An NPN transistor is used in common emitter configuration as an amplifier with 1 k$$\Omega$$ load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15 $$\mu$$A change in the base current of the amplifier. The input resistance and voltage gain are:

We have an NPN transistor working in the common-emitter mode. The numerical data given are:

$$\Delta V_{BE}=10\ \text{mV}=10\times10^{-3}\ \text{V}=0.01\ \text{V}$$ $$\Delta I_B = 15\ \mu\text{A}=15\times10^{-6}\ \text{A}$$ $$\Delta I_C = 3\ \text{mA}=3\times10^{-3}\ \text{A}$$ $$R_L = 1\ \text{k}\Omega = 1\times10^{3}\ \Omega$$

First, let us find the input resistance seen at the base-emitter junction. For a small-signal analysis the input resistance is defined by the ratio of the small change in input voltage to the corresponding small change in input current:

$$r_{in} = \frac{\Delta V_{BE}}{\Delta I_B}$$

Substituting the given values, we get

$$r_{in} = \frac{0.01\ \text{V}}{15\times10^{-6}\ \text{A}} = \frac{0.01}{0.000015}$$

Carrying out the division step by step,

$$0.01 = 1.0\times10^{-2},\quad 0.000015 = 1.5\times10^{-5}$$ $$\frac{1.0\times10^{-2}}{1.5\times10^{-5}} = \frac{1.0}{1.5}\times10^{(-2)-(-5)} = \frac{2}{3}\times10^{3} = 0.6667\times10^{3}\ \Omega$$

So,

$$r_{in} \approx 6.67\times10^{2}\ \Omega = 0.67\ \text{k}\Omega$$

Now we calculate the a.c. voltage gain. The output voltage change across the collector load is obtained from Ohm’s law:

$$\Delta V_{C} = \Delta I_C \times R_L$$

Substituting,

$$\Delta V_{C} = 3\times10^{-3}\ \text{A} \times 1\times10^{3}\ \Omega = 3\ \text{V}$$

The magnitude of the voltage gain (common-emitter gain has a phase inversion, but we are asked only for magnitude) is

$$A_v = \left|\frac{\Delta V_{C}}{\Delta V_{BE}}\right| = \left|\frac{3\ \text{V}}{0.01\ \text{V}}\right| = 300$$

Thus the input resistance is $$0.67\ \text{k}\Omega$$ and the voltage gain is $$300$$.

Hence, the correct answer is Option B.