Electric Filed Strength And Electric Flux Density Engineering Essay

July 19, 2017 Engineering

All organic structures are made up of atoms, which consist of a nucleus containing protons ( +ve ) and neutrons ( impersonal ) and environing the karyon are revolving negatrons ( -ve ) .

When a organic structure is uncharged it is electrically impersonal, it has the same negative charge as positive charge.

If a music director had a shortage of negatrons it would exhibit a net positive charge and if it was to hold a excess of negatrons it would exhibit a net negative charge ( retrieve the old survey of the atom mention +ve/-ve ions ) . An instability in charge can be produced by clash ( taking or lodging negatrons utilizing stuffs such as silk and pelt, severally ) or initiation ( by pulling or driving negatrons utilizing a 2nd organic structure which is, severally, positively or negatively charged ) .

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Coulomb ‘s Law states that if charged organic structures exist at two points, the force of attractive force ( if the charges are of opposite mutual opposition ) or repulsive force ( if the charges have the same mutual opposition ) will be relative to the merchandise of the magnitude of the charges divided by the square of their distance apart. Frankincense:

+

+

+

“ Direct & A ; Inverse Proportionality ” – Mathematicss

Q1 and Q2 are the charges present at the two points ( in Coulombs ) , d is the distance dividing the two points ( in meters ) , F is the force ( in Newtons ) , and K is a mathematical invariable depending upon the medium in which the charges exist.

In a vacuity or ‘free infinite ‘ ,

Iµ0 is the permittivity of free infinite ( 8.854 x 10-12 F/m – Farad per metre ) .

The force exerted on a charged atom is a manifestation of the being of an electric field. The electric field defines the way and magnitude of a force on a charged object. The field itself is unseeable to the human oculus but can be drawn by building lines which indicate the gesture of a free positive charge within the field ; the figure of field lines in a peculiar part being used to bespeak the comparative strength of the field at the point in inquiry.

The figure above shows the electric Fieldss between charges of the same and opposite mutual opposition.

The figure below shows the field which exists between two charged parallel home bases.

Bacillus

A

As illustrated above, plates A and B are doped and charged to different potencies. If an negatron that has a negative charge is placed between the home bases, a force will move on the negatron be givening to force it off from the negative home base B and towards the positive home base A. Similarly, a positive charge would be acted on by a force be givening to travel it toward the negative home base.

The part between the home bases in which an electric charge experiences a force, is called an electrostatic field. The way of the field is defined by the force moving on a positive charge placed in the field, i.e. the way of the force is from the positive home base to the negative home base.

Such a field may be represented in magnitude and way by lines of electric force drawn between the charged surfaces. The intimacy of the lines is an indicant of the field strength. Whenever a p.d. is established between two points, an electric field will ever be.

The figure above shows two parallel carry oning home bases separated from each other by air, and are connected to opposite terminuss of a battery of electromotive force V Vs. There is therefore an electric field in the infinite between the home bases. If the home bases are close together, the electric lines of force will be straight and parallel and every bit spaced, except near the border where fringing will happen ( see old figure ) . Over the country in which there is negligible fringing,

Tocopherol is the electric field strength ( V/m ) , V is the applied possible difference across the parallel home bases ( V ) and vitamin D is the distance ( m ) .

**Note: Electric Field Strength is besides called Potential/Voltage Gradient.

A unit electric flux is defined as emanating from a positive charge of 1 C. Thus electric flux I? is measured in C, and for a charge of Q C, the electric flux I? is equal to Q C. Electric flux denseness D is the sum of flux go throughing through a defined country A that is perpendicular to the way of the flux:

I? is the electric flux measured in C, Q is the electric charge besides measured in C, and A is the country in M2 over which the flux is distributed.

Problem 1:

Two parallel rectangular home bases mensurating 20cm by 40cm carry an electric charge of 0.2 AµC. ( a ) Calculate the electric i¬‚ux denseness.

( B ) If the home bases are spaced 5mm apart and the electromotive force between them is 0.25 kilovolt find the electric field strength.

