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Question
If each diode in figure has a forward bias resistance of 25 Ω and infinite resistance in reverse bias, what will be the values of the current I1, I2, I3 and I4?
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Solution
According to the problem, forward biased resistance = 25 Ω and reverse biased resistance = ∞
As shown in the figure, the diode in branch CD is in reverse biased which has infinite resistance.
So, the current in that branch is zero, i.e. I3, = 0
Resistance in branch AB = 25 + 125 = 150 Ω say R1
Resistance in branch EF = 25 + 125 = 150 Ω, say R2
AB is parallel to EF
So, effective resistance
`1/R^' = 1/R_1 + 1/R_2 = 1/150 + 1/150 = 2/150`
⇒ R' 75 Ω
Total resistance R of the circuit = R' + 25 = 75 + 25 = 100 Ω
Current `I_1 = V/R = 5/100` = 0.05 A
According to Kirchhoff's, current law (KCL),
I1 = I4 + I2 + I3 .....(Here I3 = 0)
So I1 = I4 + I2
Here, the resistance R1 and R2 are same.
i.e., I4 = I2
∴ I1 = 2I2
⇒ `I_2 = I_1/2 = 0.05/2` = 0.025 A
And I4 = 0.025 A
Therefore, we get, I1 = 0.05 A, I2 = 0.025 A, I3 = 0 and I4 = 0.025 A.
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