In an NPN transistor the values of base current and collector current are `100muA` and `9mA` respectively the emitter current will be-

Options

A. 9.1 mA
B. 18.2 mA
C. `9.1muA`
D. `18.2muA`

Answer

To find the emitter current (Ie) in an NPN transistor given the base current (Ib) and collector current (Ic), we can follow these steps:

### Step 1: Identify the given values
- Base current (Ib) = 100 µA = 0.1 mA (since 1 mA = 1000 µA)
- Collector current (Ic) = 9 mA

### Step 2: Use the relationship between the currents
In a transistor, the emitter current (Ie) is related to the base current (Ib) and collector current (Ic) by the equation:
\[ I_e = I_b + I_c \]

### Step 3: Substitute the known values into the equation
Now, we can substitute the values we have:
\[ I_e = I_b + I_c \]
\[ I_e = 0.1 \, \text{mA} + 9 \, \text{mA} \]

### Step 4: Perform the addition
Now, we add the two currents:
\[ I_e = 0.1 \, \text{mA} + 9 \, \text{mA} = 9.1 \, \text{mA} \]

### Step 5: State the final answer
Thus, the emitter current (Ie) is:
\[ I_e = 9.1 \, \text{mA} \]

### Conclusion
The emitter current in the NPN transistor is 9.1 mA.

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