Question

In the following circuit shown in figure, three diodes are connected each with forward resistance \(40 \Omega\) and infinite backward resistance. The current through \(100 \Omega\) resistance is

• A. 18 mA
• B. 36 mA
• C. 9 mA
• D. 27 mA

Hint:

Decide which diodes are forward biased, then combine the conducting branches and apply Ohm's law.

Answer 1

The Correct Option is B

Step 1: Equivalent Circuit
Remove the reverse-biased branch of $ D_1 $. Keep $ 100\,\Omega $ in series with the parallel pair of $ 100\,\Omega $ and $ 200\,\Omega $ branches.

Step 2: Resistance
$ \dfrac{1}{R_p}=\dfrac{1}{100}+\dfrac{1}{200}=\dfrac{3}{200} $, so $ R_p=66.7\,\Omega $.
$ R=100+66.7=166.7\,\Omega $

Step 3: Ohm's Law
$ I=\dfrac{V}{R}=\dfrac{6}{166.7}\approx 36 \text{ mA} $, option (B).

Final Answer:
Using Ohm's law gives 36 mA. \[ \boxed{\text{(B) }\text{36 mA}} \]