In an n-p-n transistor circuit, the collector current ia 10 mA. If 90% of the electrons emitted reach the collector.

Options

A. the emitter current will be 9 mA
B. the base current will be 1 mA
C. the emitter current will be 11 mA
D. the base current will be -1 mA

Explanation

To solve the problem step by step, we will analyze the given information about the n-p-n transistor circuit.

### Step 1: Understand the given information
We are given:
- Collector current (IC) = 10 mA
- 90% of the electrons emitted reach the collector.

### Step 2: Relate collector current (IC) to emitter current (IE)
From the problem statement, we know that:
\[ IC = 0.90 \times IE \]
This means that the collector current is 90% of the emitter current.

### Step 3: Substitute the value of IC into the equation
We can substitute the value of IC into the equation:
\[ 10 \, \text{mA} = 0.90 \times IE \]

### Step 4: Solve for emitter current (IE)
To find IE, we rearrange the equation:
\[ IE = \frac{IC}{0.90} \]
Substituting the value of IC:
\[ IE = \frac{10 \, \text{mA}}{0.90} \]
\[ IE = 11.11 \, \text{mA} \] (approximately 11 mA)

### Step 5: Use the relationship between emitter current (IE), base current (IB), and collector current (IC)
The relationship is given by:
\[ IE = IB + IC \]
We can rearrange this to find the base current (IB):
\[ IB = IE - IC \]

### Step 6: Substitute the values of IE and IC
Substituting the values we found:
\[ IB = 11.11 \, \text{mA} - 10 \, \text{mA} \]
\[ IB = 1.11 \, \text{mA} \] (approximately 1 mA)

### Final Results
- Emitter current (IE) ≈ 11 mA
- Base current (IB) ≈ 1 mA

### Summary of Answers
1. Emitter current (IE) = 11 mA
2. Base current (IB) = 1 mA