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Since JEE Main 2020 is approaching, candidates must be preparing hard for the entrance examination. To help and assist them in their preparation, we have provided the study notes on the topic Capacitors. The first step in preparing for JEE Main should be to develop an understanding of the chapters and revising the notes before going for the exam.
Electric Potential and Capacitors are a part of Electrostatics and Magnetism. This Chapter has a weightage of 3-4% of the total chapters that are included under this Unit.
The questions from this chapter are mainly asked from the major topics such as Effect of Dielectric on Capacitance, Capacitors, and Capacitance, Van De Graff Generator, Combination of Capacitor, etc.
In this article we have covered the important topics, sample questions to help JEE Main aspirants, and to score well in the entrance examination.
These are electric-condensers that have two passive terminals. Capacitors have the energy storing ability in the form of electric charges. This ability is referred to as Capacitance (C).
Where Q refers to charge, V refers to voltage, and C is a constant that is independent of charge and depends on the alignment of conductors.
For example, when a capacitor is connected to a battery of 12 V, it will store 12 coulombs of electric charge, thus its capacitance will be as follows:
C= 12/12= 1 farad
Candidates can check the list of various uses of Capacitors mentioned below:
It helps in Storing electric potential energy such as batteries.
Used in the audio system of the vehicle.
Capacitors play a major role in filtering out unwanted frequency signals.
Delaying voltage changes when coupled with resistors
It is also used as a sensing device.
Used to separate AC and DC
Parallel Plate Capacitor consists of two large plane parallel conducting plates separated by a small distance and charge is distributed along the conducting surface of the plates. These parallel plate capacitors can be arranged in a series leading to parallel combinations. Considering C1 and C2 as two capacitors arranged in series, the net capacitance can be presented as follows:
CNet = C1 + C2
If n number of capacitors are connected in parallel, then the net capacitance will be as follows:
CNet = C1 + C2 + C3 + ....... + Cn
The capacitance of a parallel plate capacitor is
The capacitors have two electrical conductors, i.e. surface separated by a dielectric and within the metallic plates. This may be in different forms, such as metal, electrolytes, thin film, etc. The dielectric is made up of non-conducting materials like glass, paper, mica, plastic film, or ceramic, which assists in increasing the capacitor's charging ability.
Two different capacitors can have the same capacitance; however, its voltage rating will always differ. For example, taking into consideration two capacitors that have two different voltage ratings, i.e. one with a high and the other with a low voltage rating, if we substitute the small voltage rated capacitor with a high rated capacitor, there will be an increase in the voltage. This will result in damage of the capacitor. Therefore, there are common DC voltages of capacitors, such as 10, 16, 25, 35, 50, 63, 100, 160, 250, 400 and 1000V. These ratings are printed on the capacitor.
If two capacitors C1 and C2 are connected in parallel then the net Capacitance is
If n capacitors are connected in parallel then the net capacitor is
If two capacitors C1 and C2 are connected in series then the net capacitance is
If n capacitors are connected in series then the next capacitance is
Dielectric constant k, if inserted within two capacitor plates, then the net capacitance will be as follows:
Here, is the dielectric constant and Co refers to capacitance without the dielectric constant.
When work is done in moving a positive charge to a negative charge in the presence of a repulsive force, this results in the storage of energy in the capacitor's plates. This can be represented by the following equation:
Where, Q = charge and C = capacitance
Some of the important features have been listed below:
Nominal capacitance (C): The units of measurement are pico-Farads (pF), nano-Farads
(nF) and micro-Farads (µF). You can find this number printed on the capacitor.
Working voltage (WV): It is defined as the total current, whether Direct or Indirect current, passing through the capacitor without causing any failure in the whole lifetime of the capacitor.
Tolerance (±%): There is a tolerance rating assigned for all capacitors. This varied from minus to plus value. The unit of measurement is usually in ±pF in case of low-value capacitors (usually less than 100 pF). For high-value capacitors, it is usually more than 100 pF.
Leakage current: This refers to a small DC flow in Nano-Amps or nA. This is generated when the electrons pass through the dielectric medium and come in contact with it. If we remove the supply voltage to the capacitor, then the leakage current will lead to its complete discharge.
These are also known as Electric Double-Layer Capacitors (EDLC), Gold cap, or Supercap. It has a high capacitance value associated with it when compared to other capacitors. But its voltage limit is low.
They can be classified into two broad categories, i.e. Double layer and Pseudo-capacitors. In case of the double-layer capacitors, the charge is stored electrostatically and in case of pseudo-capacitors, the charge is stored electrochemically. The double-layer consists of activated carbons, carbon nanotubes (CNT) and carbon aerogels, whereas the pseudocapacitors contain conducting polymers and metal oxides. Both these capacitors can be connected into a hybrid capacitor that can store charge both electrostatically and electrochemically. It consists of asymmetric pseudo or EDLC, composite, and is the rechargeable battery type.
This is an electrostatic generator that has the ability to generate large static electric potential, which can be few million volts. The construction consists of a moving belt for storing the charge in a hollow metal. This hollow metal is like a globe that is placed over an insulating column. This belt carries a positive charge from ground to the top of the column. Then this positive charge gets transferred to the hollow metal globe using a carbon brush. This process continues until the breakdown field of air is reached.
Question: Find the total capacitance for three capacitors connected in series, given their individual capacitances are 1.000μF, 5.000μF, and 8.000μF.
1CNet=11.000 + 15.000 + 18.000 = 1.325μF
On inverting the equation, we get:
CNet = μF1.325 = 0.755 μF
Question: Find the net capacitance for three capacitors connected in parallel, given their individual capacitances are 15.0μF , 25.0μF , and 35.0μF.
C = C1 + C2 + C3 = 15.0 + 25.0 + 35.0 = 75 μF
Question: Five capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 6 μF, C4 = 5 μF, C5 = 10 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination.
Capacitors connected in parallel:
Cp = C3 + C2
= 4 + 6 = 10 µF
Capacitors connected in series:
1/Cnet = 1/C1 + 1/Cp + 1/C4 + 1/C5
= 1/2 + + 1/10 + 1/5 + 1/10
Cnet = 10/9 µF