# Voltage after connecting two capacitors

• cashmoney805
In summary, the 2\muF capacitor is initially charged by a 12 Volt battery, then is charged by a 30 Volt battery. The final charge on the 2\muF capacitor is 48 \muC.

## Homework Statement

A 2.00 $$\mu$$F capacitor is initially connected to a 12 Volt battery. In a separate circuit, a 4.00 $$\mu$$F capacitor is connected to a 30 Volt battery. The capacitors are then disconnected from their respective batteries and connected to each other (positive sides are connected together, and negative sides are connected together). What is the final charge on the 2$$\mu$$F capacitor

## Homework Equations

Qtot = Q1 + Q2
.5(C1 V12) + .5(C2 V22) = .5(C1+C2)V2

## The Attempt at a Solution

I'm not sure if you use conservation of charge or conservation of energy here. If I use conservation of charge, I get a common V of 24V and charge of 48$$\mu$$C on the 2$$\mu$$F capacitor.
If I use conservation of energy I get a common V of 25.5 V and a charge of 51$$\mu$$C on the 2$$\mu$$F capacitor.
The question has some words italicized which means they are probably key words, but I don't really get it. As always, any help is greatly appreciated!

You have a conservation of charge.

Figure the total charge.

Figure the equivalent C.

Figure the new Voltage, from the equivalent C and total charge.

Figure the charge across the C at the new voltage calculated.

LowlyPion said:
You have a conservation of charge.

Figure the total charge.

Figure the equivalent C.

Figure the new Voltage, from the equivalent C and total charge.

Figure the charge across the C at the new voltage calculated.
Qtot = (C1*V1) + (C2*V2) = 1.44 *10-4C
Ceq = 6$$\mu$$F
Qtot/Ceq = V = 24V

Q = C1*V = 4.8 * 10-5

I can do the math, but I don't understand it conceptually. When would you use conservation of energy?

cashmoney805 said:
I can do the math, but I don't understand it conceptually. When would you use conservation of energy?

There are only so many charges once everything is charged. You can't create them. They got to go somewhere. So apply the conservation of mass - as the conservation of charges.

Can't think of an example for using conservation of energy for this kind of problem. Voltage sources are usually doing work to keep things charged, and current flowing and resistors are dissipating power.

Hm. In my textbook there is a similar problem. One capacitor is charged by one source, and a second one is charged by a second source. They are disconnected from their batteries and then connected to each other. Find potential difference across the capacitors and charge on each capacitor.

The book solves that problem by using conservation of energy. Anyone see the difference?

## 1. What happens to the voltage after connecting two capacitors in series?

When two capacitors are connected in series, the total voltage across both capacitors is equal to the sum of the individual voltages. This means that the voltage after connecting two capacitors in series will be less than the voltage across each capacitor individually.

## 2. What is the equation for calculating the total voltage after connecting two capacitors in parallel?

The equation for calculating the total voltage after connecting two capacitors in parallel is Vtotal = V1 + V2, where V1 and V2 are the individual voltages across each capacitor.

## 3. How does the capacitance of the two capacitors affect the total voltage after connecting them?

The total voltage after connecting two capacitors is not affected by the capacitance of the individual capacitors. It is solely determined by the sum of the individual voltages.

## 4. Why does the voltage decrease when two capacitors are connected in series?

When two capacitors are connected in series, the total voltage is shared between the two capacitors. This is because the capacitors act like a voltage divider, with the voltage being divided between them based on their capacitance values. As a result, the overall voltage decreases.

## 5. Can the total voltage after connecting two capacitors ever be greater than the individual voltages?

No, the total voltage after connecting two capacitors will always be less than or equal to the individual voltages. This is a fundamental property of capacitors in series, where the total voltage is equal to the sum of the individual voltages.