Question

In the diagram shown, the resistance between points A and B is '\(R_1\)' when an ideal diode D is forward biased and is '\(R_2\)' when ideal diode D is reverse biased. The ratio \(R_1/R_2\) is

• A. \(2\)
• B. \(1\)
• C. \(\frac{1}{2}\)
• D. \(\frac{1}{4}\)

Hint:

Two 30 ohm resistors are in parallel; the diode is in series with one of them.

Answer 1

The Correct Option is C

Step 1: Reading the circuit
Between A and B there are two branches in parallel. The top branch has a 30 ohm resistor in series with an ideal diode D. The bottom branch has a 30 ohm resistor.

Step 2: Diode forward biased
An ideal forward biased diode has zero resistance, so both branches conduct:
\[ R_1 = \frac{30 \times 30}{30 + 30} = 15\,\Omega \]

Step 3: Diode reverse biased
An ideal reverse biased diode has infinite resistance, so the top branch carries no current. Only the bottom resistor remains:
\[ R_2 = 30\,\Omega \]
\[ \frac{R_1}{R_2} = \frac{15}{30} = \frac12 \]
This is option (C).

Final Answer:
The ratio \(R_1/R_2\) is \(\frac12\), option (C). \[ \boxed{\frac{1}{2}} \]