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Chapter 19: Problem 2599

The A.C. current gain of a transistor is 100 . If the base current changes by \(100 \mu \mathrm{A}\), What is the charge in collector current? (A) \(20 \mathrm{~mA}\) (B) \(30 \mathrm{~mA}\) (C) \(10 \mathrm{~mA}\) (D) \(10 \mu \mathrm{A}\)

Short Answer

Expert verified
The change in collector current (\(\Delta I_C\)) is \(10 mA\). Therefore, the correct answer is (C) \(10 \mathrm{~mA}\).

Step by step solution

01

Write the Formula for the A.C. Current Gain

The A.C. current gain of a transistor is denoted by \(\beta\) and is given by the formula: \(\beta = \frac{\Delta I_C}{\Delta I_B}\) where \(\Delta I_C\) is the change in the collector current, and \(\Delta I_B\) is the change in the base current.
02

Plug in the Given Information

We are given the A.C. current gain \(\beta = 100\) and the change in base current \(\Delta I_B = 100 \mu A\). Plugging these values into our formula, we get: \(100 = \frac{\Delta I_C}{100 \mu A}\)
03

Solve for the Change in Collector Current

Now, we can solve for \(\Delta I_C\) by multiplying both sides of the equation by \(100 \mu A\): \(\Delta I_C = 100 \cdot 100 \mu A\) \(\Delta I_C = 10,000 \mu A\)
04

Convert to Milliamperes (mA)

Since our answer choices are given in milliamperes (mA), we need to convert our answer from microamperes (\(\mu A\)) to milliamperes (mA). We know that \(1000 \mu A = 1 mA\), so: \(\Delta I_C = \frac{10,000 \mu A}{1,000 \frac{\mu A}{mA}}\) \(\Delta I_C = 10 mA\)
05

Choose the Correct Option

Now we can see that the charge in collector current is \(10 mA\), which corresponds to option (C). Therefore, the correct answer is: (C) \(10 \mathrm{~mA}\)

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Most popular questions from this chapter

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A n-p-n transistor is used in common emitter made in an amplifier it. A change of \(40 \mu \mathrm{A}\) in the base current changes the output current by $2 \mathrm{~mA}\( and \)0.04 \mathrm{~V}$ in input voltage. The current amplification factor is (A) 20 (B) 30 (C) 50 (D) 400
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