Let $$R_C=600Omega$$, and let $$R_E=500Omega$$.  When $$R_1=2.00kOmega$$ and $$R_2=1.00kOmega$$, the value $$I_C$$ (mA) is most nearly:

$$R_g=100Omega$$ , $$C_o=50.0muF$$,  $$C_1=10muF$$, $$C_2=10muF$$, $$V_(BE)=0.7V$$

A.  10
B.  19
C.  24
D.  27

We will use DC analysis to find the DC current, $I_C$. In DC, capacitors act as open circuit, leaving us with a circuit that looks like this:

To solve this equation, we can use the following equations to solve for $I_{EQ}$. Since our beta value is not given we will assume it is large. Since beta is large, and since we know $I_B+I_C=I_E, I_C=\beta I_B$, we can conclude that $I_C\approx I_{EQ}$:

$$\begin{gathered} I_{EQ}=\frac{V_{BB}-V_{BE}}{{\displaystyle \frac{R_B}{\beta +1}}+R_E} \\ {V}_{BB}=V_{CC}\left(\frac{R_2}{R_2+R_1}\right) \end{gathered}$$

Plugging our values from our problem in:

$$\begin{gathered} V_{BB}=\left(30\right)\left(\frac{1k}{1k+2k}\right) \\ {V}_{BB}=10V \end{gathered}$$ 

$$\begin{gathered} I_{EQ}=\frac{V_{BB}-V_{BE}}{{\displaystyle \frac{R_B}{\beta +1}+R_E}} \\ {I}_{EQ}=\frac{10-0.7}{{\displaystyle \frac{R_B}{\beta +1}+500}} (\beta \approx \infty ) \\ {I}_{EQ}=\frac{9.3}{{\displaystyle 500}} =18.6mA \end{gathered}$$

$I_{EQ}=I_C=18.6mA$

Therefore, our collector current is mostly nearly 19 mA, answer B.