In a n-p-n transister circuit the collector current is `9 mA`. If `90%` of the electrons emitted reach the collector, find emitter current and base current

Options

A. `I_E=-1mA,I_B = 9 mA`
B. `I_E = 9 mA,I_B=-1mA`
C. `I_E=1mA,I_B=11mA`
D. `I_E=11mA, I_B = 1 mA`

Answer

To solve the problem step by step, we will use the relationships between the collector current (Ic), emitter current (Ie), and base current (Ib) in an NPN transistor.

### Step 1: Understand the relationship between currents
In an NPN transistor:
- The collector current (Ic) is the current that flows from the collector to the emitter.
- The emitter current (Ie) is the total current flowing out of the emitter.
- The base current (Ib) is the current flowing into the base.

The relationship between these currents can be expressed as:
\[ I_e = I_c + I_b \]

### Step 2: Given data
From the problem:
- Collector current, \( I_c = 9 \, \text{mA} \)
- 90% of the emitted electrons reach the collector.

### Step 3: Relate collector current to emitter current
Since 90% of the emitter current (Ie) reaches the collector (Ic), we can express this as:
\[ I_c = 0.9 \times I_e \]

### Step 4: Substitute the known value of Ic
Substituting the value of \( I_c \) into the equation:
\[ 9 \, \text{mA} = 0.9 \times I_e \]

### Step 5: Solve for emitter current (Ie)
To find \( I_e \), we rearrange the equation:
\[ I_e = \frac{I_c}{0.9} = \frac{9 \, \text{mA}}{0.9} = 10 \, \text{mA} \]

### Step 6: Find the base current (Ib)
Now that we have the emitter current, we can find the base current. Since 10% of the emitter current goes to the base:
\[ I_b = 0.1 \times I_e \]
\[ I_b = 0.1 \times 10 \, \text{mA} = 1 \, \text{mA} \]

### Final Results
- Emitter current, \( I_e = 10 \, \text{mA} \)
- Base current, \( I_b = 1 \, \text{mA} \)

### Summary
Thus, the emitter current is 10 mA and the base current is 1 mA.

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