# Charging a Capacitor with Another Capacitor

http://img230.imageshack.us/img230/8480/circuitbb5.th.gif [Broken]

Switch 1 is turned on until the 12 microF capacitor is completely charged. Then Switch 1 is switched off, while switch 2 is switched on.

How do you write a differential equation that may be used to find charge on the 4 mircroF capacitor as a function of time?

My attempt:

Applying Kirchhoff's Loop Rule: $$0 = V_1 - R_1 \frac{dq}{dt} - R_2 \frac{dq}{dt} - \frac{q}{C_2}$$, where q is the charge on the 4 microF capacitor.

This can be rewritten as
$$0 = \frac{Q-q}{C_1} - \frac{dq}{dt} (R_1 + R_2)- \frac{q}{C_2}$$

In this last step I assumed the charged discharged from the 12 microF capacitor is the charge that appears on the 4 microF capacitor.

(loops used are clockwise)

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## Answers and Replies

Defennder
Homework Helper
What does the $$\mbox{20M} \Omega$$ refer to? How can a voltage source have such a high resistance? The voltage provided by the source isn't given as well. Looks like there's something wrong with the picture.

Yeah that's a typo in the book. It should be 20MV.

The exact question asks:

a. What is the charge on the 4 microF capacitor after switch 2 is closed for 1 minute?

b. What is the final charge on the 2 capacitors?

alphysicist
Homework Helper
Hi breez,

I think your differential equation is okay; and you don't need the differential equation to solve for part b.

Oh thanks. Are these 2 capacitors in series or parallel? I think they should be in series right?

EDIT: Actually after solving this problem the capacitors end up with different charges and different voltages I find they are in fact parallel...

I guess as long as 2 capacitors are in a closed loop, they must be in parallel since their 2 ends are at the same 2 voltages.

This is kind of counterintuitive because they seem to be in series...

So another requirement for them to be in series is not only for them to flow into one another, but for their ends to be at a different voltage right?

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alphysicist
Homework Helper
They are not in series; in series capacitors have the positive plate of one capacitor attached to the negative plate of another.

Normally they are also not in parallel, since their voltages are not the same. When the current goes to zero, however, the potential differences across the resistors will vanish and at that point they can be treated as if they are in parallel.