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Chapter 19: Problem 2634

A n-p-n transistor circuit has \(\alpha=0.985 .\) If $\mathrm{I}_{\mathrm{c}}=9 \mathrm{~mA}\( then the value of \)\mathrm{I}_{\mathrm{b}}$ is (A) \(0.003 \mathrm{~mA}\) (B) \(0.66 \mathrm{~mA}\) (C) \(0.015 \mathrm{~mA}\) (D) \(0.13 \mathrm{~mA}\)

Short Answer

Expert verified
The short answer is: \( I_{b} \approx 0.13 \mathrm{~mA} \). The correct option is (D).

Step by step solution

01

Rearranging the equation to solve for Ib

First, we will rearrange the equation to isolate Ib: α = (Ic - Ib) / Ic Multiplying both sides by Ic, we get: α * Ic = Ic - Ib Adding Ib to both sides and subtracting α * Ic from both sides, we get: Ib = Ic - α * Ic
02

Plug in the given values

Now, we can substitute the given values of α and Ic: Ib = 9 mA - 0.985 * 9 mA
03

Calculate Ib

Performing the arithmetic, we find the value of Ib: Ib = 9 mA - 8.865 mA Ib ≈ 0.135 mA
04

Choose the correct answer

Comparing the calculated value of Ib with the given options, we find that the closest match is: (D) 0.13 mA So, the correct answer is option (D).

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