The current gain of a common emitter amplifier is 69. If the emitter current is 7.0 mA, collector current is:

The problem gives the current gain of a common-emitter (CE) amplifier and the emitter current, and asks us to find the collector current.

First, we recall the definition of current gain for a CE configuration. The small-signal current gain, usually denoted by $$\beta$$ or $$h_{FE}$$, is defined as

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

where $$I_C$$ is the collector current and $$I_B$$ is the base current.

We are told that $$\beta = 69$$. Hence, from the definition,

$$I_B \;=\; \frac{I_C}{\beta} \;=\; \frac{I_C}{69}$$

Next, we use Kirchhoff’s current relation at the transistor terminals. In any transistor, the emitter current $$I_E$$ is the sum of the collector current and the base current:

$$I_E \;=\; I_C + I_B$$

Substituting $$I_B = \dfrac{I_C}{69}$$ into the above relation, we get

$$I_E \;=\; I_C \;+\; \frac{I_C}{69} \;=\; I_C\!\left(1 + \frac{1}{69}\right)$$

Adding the fractions inside the brackets,

$$1 + \frac{1}{69} \;=\; \frac{69}{69} + \frac{1}{69} \;=\; \frac{70}{69}$$

So,

$$I_E \;=\; I_C \times \frac{70}{69}$$

We are given the numerical value $$I_E = 7.0 \text{ mA}$$. Therefore,

$$7.0 \text{ mA} \;=\; I_C \times \frac{70}{69}$$

To isolate $$I_C$$, we multiply both sides by $$\dfrac{69}{70}$$:

$$I_C \;=\; 7.0 \text{ mA} \times \frac{69}{70}$$

Now, observe that $$\dfrac{69}{70} = 0.987142\dots$$, but we can recognize an easier path: the factor $$\dfrac{69}{70}$$ is the same as multiplying by $$69$$ and then dividing by $$70$$. Doing that directly,

$$I_C \;=\; 7.0 \times \frac{69}{70} \text{ mA} \;=\; \left( \frac{7 \times 69}{70} \right) \text{ mA}$$

The product in the numerator is $$7 \times 69 = 483$$, so

$$I_C \;=\; \frac{483}{70} \text{ mA}$$

Dividing,

$$\frac{483}{70} = 6.9$$

Thus,

$$I_C = 6.9 \text{ mA}$$

This matches Option C in the given list.

Hence, the correct answer is Option C.