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Examples of Capacitance | Definition | Units | Calculation - ReadPhysics
Examples

Examples of Capacitance

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We express that what are the examples of capacitance? with definition, capacitance units and calculation of capacitance.

Capacitance examples of capacitance

The Capacitance is one of the properties of electric capacitors. Capacitors are devices that store electrical charge , and are commonly used in a wide variety of electrical circuits.

The capacitors are used, for example, to tune the frequency in radio receivers . Also, as filters in power sources . To eliminate sparking in automobile ignition systems . As energy storage devices in electronic flash units.

Basically, a capacitor consists of two conductors that have equal charges but of opposite sign. The capacitance given in a device depends on its geometry and the material that separates the conductors , called Dielectric.

Dielectric is an insulating material that has distinctive electrical properties, which can best be understood by considering the properties of atoms.

What is Capacitance?

To better understand Capacitance, two conductors having a Potential Difference V between them are considered, and the two conductors are assumed to have equal charges and opposite sign. examples of capacitance

This can be accomplished by connecting the two discharged conductors to the terminals of a battery. A combination of conductors thus charged is a device known as a Capacitor.

Capacitance applied in Radio-receivers

The Potential Difference V is proportional to the charge Q on the capacitor. The Capacitance C of a capacitor is defined as the ratio of the magnitude of the charge in either of the two conductors to the Potential Difference between them .

C = Q / V

By definition, Capacitance is always a positive quantity. Also, since the Potential Difference increases with increasing charge stored in the capacitor, the Q / V ratio is a constant for a given capacitor.

Therefore, the capacitance of a device is the measure of its ability to store charge and electrical potential energy.

Capacitance Units examples of capacitance

The Units of Capacitance in the International System are the Coulomb per Volt (C / V). The unit in the International System for Capacitance is the Farad (F), in honor of Michael Faraday. Namely:

Capacitance = 1 F = 1 C / V

The Farad is a very large unit of capacitance; In practice, typical devices have capacitances ranging from microfarads (1 μF = 10 -6 F) to picofarad (1 pF = 10 -12 F). As a practical note, the capacitors have labels that mark mF for microfarad and mmF for micromicrofarad (or picofarad).

Capacitance depends on the geometric arrangement of the conductors. To show this point, the Capacitance of an isolated spherical conductor of radius R and charge Q is determined. The second conductor can be considered as a concentric hollow spherical shell of infinite radius. Since the potential of the sphere is kQ / R (where V = 0 at infinity) its capacitance is given by:

C = Q / V

C = Q / (kQ / R)

C = R / k = 4πε 0 R

This shows that the capacitance of an isolated charged sphere is proportional to its radius and that it is independent of both charge and potential difference. For example, an insulated metallic sphere of radius 0.15m has a capacitance of:

C = 4πε 0 R = 4π (8.85×10 -12 C 2 / Nm 2 ) (0.15 m) = 17 pF

Calculation of Capacitance

The capacitance of a pair of conductors charged with opposite charges can be calculated as follows. A charge of magnitude Q is assumed, and the Potential Difference V is determined by experimental techniques.

Thus, C = Q / V is simply used to evaluate capacitance. As might be expected, calculating capacitance is relatively easy if the geometry of the capacitor is simple.

Three geometries are common in conductors: Two parallel plates, Two concentric cylinders and two concentric spheres. In these examples it will be assumed that the charge on the conductors is separated by vacuum, which is the dielectric.

Capacitor in physical form

Examples of Capacitance

1.- Calculate the Capacitance in a Capacitor whose conductors have a charge of 1.15×10 -10 Coulomb and a Potential Difference of 110V. examples of capacitance

C = Q / V

C = (1.15×10 -10 C) / (110 V)

C = 1.0454×10 -12 F

C = 1.0454 pF

2.- Calculate the Capacitance in a Capacitor whose conductors have a charge of 1.64×10 -10 Coulomb and a Potential Difference of 110V.

C = Q / V

C = (1.64×10 -10 C) / (110 V)

C = 1.4909×10 -12 F

C = 1.4909 pF

3.- Calculate the Capacitance in a Capacitor whose conductors have a charge of 1.38×10 -10 Coulomb and a Potential Difference of 110V.

C = Q / V

C = (1.38×10 -10 C) / (110 V)

C = 1.2545×10 -12 F

C = 1.2545 pF

4.- Calculate the Capacitance in a Capacitor whose conductors have a charge of 1.42×10 -10 Coulomb and a Potential Difference of 110V.

C = Q / V

C = (1.42×10 -10 C) / (110 V)

C = 1.2909×10 -12 F

C = 1.2909 pF

5.- Calculate the Capacitance in a Capacitor whose conductors have a charge of 1.08×10 -10 Coulomb and a Potential Difference of 110V.

C = Q / V

C = (1.08×10 -10 C) / (110 V)

C = 0.9818×10 -12 F

C = 0.9818 pF

6.- Calculate the Capacitance in a Capacitor whose conductors have a charge of 1.65×10 -10 Coulomb and a Potential Difference of 110V.

C = Q / V

C = (1.65×10 -10 C) / (110 V)

C = 1.5×10 -12 F

C = 1.5 pF

7.- Calculate the Capacitance in a Capacitor whose conductors have a charge of 2.02×10 -10 Coulomb and a Potential Difference of 110V.

C = Q / V

C = (2.02×10 -10 C) / (110 V)

C = 1.8363 x10 -12 F

C = 1.8363 pF

8.- Calculate the Capacitance in a Capacitor whose conductors have a charge of 2.14×10 -10 Coulomb and a Potential Difference of 110V.

C = Q / V

C = (2.14×10 -10 C) / (110 V)

C = 1.9454 x10 -12 F

C = 1.9454 pF

9.- Calculate the Capacitance in a Capacitor whose conductors have a charge of 2.56×10 -10 Coulomb and a Potential Difference of 110V.

C = Q / V

C = (2.56×10 -10 C) / (110 V)

C = 2.3272×10 -12 F

C = 2.3272 pF

10.- Calculate the Capacitance in a Capacitor whose conductors have a charge of 2.80×10 -10 Coulomb and a Potential Difference of 110V.

C = Q / V

C = (2.80×10 -10 C) / (110 V)

C = 2.5454×10 -12 F

C = 2.5454 pF

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