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In short.

$$P_Q = I_{CQ} \times V_{CEQ} \approx 2\text{mA} \times 6.97\text{V} \approx 13.94 \text{mW}$$

And the AC collector current is equal to

$$i_c = g_m v_{\pi} = 1/13\Omega \times 18 \text{mV} \approx 1.38 \text{mA}$$

Therefore power dissipated in the transistor is:

$$P_Q = I_{CQ}V_{CEQ}-\frac{i^2_c \left(R_C||R_L\right)}{2} = 13.94\text{mW} - \frac{1.38\text{mA}^2 \times 1\text{k}\Omega}{2} \approx 13\text{mW}$$

In short.

$$P_Q = I_{CQ} \times V_{CEQ} \approx 2\text{mA} \times 6.97\text{V} \approx 13.94 \text{mW}$$

And the AC component of a collector current is equal to

$$i_c = g_m v_{\pi} = 1/13\Omega \times 18 \text{mV} \approx 1.38 \text{mA}$$

Therefore power dissipated in the transistor is:

$$P_Q = I_{CQ}V_{CEQ}-\frac{i^2_c \left(R_C||R_L\right)}{2} = 13.94\text{mW} - \frac{1.38\text{mA}^2 \times 1\text{k}\Omega}{2} \approx 13\text{mW}$$

In short.

$$P_Q = I_{CQ} \times V_{CEQ} \approx 2mA \times 6.97V \approx 13.94mW$$

And the AC collector current is equal to

$$i_c = g_m v_{\pi} = 1/13\Omega \times 18mV \approx 1.38mA$$

Therefore power dissipated in the transistor is:

$$P_Q = I_{CQ}V_{CEQ}-\frac{i^2_c \left(R_C||R_L\right)}{2} = 13.94mW - \frac{1.38mA^2 \times 1k\Omega}{2} \approx 13mW$$

In short.

$$P_Q = I_{CQ} \times V_{CEQ} \approx 2\text{mA} \times 6.97\text{V} \approx 13.94 \text{mW}$$

And the AC collector current is equal to

$$i_c = g_m v_{\pi} = 1/13\Omega \times 18 \text{mV} \approx 1.38 \text{mA}$$

Therefore power dissipated in the transistor is:

$$P_Q = I_{CQ}V_{CEQ}-\frac{i^2_c \left(R_C||R_L\right)}{2} = 13.94\text{mW} - \frac{1.38\text{mA}^2 \times 1\text{k}\Omega}{2} \approx 13\text{mW}$$

In short.

$$P_Q = I_{CQ} \times V_{CEQ} \approx 2mA \times 6.93V \approx 13.9mW$$

And the AC collector current is equal to

$$i_c = g_m v_{\pi} = 1/13\Omega \times 18mV \approx 1.38mA$$

Therefore power dissipated in the transistor is:

$$P_Q = I_{CQ}V_{CEQ}-\frac{i^2_c \left(R_C||R_L\right)}{2} = 13.9mW - \frac{1.38mA^2 \times 1k\Omega}{2} \approx 13mW$$

In short.

$$P_Q = I_{CQ} \times V_{CEQ} \approx 2mA \times 6.97V \approx 13.94mW$$

And the AC collector current is equal to

$$i_c = g_m v_{\pi} = 1/13\Omega \times 18mV \approx 1.38mA$$

Therefore power dissipated in the transistor is:

$$P_Q = I_{CQ}V_{CEQ}-\frac{i^2_c \left(R_C||R_L\right)}{2} = 13.94mW - \frac{1.38mA^2 \times 1k\Omega}{2} \approx 13mW$$