Question

In common emitter configuration of transistor amplifier, \(r_i\), \(R_L\) and \(β\) represent the input resistance, load resistance and the a.c. current gain respectively. The voltage gain \(A_V\) and power gain \(A_P\) are represented in magnitude respectively by

• A. \(β(\frac{r_i}{R_L}),β^2(\frac{r_i}{R_L})\)
• B. \(β(\frac{R_L}{r_i}),β(\frac{R_L}{r_i})^2\)
• C. \(β(\frac{R_L}{r_i}),β^2(\frac{R_L}{r_i})\)
• D. \(β(\frac{r_i}{R_L}),β(\frac{r_i}{R_L})^2\)

Hint:

A forward biased ideal diode acts as a wire and a reverse biased one acts as an open switch.

Answer 1

The Correct Option is C

Step 1: Understanding the Circuit:
Between A and B there are two parallel branches. The upper branch has a 30 ohm resistor in series with the ideal diode D. The lower branch has a 30 ohm resistor.

Step 2: Forward biased:
An ideal forward biased diode has zero resistance, so both branches carry current: \(R_1 = \frac{30\times30}{30 + 30} = 15\ \Omega\).

Step 3: Reverse biased:
An ideal reverse biased diode has infinite resistance, so the upper branch carries no current. Only the lower resistor is left: \(R_2 = 30\ \Omega\).

Step 4: Ratio:
\[ \frac{R_1}{R_2} = \frac{15}{30} = \frac12 \]

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