# RC parallel branch behaviour

Understanding the overall function and various sections in this simple timing circuit are clear to me, however it's apparent I have holes in my foundation of branch behavior with some components. I'm someone who visualizes what's happening to get a clear idea. I'd like to see the current flow and voltage drops before applying hard calculations. Since RC circuits change over time I'm trying to see what the voltage and current are over R1, R2 and C1 at start up, around half of cap charge and final steady state. The R2 and cap interaction is where I'm unclear.

When the switch is closed I assume R1 has the full 12v immediately with cap at 0v to ground. R2 has minimal to no voltage so Q is off. When cap is charging- this is what I need to understand- Does R2 have a voltage drop directly tied to cap voltage over time -Vbe? Does R2 and the cap form a type of equivalent voltage divider with R1 that can make the picture more clear? And finally in steady state, cap is at 12v with R2 and R1 in series powering Q. Kindly help give me clearer picture over time so I can plug in equations properly.

Edit: Cap steady state is a product of voltage divider.

Cheers.

• To be able to find the final steady state capacitor voltage. Remove the capacitor and try to find the voltage at this point. – G36 May 27 '17 at 14:16
• Would it be the voltage divider Vs(1k/28k)? – Archaeus May 27 '17 at 14:34
• Yes, but Vs = (Vbat - Vbe) – G36 May 27 '17 at 14:41
• Yes sorry, I was about to edit for that. Thank you. – Archaeus May 27 '17 at 14:42
• To be more precise, at the beginning Vc = 0V we have I2 = 0V, hence all supply voltage is present across R1 and I1 = 444uA. As the charging process goes on and Vc reach about 0.6V the R2 and base current will start to flow. And all this current ( for R2 ) is provided by R1 resistor I_R1 = I _C1 + I_R2. – G36 May 27 '17 at 15:27