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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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प्रश्न

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?

दीर्घउत्तर
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उत्तर

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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अध्याय 14: Semiconductor Electronics - Exercises [पृष्ठ ९३]

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एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 12
अध्याय 14 Semiconductor Electronics
Exercises | Q 14.31 | पृष्ठ ९३

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