Question

If an emitter current is changed by $4 mA ,$ the collector current changes by $3.5 mA$. The value of $\beta$ will be

• A. 7
• B. 0.5
• C. 0.875
• D. 3.5

Answer 1

The Correct Option is A

To find the value of $\beta$ in a bipolar junction transistor (BJT), we use the relationship between the emitter current ($I_E$), base current ($I_B$), and collector current ($I_C$). The formula is given by:

$I_E = I_C + I_B$

The current gain $\beta$ is defined as the ratio of the collector current to the base current:

$\beta = \frac{I_C}{I_B}$

Since $I_B = I_E - I_C$, we can substitute this into the equation for $\beta$:

$\beta = \frac{I_C}{I_E - I_C}$

Given that an increase in the emitter current by $4 \, \text{mA}$ results in an increase in the collector current by $3.5 \, \text{mA}$, the changes in respective currents are:

• Change in $I_E$: $\Delta I_E = 4 \, \text{mA}$
• Change in $I_C$: $\Delta I_C = 3.5 \, \text{mA}$

Assuming the transistor is operating in active mode, and the changes in current are proportionate, the change in equations can be used:

$\Delta I_E = \Delta I_C + \Delta I_B$

Using the known values:

$4 \, \text{mA} = 3.5 \, \text{mA} + \Delta I_B$

Rearranging gives:

$\Delta I_B = 4 \, \text{mA} - 3.5 \, \text{mA} = 0.5 \, \text{mA}$

Now we can find $\beta$:

$\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{3.5 \, \text{mA}}{0.5 \, \text{mA}} = 7$

Thus, the correct answer is 7.