An NPN transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance is 100 Ω and the output load resistance is 10 kΩ. The common emitter current gain β is:

We are told that the common-emitter amplifier has a power gain of 60 dB, an input resistance of $$R_{\text{in}} = 100\;\Omega$$ and a load resistance of $$R_L = 10\;\text{k}\Omega = 10\,000\;\Omega$$. The task is to find the common-emitter current gain $$\beta = \dfrac{I_C}{I_B}$$.

First, we translate the power gain from decibels to an ordinary (dimensionless) ratio. The definition of power gain in decibels is

$$G_{\text{dB}} = 10\;\log_{10}\!\left(\dfrac{P_{\text{out}}}{P_{\text{in}}}\right)\!.$$

Here $$G_{\text{dB}} = 60$$, so

$$60 = 10\;\log_{10}\!\left(\dfrac{P_{\text{out}}}{P_{\text{in}}}\right).$$

Dividing both sides by 10 gives

$$6 = \log_{10}\!\left(\dfrac{P_{\text{out}}}{P_{\text{in}}}\right).$$

Removing the logarithm by raising 10 to the power of each side we obtain

$$\dfrac{P_{\text{out}}}{P_{\text{in}}} = 10^6.$$

So the numerical power gain is

$$G_P = 10^6.$$

Next, we connect this power gain to the current gain $$\beta$$. For a common-emitter stage, the input power is the power in the base circuit, and the output power is the power delivered to the collector load. Assuming all the base current $$I_B$$ flows through $$R_{\text{in}}$$ and all the collector current $$I_C$$ flows through $$R_L$$, we write

$$P_{\text{in}} = I_B^{\,2}\;R_{\text{in}}, \qquad P_{\text{out}} = I_C^{\,2}\;R_L.$$

Taking their ratio gives

$$\dfrac{P_{\text{out}}}{P_{\text{in}}} = \dfrac{I_C^{\,2}\;R_L}{I_B^{\,2}\;R_{\text{in}}}.$$

Now we use the definition $$\beta = \dfrac{I_C}{I_B}$$. Hence $$I_C^{\,2} = \beta^{\,2}\,I_B^{\,2}$$, and substituting this into the expression above yields

$$\dfrac{P_{\text{out}}}{P_{\text{in}}} = \beta^{\,2}\;\dfrac{R_L}{R_{\text{in}}}.$$

But we have already shown that $$\dfrac{P_{\text{out}}}{P_{\text{in}}} = 10^6$$, so we write

$$10^6 = \beta^{\,2}\;\dfrac{R_L}{R_{\text{in}}}.$$

The resistance ratio is

$$\dfrac{R_L}{R_{\text{in}}} = \dfrac{10\,000}{100} = 100 = 10^2.$$

Substituting this value in, we get

$$10^6 = \beta^{\,2}\;(10^2).$$

Dividing both sides by $$10^2$$ produces

$$\beta^{\,2} = \dfrac{10^6}{10^2} = 10^{6-2} = 10^4.$$

Taking the positive square root (current gain is positive) gives

$$\beta = \sqrt{10^4} = 10^2 = 100.$$

Hence, the correct answer is Option B.