KVL in a BJT circuit
I think you've overcomplicated your analysis.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1.

The collector voltage is just the supply voltage minus the voltage drop across Rc.

$$V_C = 10 V - V_{RC}$$

The voltage drop across Rc is given by Ohm's Law (as you stated) using the collector current.

$$ V_{RC} = I_C \cdot R_C$$

and substituting this gives you

$$V_C = 10 V - I_C \cdot R_C$$



Since the base is held at 4 V the emmitter will be held at about 3.3 V and this will determine the current through Re. The current through Rc will be the current through Re less the base current.