Question

A n-p-n transistor is used in common emitter mode in an amplifier it. A change of \(40 \mu \mathrm{A}\) in the base current changes the output current by $2 \mathrm{~m} \mathrm{~A}\( and \)0.04 \mathrm{~V}$ in input voltage. The input resistance is (A) \(1 \mathrm{k} \Omega\) (B) \(10 \Omega\) (C) \(10 \mathrm{k} \Omega\) (D) \(100 \Omega\)

Short answer

The input resistance (Rin) of the n-p-n transistor operating in common emitter mode is calculated using the relationship Rin = ∆Vin / ∆Ib. Given ∆Ib = \(40 \times 10^{-6}\) A and ∆Vin = 0.04 V, we can find Rin = \( \frac{0.04}{40 \times 10^{-6}} \) = 1000 Ω or 1 kΩ. Thus, the correct option is (A) 1 kΩ.

Step-by-step solution

Understand the relationship between input voltage, base current, and input resistance in an n-p-n transistor

An n-p-n transistor operating in the common emitter mode has the following relationship between input resistance (Rin), change in input voltage (∆Vin), and change in base current (∆Ib): Rin = ∆Vin / ∆Ib We are given the values of ∆Ib and ∆Vin, and our goal is to calculate the input resistance (Rin) of the transistor.

Convert given values to standard units

The given values for the change in base current and the change in input voltage are provided in microamperes (µA) and milliamperes (mA), respectively. Convert these values to amperes (A) before proceeding with the calculation: ∆Ib = 40 µA = 40 × 10^(-6) A ∆Io = 2 mA = 2 × 10^(-3) A

Calculate the input resistance of the transistor

Plug the given values of ∆Ib and ∆Vin into the formula for input resistance: Rin = ∆Vin / ∆Ib Rin = (0.04 V) / (40 × 10^(-6) A)

Solve for Rin

Perform the division to calculate the input resistance: Rin = (0.04 V) / (40 × 10^(-6) A) = 1000 Ω = 1 kΩ

Identify the correct option

Now that we have calculated the input resistance, we can identify the correct option: The input resistance is (A) 1 kΩ.