# Two Capacitors in series for two circuits

1. Nov 10, 2014

### andre220

1. The problem statement, all variables and given/known data
A capacitor of capacitance $C_1 = C$ is charged by a battery of potential difference $V_0$. After fully charged, it is disconnected from the batter and reconnected in series to a second, uncharged capacitor of capacitance $C_2 = C/2$ and another battery of potential difference $10V_0$. The positive side of the first capacitor is connected to the positive terminal of the battery. Calculate the final potential difference across each of the capacitors.

2. Relevant equations
For the second circuit, $C_{\mathrm{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3} C$ and also $Q = C_{\mathrm{eq}} 10V_0 = \frac{10}{3}V_0 C$.

3. The attempt at a solution
So then, normally from here it would be straightforward, however, the first part about $C_1$ being charged to only $V_0$ is throwing me off. I know it would affect the potential difference across $C_1$, but I can't quite see where that would fit in equation-wise.

2. Nov 10, 2014

### Staff: Mentor

Hint: For purposes of analysis you can replace the capacitor that has the initial charge and voltage $V_o$ with an uncharged capacitor of the same value in series with a voltage source with value $V_o$. This becomes the equivalent circuit model for that initially charged capacitor.

Alternatively you could determine the expression for the charge on the first capacitor and then work out what charge movements are required to satisfy KVL around the new loop (whatever change in charge occurs to one capacitor must occur to the other as well since they are in series).

3. Nov 10, 2014

### andre220

Okay so here is what I have: $10V_0 = V_1 + V_2$, and $V_2 = Q/C_2 = \frac{10}{3}V_0 C\frac{2}{C} = \frac{20}{3}V_0$ and thus because of KVL $V_1 = \frac{10}{3}V_0$. Doesn't feel like that's right though.

4. Nov 10, 2014

### Staff: Mentor

Doesn't look right to me either.

Start with the initial charge on the first capacitor. Call it q (and you should have an expression for q based upon the first capacitance and initial voltage). Then you're going to add some charge $\Delta q$ to each capacitor such that the total voltage of the two capacitors yields your new total potential difference. If you can find this $\Delta q$ you can work out the new potentials.

5. Nov 10, 2014

### andre220

Okay. I have for the first circuit $Q = CV_0$, then the second circuit (call it the primed circuit) $Q' = 10V_0 C_{\mathrm{eq}} = \frac{10}{3}C V_0$, Then the first capacitor is carrying $Q$ from the first circuit so that $\Delta Q = Q + Q' = \frac{13}{3} C V_0$.

6. Nov 10, 2014

### Staff: Mentor

No, ΔQ is not Q + Q'. ΔQ is the charge added to the first and second capacitor. Forget the equivalent capacitance, work with the individual capacitors. The new charge on the first capacitor is Q + ΔQ. The new charge on the second capacitor is just ΔQ. Write the expression for the total voltage across the series capacitors. This total must equal the new battery's potential difference.

7. Nov 10, 2014

### andre220

Okay, I think I have it.
$$\begin{eqnarray} 10V_0 &=& V_1 + V_2\\ &=&\frac{Q + \Delta Q}{C_1} + \frac{\Delta Q}{C_2} \\ &=& V_0 + \frac{\Delta Q}{C} + \frac{2 \Delta Q}{C} \\ & = & \frac{3\Delta Q}{C} + V_0 \end{eqnarray}$$
Then, $\Delta Q = 3 V_0 C$, then $V_1 = 4V_0$ and $V_2 = 6V_0$. And $V_1 + V_2 = 10V_0$ as a check.

8. Nov 10, 2014

### Staff: Mentor

Huzzah!

9. Nov 10, 2014