JEE Physics 2023 Question with Solution

Given: From the graph, $I_C = 50 \, \text{mA}$ for $I_B = 500 \, \mu\text{A}$. Also, $R_B = 10 \, k\Omega$ and $R_C = 1 \, k\Omega$.

Find: The value of $x$ if power gain $= 10^x$.

In CE configuration,

$$\text{Power gain} = \beta \times \frac{R_C}{R_B}$$

where

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

From the graph,

$$\beta = \frac{50 \times 10^{-3}}{500 \times 10^{-6}} = 100$$

Now,

$$\text{Power gain} = 100 \times \frac{1}{10} = 10$$

So,

$$10^x = 10 = 10^1$$

Hence,

$$x = 1$$

Therefore, the required numerical value is $1$.

Given: The transfer characteristic is a straight line passing through points such as $\left(500 \, \mu\text{A}, 50 \, \text{mA}\right)$.

Find: The value of $x$.

The current gain is obtained from the ratio of collector current to base current:

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

Using the marked point,

$$I_C = 50 \, \text{mA} = 50 \times 10^{-3} \, \text{A}$$ $$I_B = 500 \, \mu\text{A} = 500 \times 10^{-6} \, \text{A}$$

Therefore,

$$\beta = \frac{50 \times 10^{-3}}{500 \times 10^{-6}} = 10^2 = 100$$

Now use the power gain relation for CE configuration:

$$\text{Power gain} = \beta \frac{R_C}{R_B}$$

Substituting $R_C = 1 \, k\Omega$ and $R_B = 10 \, k\Omega$,

$$\text{Power gain} = 100 \times \frac{1 \, k\Omega}{10 \, k\Omega} = 100 \times \frac{1}{10} = 10$$

Since the power gain is written as $10^x$,

$$10^x = 10$$

Thus,

$$x = 1$$

The correct answer extracted from the working is $1$. The raw listed answer showing $3$ disagrees with the solution working.