While a collector-emitter voltage is constant in a transistor, the collector current changes by 8.2mA when the emitter current changes by 8.3mA. The change in base current is

Options

A. `100 mu A `
B. `200 mu A`
C. `300 mu A `
D. `400 mu A `

Answer

To solve the problem, we can follow these steps:

### Step 1: Understand the relationship between emitter current, base current, and collector current.
In a transistor, the relationship between the emitter current (IE), base current (IB), and collector current (IC) is given by the equation:
\[ I_E = I_B + I_C \]

### Step 2: Write the equation for changes in currents.
When we consider small changes in these currents, we can express this relationship in terms of changes:
\[ \Delta I_E = \Delta I_B + \Delta I_C \]
where:
- \(\Delta I_E\) is the change in emitter current,
- \(\Delta I_B\) is the change in base current,
- \(\Delta I_C\) is the change in collector current.

### Step 3: Rearrange the equation to find the change in base current.
From the equation above, we can rearrange it to solve for the change in base current:
\[ \Delta I_B = \Delta I_E - \Delta I_C \]

### Step 4: Substitute the given values into the equation.
We are given:
- \(\Delta I_E = 8.3 \, \text{mA}\)
- \(\Delta I_C = 8.2 \, \text{mA}\)

Now, substituting these values into the equation:
\[ \Delta I_B = 8.3 \, \text{mA} - 8.2 \, \text{mA} \]

### Step 5: Calculate the change in base current.
Now, performing the subtraction:
\[ \Delta I_B = 0.1 \, \text{mA} \]

### Step 6: Convert the result to microamperes.
Since \(1 \, \text{mA} = 1000 \, \mu\text{A}\), we can convert \(0.1 \, \text{mA}\) to microamperes:
\[ \Delta I_B = 0.1 \, \text{mA} = 100 \, \mu\text{A} \]

### Final Answer:
The change in base current is \(100 \, \mu\text{A}\).

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