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Chapter 19: Problem 2592

In an P.N.P transistor circuit, the collector current is \(10 \mathrm{~mA}\). If \(90 \%\) of the electrons emitted reach the collector: (A) \(\mathrm{I}_{\mathrm{E}}=9 \mathrm{~m} \mathrm{~A}\) (B) \(\mathrm{I}_{\mathrm{E}}=10 \mathrm{~mA}\) (C) \(\mathrm{I}_{\mathrm{B}}=1 \mathrm{~mA}\) (D) \(\mathrm{I}_{\mathrm{B}}=-1 \mathrm{~mA}\)

Short Answer

Expert verified
(C) $\mathrm{I}_{\mathrm{B}}=1 \mathrm{~mA}$

Step by step solution

01

Review the transistor current relationships

In a PNP transistor, the following relationships between the currents exist: - Emitter current (I_E) = Collector current (I_C) + Base current (I_B)
02

Calculate the percentage of electrons that reach the base

We are given that 90% of the emitted electrons reach the collector. Consequently, the remaining 10% of the electrons reach the base.
03

Determine the relationship between I_C and I_B

Since 90% of the emitted electrons reach the collector, and 10% reach the base, the ratio of I_C to I_B is 9:1 or, \(I_{C} = 9 \times I_{B}\)
04

Use the given I_C value to find I_B

We are given I_C = 10 mA. We can now use the relationship from Step 3 to find I_B: \(I_{C} = 9 \times I_{B}\) \(10\:mA = 9 \times I_{B}\) \(I_{B} = \frac{10\:mA}{9} = 1.11 \:mA\)
05

Calculate I_E using the transistor current relationship

Now we can find I_E using the transistor current relationship from Step 1: \(I_{E} = I_{C} + I_{B}\) \(I_{E} = 10\:mA + 1.11\:mA = 11.11 \:mA\) Looking at the given options, we can see that there is no exact match. However, the closest option that approximately matches our calculated values is: (C) I_B = 1 mA

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