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?

Answer 1

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.