# Voltage between collector and emitter (in transistor circuit) [2N3904]

Let's say we have the following circuit with silicon transistor 2N3904 (the transistor has $\beta = 125$):

I am trying to calculate the DC voltage between the collector and the emitter. One of the solutions I found is the following:

$$Ι_b = \frac{V_{bb} - V_{be}}{R_b} = \frac{10\text{V} - 0.7\text{V}}{1.5\text{M}\Omega} = 6.2\mu \text{Α}$$ $$I_c = \beta \times I_b = 125 \times 6.2\mu \text{Α} = 0.775\text{mA}$$ $$V_{ce} = V_{cc} - I_c \times R_c = 20\text{V} - 0.775\text{mA} \times 5\text{k}\Omega = 16.1\text{V}$$

This solution is not mine and I am not sure that it's correct. I have two questions:

1. $Ι_b = \dfrac{V_{bb} - V_{be}}{R_b}$. Is this true? Shouldn't it be $Ι_b = \dfrac{V_{bb} - V_b}{R_b}$?

2. $V_{ce} = V_{cc} - I_c \times R_c$. Is this true? Shouldn't it be $V_c = V_{cc} - I_c \times R_c$?

$V_{BE}$ and $V_{CE}$ are equal to $V_{B}$ and $V_{C}$ respectively when the emitter is grounded, however no ground is shown in your circuit, so it's not 100% clear what $V_{B}$ and $V_{C}$ would mean.