Solution 1:

PERMITTIVITY

At any point in an electric field, the electric field strength E maintains the electric flux and produces a peculiar value of electric flux denseness D at that point. For a field established in vacuity ( or for practical intents in air ) , the ratio D/E is a changeless Iµ0, i.e.

Iµ0 is called the permittivity of free infinite or the free infinite invariable.

The value of Iµ0 is 8.854 ten 10-12 F/m – Farad per metre.

When a insulator ( i.e. insulating medium dividing charged surfaces ) , such as isinglass, paper, plastic or ceramic is introduced into the part of an electric field, the ratio of D/E is modified.

Iµr is called the comparative permittivity of the insulating stuff and indicates its insulating power compared with that of vacuity.

Iµr has no units and typical belongingss of some common insulating dielectric stuffs are shown below.

The merchandise of Iµ0 Iµr is called the absolute permittivity, Iµ , i.e.

As discussed earlier, the insulator is an insulating medium dividing charged surfaces and has the belongings of really high electric resistance. They are hence used to separate music directors at different potencies, such as capacitance home bases or electric power lines.

The dielectric strength of an insulating insulator is the maximal electric field strength that can safely be applied to it before dislocation ( conductivity ) occurs.

The sum of charge produced for a given applied electromotive force on the two analogue home bases shown before will depend non merely on the physical dimensions but besides on the insulating dielectric stuff that appears between the home bases. Such stuffs need to hold a really high value of electric resistance ( i.e. they must non carry on charge ) coupled with an ability to defy high electromotive forces without interrupting down.

A more practical agreement of parallel home bases with an insulating dielectric stuff is shown.

In this agreement the ratio of charge, Q, to the possible difference, V, is given by the undermentioned relationship.

A = country of one on the home bases, in M2

D = thickness of the insulator in m

Iµ = absolute permittivity of the dielectric stuff

*Later acquisition, i.e. the parallel home base capacitor/capacitance and physical dimensions.

aˆ¦aˆ¦aˆ¦ individual brace of home bases

aˆ¦aˆ¦aˆ¦ agreement of ‘n ‘ home bases

Problem 1:

The i¬‚ux denseness between two home bases separated by isinglass of comparative permittivity 5 is 2AµC/m2. Find the electromotive force gradient between the home bases.

Solution 1:

Problem 2:

Two parallel home bases holding a p.d. of 200V between them are separated 0.8mm apart.

What is the electric i¬?eld strength?

Find besides the electric i¬‚ux denseness when the insulator between the home bases is ( a ) air, and ( B ) polyethylene of comparative permittivity 2.3

Solution 2:

SELF ASSESSMENT ( 1-2 )

Note: Where appropriate take Iµ0 as 8.85 ten 10-12 F/m

A capacitance uses a dielectric 0.04mm midst and operates at 30V. What is the electric field strength across the insulator at this electromotive force? [ Answer: 750kV/m ]

A two-plate capacitance has a charge of 25C. If the effectual country of each home base is 5cm2 determine the electric i¬‚ux denseness of the electric field. [ Answer: 50 kC/m2 ]

A charge of 1.5AµC is carried on two parallel rectangular home bases each mensurating 60mm by 80mm. ( a ) Calculate the electric i¬‚ux denseness. ( B ) If the home bases are spaced 10mm apart and the electromotive force between them is 0.5kV determine the electric i¬?eld strength.

[ Answer: ( a ) 312.5AµC/m2, ( B ) 50kV/m ]

Two parallel home bases are separated by a insulator and charged with 10AµC. Given that the country of each home base is 50cm2, calculate the electric i¬‚ux denseness in the dielectric dividing the home bases. [ Answer: 2mC/m2 ]

The electric i¬‚ux denseness between two home bases separated by polystyrene of comparative permittivity 2.5 is 5AµC/m2. Find the electromotive force gradient between the home bases.

[ Answer: 226kV/m ]

Two parallel home bases holding a p.d. of 250V between them are separated 1mm apart.

( a ) Determine the electric i¬?eld strength.

( B ) Find besides the electric i¬‚ux denseness when the insulator between the home bases is

( I ) air and ( two ) isinglass of comparative permittivity 5.

