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## Homework Statement

For the circuit above, draw the DC equivalent circuit.

Then compute ##I_B, V_C, V_{CE}##.

## Homework Equations

## The Attempt at a Solution

So the DC equivalent circuit was easy to draw. My concern is about the final answer. The answer given does not match the answer I've obtained and I'm wondering who is correct.

After drawing the DC circuit, I obtained the following relationships:

$$I_C = \frac{V_{CC} - V_C}{R_C} = \frac{21 - V_C}{2k}$$

$$I_B = \frac{V_C - V_{BE}}{R_1 + R_2} = \frac{V_C - 0.7}{81k}$$

$$I_E = \frac{V_E}{R_E} = \frac{V_E}{1k}$$

Now we know ##I_C = \beta I_B##, so the first and second equations can be related:

$$40I_B = \frac{21 - V_C}{2k}$$

$$I_B = \frac{V_C - 0.7}{81k}$$

Simplifying we get:

$$I_B = \frac{21 - V_C}{80k}$$

$$I_B = \frac{V_C - 0.7}{81k}$$

Solving we get:

$$\frac{21 - V_C}{80k} = \frac{V_C - 0.7}{81k}$$

$$V_C = 10.9V$$

Going back to the original equations, we see:

$$I_C = \frac{21 - V_C}{2k} = \frac{21 - 10.9}{2k} = 5.05 mA$$

$$I_B = \frac{V_C - 0.7}{81k} = \frac{10.9 - 0.7}{81k} = 0. 126 mA$$

Then we know:

$$I_E = I_C + I_B = 5.05 mA + 0.126 mA = 5.176 mA$$

From which we find:

$$I_E = \frac{V_E}{1k}$$

$$V_E = (5.176 mA)(1k) = 5.176 V$$

Hence we know:

$$V_{CE} = V_C - V_E = 10.9V - 5.176V = 5.72V$$

Now all of my analysis seems reasonable to me, but for some reason the answer is given:

$$I_B = 0.0995 mA$$

$$V_C = 12.8 V$$

$$V_{CE} = 8.76 V$$

This answer does not match mine whatsoever. Have I done something wrong?

EDIT: Here is a picture of the analysis given in the answer: