In the given figure, each diode has a forward bias resistance of 30 $Ω$ and infinite resistance in reverse bias. The current I₁ will be:

Option 1 - 3.75 A

Option 2 - 2.73 A

Option 3 - 2.35 A

Option 4 - 2 A

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Answered by

Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 4

Detailed Solution:

In give diagram Diode 1 and 2 are in forward bias with R = $30 Ω$ and Diode 3 is reverse bias with R = infinite l₁ current is flowing through 20 $Ω$

So $\frac{l_{1}}{2} a n d \frac{l_{1}}{2}$ current will flow through Diode 1 and 2. As resistance is same applying KCl in ABCD Loop $- \frac{l_{1}}{2} * 130 - \frac{l_{1}}{2} * 130 - l_{1} * 20 + 200 = 0$

-100l₁ + 200 = 0

l₁ = 2

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Physics Ncert Solutions Class 12th 2023

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In the given figure, each diode has a forward bias resistance of 30 Ω and infinite resistance in reverse bias. The current I1 will be:

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