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In an npn transistor circuit, the collector current is 8 mA. If 90% of the electrons emitted reach the collector, the emitter current (IE) and base current (IB) are given bySelect one:a. IE = – 8 mA, IB = 9 mAb. IE = 9 mA, IB = – 8 mAc. IE = 1 mA, IB = 9 mAd. IE = 9 mA, IB = 1 mA

Question

In an npn transistor circuit, the collector current is 8 mA. If 90% of the electrons emitted reach the collector, the emitter current (IE) and base current (IB) are given bySelect one:a. IE = – 8 mA, IB = 9 mAb. IE = 9 mA, IB = – 8 mAc. IE = 1 mA, IB = 9 mAd. IE = 9 mA, IB = 1 mA

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Solution

The answer is d. IE = 9 mA, IB = 1 mA.

Here's the step-by-step explanation:

  1. In an npn transistor, the collector current (IC) is made up of the majority of the emitter current (IE) and a small portion of the base current (IB). This is represented by the equation IC = IE - IB.

  2. In this question, it's given that IC = 8 mA and 90% of the electrons emitted reach the collector. This means that the emitter current (IE) is 8 mA / 0.9 = 8.89 mA, which is approximately 9 mA.

  3. To find the base current (IB), we can rearrange the equation from step 1 to get IB = IE - IC. Substituting the values we have, IB = 9 mA - 8 mA = 1 mA.

So, the emitter current is 9 mA and the base current is 1 mA.

This problem has been solved

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