Capacitors are said to be connected together “in parallel” when both of their terminals are respectively connected to each terminal of the other capacitor or capacitors. The voltage ( Vc ) connected across all the capacitors that are connected in parallel is **THE SAME**. Then, **Capacitors in Parallel** have a “common voltage” supply across them giving:

V_{C1} = V_{C2} = V_{C3} = V_{AB} = 12V

In the following circuit the capacitors, C_{1}, C_{2} and C_{3} are all connected together in a parallel branch between points A and B as shown.

When capacitors are connected together in parallel the total or equivalent capacitance, C_{T} in the circuit is equal to the sum of all the individual capacitors added together. This is because the top plate of capacitor, C_{1} is connected to the top plate of C_{2} which is connected to the top plate of C_{3} and so on.

The same is also true of the capacitors bottom plates. Then it is the same as if the three sets of plates were touching each other and equal to one large single plate thereby increasing the effective plate area in m^{2}.

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Since capacitance, C is related to plate area ( C = ε A/d ) the capacitance value of the combination will also increase. Then the total capacitance value of the capacitors connected together in parallel is actually calculated by adding the plate area together. In other words, the total capacitance is equal to the sum of all the individual capacitance’s in parallel. You may have noticed that the total capacitance of parallel capacitors is found in the same way as the total resistance of series resistors.

The currents flowing through each capacitor and as we saw in the previous tutorial are related to the voltage. Then by applying Kirchoff’s Current Law, ( KCL ) to the above circuit, we have

and this can be re-written as:

Then we can define the total or equivalent circuit capacitance, C_{T} as being the sum of all the individual capacitance’s add together giving us the generalized equation of

When adding together capacitors in parallel, they must all be converted to the same capacitance units, whether it is uF, nF or pF. Also, we can see that the current flowing through the total capacitance value, C_{T} is the same as the total circuit current, i_{T}

We can also define the total capacitance of the parallel circuit from the total stored coulomb charge using the Q = CV equation for charge on a capacitors plates. The total charge Q_{T} stored on all the plates equals the sum of the individual stored charges on each capacitor therefore,

As the voltage, ( V ) is common for parallel connected capacitors, we can divide both sides of the above equation through by the voltage leaving just the capacitance and by simply adding together the value of the individual capacitances gives the total capacitance, C_{T}. Also, this equation is not dependent upon the number of **Capacitors in Parallel** in the branch, and can therefore be generalized for any number of N parallel capacitors connected together.

So by taking the values of the three capacitors from the above example, we can calculate the total equivalent circuit capacitance C_{T} as being:

C_{T} = C_{1} + C_{2} + C_{3} = 0.1uF + 0.2uF + 0.3uF = 0.6uF

One important point to remember about parallel connected capacitor circuits, the total capacitance ( C_{T} ) of any two or more capacitors connected together in parallel will always be **GREATER** than the value of the largest capacitor in the group as we are adding together values. So in our example above C_{T} = 0.6uF whereas the largest value capacitor is only 0.3uF.

When 4, 5, 6 or even more capacitors are connected together the total capacitance of the circuit C_{T} would still be the sum of all the individual capacitors added together and as we know now, the total capacitance of a parallel circuit is always greater than the highest value capacitor.

This is because we have effectively increased the total surface area of the plates. If we do this with two identical capacitors, we have doubled the surface area of the plates which in turn doubles the capacitance of the combination and so on.

Calculate the combined capacitance in micro-Farads (uF) of the following capacitors when they are connected together in a parallel combination:

- a) two capacitors each with a capacitance of 47nF
- b) one capacitor of 470nF connected in parallel to a capacitor of 1uF

a) Total Capacitance,

C_{T} = C_{1} + C_{2} = 47nF + 47nF = 94nF or 0.094uF

b) Total Capacitance,

C_{T} = C_{1} + C_{2} = 470nF + 1uF

therefore, C_{T} = 470nF + 1000nF = 1470nF or 1.47uF

So, the total or equivalent capacitance, C_{T} of an electrical circuit containing two or more **Capacitors in Parallel** is the sum of the all the individual capacitance’s added together as the effective area of the plates is increased.

In our next tutorial about capacitors we look at connecting together Capacitors in Series and the affect this combination has on the circuits total capacitance, voltage and current.

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How does the internal energy of the capacitor changes as the voltage changes within it? THANK YOU FOR THE GOOD SERVICES

It’s a great service to know engineering solution.thanks a lot guy.pls some math will be publish about capacitor.

I need the derivtation of parallel plate capacitor

This is very fantastic site to be as far as excellence in physics is concerned……….

Im a little confused

1. if you hook identical capacitors in parallel….the total capacitance adds…is the working voltage additive also?

2. if you hook identical capacitors in series …the total capacitance is the reciprical…is the working voltage additive or what

ok 2 each 50 micro farad capacitors rated at 400 volts

in parallel they equal 100 micro farad at what voltage rating ? (400V)

in series they equal 25 micro farad at what voltage rating ? (800V)

Good site…. I missed my class bt now I’m perfectly know… What happen… Thnx..

Just want to know more about calculating capacitance.

Is it possible to connect capacitor of different voltage rating ie 2200u(25v)+220u(25v)+100u(40v) in parallel to make 4500u in order to replace the original circuit ,1500u(16v)+1500u(16v)+1500u(16v).

Please reply

Thanks.

Sorry it’s 2200uF(25v) not 220uF(25v) missed that zero.

Hello Rohan, Yes, the as long as the working voltage is greater, which in your case they are, 25VDC and 40VDC.

Really easy

fantastic guys what a great site,missed my class on inductance last week, but ive caught up with all the formulas and procedures….thanks