# Homework Help: Archived Calculating charge on a capacitor in a DC circuit

1. Jan 27, 2009

### dancavallaro

1. The problem statement, all variables and given/known data
See attached image for a circuit diagram. The assumptions are that transients have died out and the currents and charges have reached their equilibrium values. I have to calculate the charge on the capacitor.

2. Relevant equations
$$Q_C = C \Delta V$$
$$V_C(t) = Q(t)/C = V_0(1-e^{-t/RC})$$

3. The attempt at a solution
I honestly am not really sure where to start with this. Given the formula above, it's clear that I need to calculate the voltage across the capacitor. But I only know how to calculate the voltage across something with a known resistance, using Ohm's Law. I guess my first question would be, how can I calculate the voltage across a capacitor? And once I have that, I can just use $$Q_C = C \Delta V$$ to calculate the charge.

edit: nevermind, I figured this out on my own.

#### Attached Files:

• ###### Picture1.png
File size:
4.3 KB
Views:
175
Last edited: Jan 28, 2009
2. Feb 7, 2016

### Staff: Mentor

The idea is to find the potential difference across the capacitor. From that the charge on it can be found.

At steady state there will be no current flowing to or from the capacitor; it is effectively an open circuit. So remove the capacitor from the circuit and determine the potential across the points where it was connected. This is easily done if we find them with respect to the reference node.

On the left we have a potential divider comprising a 10 V battery and two equal resistors. So the potential at their junction will be half the battery voltage, or 5 V. On the right side there is a battery in series with a resistor. Since there's no current flowing ("open" capacitor), there's no potential change across the resistor. So the potential at that capacitor terminal is equal to that battery voltage, or 3 V.

The capacitor "sees" the difference between those potentials, or 2 V. So the charge on the capacitor is, given by Q = CV, is

$Q = (0.01~μF)(2~V) = 20~nC$