R_(2)`
AB is parallel to EF.
So, resutant resistance `(1)/(R') = (1)/(R_(1)) + (1)/(R_(2)) = (1)/(150) + (1)/(150) = (2)/(150)`
`rArr R' = 75 Omega`
Total resistance `R = R' + 25 = 75 + 25 = 100 Omega`
Current `I_(1) = (V)/(R) = (5)/(100) = 0.05 A`
`I_(1) = I_(4) + I_(2) + I_(3)`
So, `I_(1) = I_(4) + I_(2)`
Here, the resistances `R_(1) " and " R_(2)` is same.
i.e., `I_(4) = I_(2)`
`:. I_(1) = 2I_(2)`
`rArr I_(2) = (I_(1))/(2) = (0.05)/(2) = 0.025 A`
and `I_(4) = 0.025 A`
Thus, `I_(1) = 0.05 A, I_(2) = 0.025 A, I_(3) = 0 " and " I_(4) = 0.025 A`