RC Circ, Capacitor charging Q....very lost

[kostoglotov]

Homework Statement

Homework Equations

Series: R_{eq} = R_1 + ... + R_n

Parallel: R_{eq} = \left(\frac{1}{R_1} + ... + \frac{1}{R_n} \right)^{-1}

Charging Capacitor: I = I_0 e^{-t/RC}

Charging Capacitor: \Delta V_C = \varepsilon (1- e^{-t/RC})

Charge: Q = C \Delta V_C

The Attempt at a Solution

I'm really not sure how to approach this. I know that the approach I've taken neither agrees with the answer in the back and doesn't make much logical sense in terms of everything else I know.

I made an assumption (a bad one I'm guessing) to treat, at least at first, the whole circuit like a parallel/series resistors problem to find the overall current through the circuit and then the voltage drops across the various components, in an attempt to find the voltage drop across the capacitor. The only problem with doing it that way is that the R_{eq} across the circuit could be calculated to be 68 ohm (assuming the cap. is a 0 ohm component...I know, probably a very wrong assumption), which puts 1.5 Amp through the circuit overall, giving a 90 Volt drop across the first resistor, leaving 10 Volts to do work on the remaining components in parallel. But then if you continue from there, turns out the 40 Ohm resistor gets 0.25 amps, the resistor/capacitor combo should be getting 1.25 Amps, which puts a 12.5 Volt drop across the 10 Ohm resistor, which is too high.

What do I need to do to solve this problem? Do I need to use those logarithmic functions? How? Are there cut off points for how much voltage will get through that 10 Ohm resistor and onto the capacitor?

 

[gneill]

For a DC circuit, when you want to find the steady state voltage across capacitors, remove the capacitors from the circuit and find the potential across the open terminals where they were connected.

In this case, remove the 2.0 μF capacitor and find the potential between the open end of the 10 Ω resistor and the bottom node.

 

[Herman Trivilino]

An equivalent way of looking at it is to realize that no current flows through the 10 ohm resistor (as long as the switch has been closed, or open for that matter, for a long time).

 

[Herman Trivilino]

Note that this part of the circuit ...

is called a "voltage divider". It essentially "divides" the voltage supplied by the battery, so that only a part of it is available to something connected across points A and B. By varying the sizes of the two resistors the voltage between points A and B can be varied to anything between zero and 100 V.