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## Physics library

### Course: Physics library>Unit 12

Lesson 2: Circuits with capacitors

# Capacitors in series

When capacitors are connected one after another, they are said to be in series. For capacitors in series, the total capacitance can be found by adding the reciprocals of the individual capacitances, and taking the reciprocal of the sum. Therefore, the total capacitance will be lower than the capacitance of any single capacitor in the circuit. . Created by David SantoPietro.

## Want to join the conversation?

• How can charges move between the two sides of a capacitor if they are separated? •  They don't move between the two sides.
A build up of negative charges on one side of the capacitor causes the negative charges on the other side to be pushed away or repelled from the edge of the capacitor. As more and more negative charges build up, more and more negative charges get repelled away from the other side and soon that side is very positive with little electrons. The electrons or negative charges keep on going in a circle.
• Why is the sum of the voltages of the capacitors equal to the voltage of the battery? • At , why is the charge stored on each of the individual capacitors equal to the charge stored on the equivalent capacitor? Why aren't the charges divided between the four- like each one has 192/4 C of charge? • NO, remember that the Capacitance unit is F, not C, So basically you messed up, you should NOT sum like this, they have the same amount of Charge NOT Capacitance.
So you add (1/48F) + (1/16F) + (1/96F) + (1/32F) = 0.125F, Then taking the reciprocal you get 8F which is the equivalent of CAPACITANCE.
When you try to find the Voltage you do this ( 192/48 ) + ( 192/16 ) + ( 192/96 ) + ( 192/32 ) = 24v which is the same voltage of the battery.
I Hope that helped!
(1 vote)
• If the voltage increases as charge increases, which increases as time goes on. How does the loop rule apply when the battery is just connected to the circuit and the capacitors haven't had enough time to build up charge yet (i.e. the sum of voltages across the capacitors wouldn't equal the batter voltage)? • I have a slightly off topic question, about Resistors being in series with a capacitor.
If the source Pd = the resistor Pd + the capacitor Pd, can it be said that the voltage across the resistor decreases as the charge increases on the capacitor (since Q is proportional to V) and as this is for charging, will discharging be : source Pd =resistor Pd - Capacitor Pd ? • nice question.

Lets take a step back a bit. The capacitor consists (as you know) of two plates separated by a dielectric...insultor. So, in a DC circuit (which you have here..) no current will flow once the capacitor is charged.
This means there will be zero voltage drop across the resistor when the capacitor is fully charged. and the the voltage across the charged resistor = source voltage.

during the charging process, the voltage drop across the resistor will be equal to the current at any time t multiplied by the resistance.
I would say that the The voltage across the capacitor will be source voltage - voltage drop across resistor.

Hope that helps

IM
• Does this mean that the higher the capacitance, the lower the voltage of a capacitor hooked up in a series? • At , why is the charge on the equivalent capacitor equal to the charge on EACH of the 3 capacitors? Why is it not divided by 3, so each capacitor holds 18/3= 6 C of charge? • Because charge is conserved. If you connect one capacitor to a battery, what happens? You get some positive charge on one plate and some negative charge on the other. The sum of the + and the - is 0. Now if you add another capacitor in series with the first one, the net charge is still going to be zero. You are going to have + charge on top plate of top capacitor, and - charge on bottom plate of bottom capacitor. What happens to the plates "in the middle". They have to have net charge of zero, too.

Note that it is a bit of a misconception to say that the capacitor "stores charge". It doesn't. The net charge on the capacitor is zero. What it does is hold separated charges separate.
(1 vote)
• why 1/Cequ = 1/c+1/c+1/c? why we do not use Cequ=c+c+c?
(1 vote) • So are capacitors what are used in backup generators, like when the power goes out? But obviously much larger so they can store more energy.
(1 vote) • I get mathematically why the charge on each of the capacitors is 18 but why wouldn't it conceptually be 18/3=6C? 