Capacitors are connected together in parallel when both of its terminals are connected to each terminal of another capacitor
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:
VC1 = VC2 = VC3 = VAB = 12V
In the following circuit the capacitors, C1, C2 and C3 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, CT in the circuit is equal to the sum of all the individual capacitors added together. This is because the top plate of capacitor, C1 is connected to the top plate of C2 which is connected to the top plate of C3 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 m2.
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, CT 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 μF, nF or pF. Also, we can see that the current flowing through the total capacitance value, CT is the same as the total circuit current, iT
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 QT 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, CT. 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 CT as being:
CT = C1 + C2 + C3 = 0.1uF + 0.2uF + 0.3uF = 0.6uF
One important point to remember about parallel connected capacitor circuits, the total capacitance ( CT ) 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 CT = 0.6μF whereas the largest value capacitor is only 0.3μF.
When 4, 5, 6 or even more capacitors are connected together the total capacitance of the circuit CT 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 (μF) of the following capacitors when they are connected together in a parallel combination:
a) Total Capacitance,
CT = C1 + C2 = 47nF + 47nF = 94nF or 0.094μF
b) Total Capacitance,
CT = C1 + C2 = 470nF + 1μF
therefore, CT = 470nF + 1000nF = 1470nF or 1.47μF
So, the total or equivalent capacitance, CT 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.
What is the charge of the second capacitor (C2)
C1=6mf
C2=8mf
C3=4mf
V=12v
Please read our tutorial about Capacitance and Charge. Therefore: Q = CV = 8uF x 12V = 96uC
In a parallel circuit with capaciters C1 and C2. if C1 is greater than C2,what is the approximate total capacitance (CT)?
It is very educating
Happy and willing to learn with you
If you want to increase more current are capacitors connected should be in parallel or in series?
Capacitors are passive components, they do not increase current. For a DC voltage supply they will block DC current once fully charged. For an AC voltage supply their reactance, which is inversely proportional to the supply frequency, will determine the amount of AC current passed
This is the best for Nepalese students.
Thanks so much I love the explaination
Kisan kumar ganjam sural
Good.
Such a wonderful tutorials for anybody who’s confused to understand the lesson! Well done.
this site I a site which help students for learning and give them experience
What is the difference of Capacitor 25v,680uf between 25v,220uf 3 capacitors in parall.it is same funtion or not.pls,reply
3 x 220uF capacitors in parallel would give an equivalent capacitance of 660uF. Then you would effectively have a 25V, 660uF capacitor.
Types of parallel and series capacitors with examples
Thank you for sharing ideas
Great works
Thank you so much for sharing knowledge.Your article really help me.
8766652620
Study
A potential difference of 2v is connected to a uniform resistance wire of length 20m and cross sectional area of 8×10^-9m².A current of 0.10A flow in the wire calculate.
(1)the resistivity of the material
(2)the conductivity of the material.
Really!!, in a tutorial about Capacitors!. Make an effort to read and understand the tutorial about RESISTIVITY first.
Good job