[ Answer: ( a ) 250kV/m ( Bi ) 2.213AµC/m2 ( bii ) 11.063AµC/m2 ]

CAPACITORS & A ; CAPACITANCE

A capacitance is a device for hive awaying electric charge. In consequence, it is a reservoir into which charge can be deposited and so subsequently extracted. In its simplest signifier a capacitance consists of two parallel metal home bases which are separated by an insulating stuff known as a insulator.

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Because of the insulator, current can non flux from one home base to the other. When the capacitance is connected to a District of Columbia beginning, negatrons accumulate on the home base connected to the negative supply terminus. The negative charge repels negatrons from the atoms of the other home base. These negatrons flow off to the positive terminus of the District of Columbia beginning ; this leaves the home base positively charged.

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If the capacitance is disconnected from the supply, the charges remain. The capacitance shops the electric charge indefinitely.

The symbols for a fixed capacitance and a variable capacitance used in electrical circuit diagrams are shown below.

Typical applications include reservoir and smoothing capacitances for usage in power supplies, matching a.c. signals between the phases of amplifiers, and uncoupling supply tracks ( i.e. efficaciously anchoring the supply rails every bit far as a.c. signals are concerned ) .

The undermentioned figures illustrate what happens to a capacitance when it is bear downing and dispatching.

If the switch is left unfastened ( place A ) , no charge will look on the home bases and in this status there will be no electric field in the infinite between the home bases nor will at that place be any charge stored in the capacitance.

When the switch is moved to place B, negatrons will be attracted from the positive home base to the positive terminus of the battery. At the same clip, a similar figure of negatrons will travel from the negative terminus of the battery to the negative home base. This sudden motion of negatrons will attest itself in a fleeting rush of current ( conventional current will flux from the positive terminus of the battery towards the positive terminus of the capacitance ) .

Finally, adequate negatrons will hold moved to do the e.m.f. between the home bases the same as that of the battery. In this province, the capacitance is said to be to the full charged and an electric field will be present in the infinite between the two home bases.

If, at some ulterior clip the switch is moved back to place A, the positive home base will be left with a lack of negatrons whilst the negative home base will be left with a excess of negatrons. Furthermore, since there is no way for current to flux between the two plates the capacitance will stay charged and a possible difference will be maintained between the home bases.

Now assume that the switch is moved to place C. The extra negatrons on the negative home base will flux through the resistance to the positive home base until a impersonal province one time once more exists ( i.e. until there is no extra charge on either home base ) . In this province the capacitance is said to be to the full discharged and the electric field between the home bases will quickly fall in. The motion of negatrons during the discharging of the capacitance will once more ensue in a fleeting rush of current ( current will flux from the positive terminus of the capacitance and into the resistance ) .

The figure below shows the way of current flow during bear downing ( i.e. the switch in place B ) and dispatching ( i.e. the switch in place C ) . It should be noted that current flows momently in both circuits even though you may believe that the circuit is broken by the spread between the capacitance plates!

The charge Q ( in C ) stored in a capacitance is given by:

I is the current in amperes and T is the clip in seconds.

Charge Q on a capacitance is relative to the applied electromotive force V, i.e. Q V.

“ Direct & A ; Inverse Proportionality ” – Mathematicss

Q = CV

The invariable of proportionality C is the electrical capacity.

The unit of electrical capacity C is the farad F ( or more normally AµF =10-6F or pF =10-12F ) , and is defined as the electrical capacity when a p.d. of one V appears across the home bases when charged with one C.

Capacitance is the ability of a circuit or object ( i.e. in this instance a capacitance ) to hive away electric charge.

Problem 1:

( a ) Determine the p.d. across a 4 AµF capacitance when charged with 5 megahertzs

( B ) Find the charge on a 50 pF capacitance when the electromotive force applied to it is 2 kilovolt.

Solution 1:

Problem 2:

A direct current of 4A flows into a antecedently uncharged 20 AµF capacitance for 3 MS. Determine the p.d. between the home bases.

Solution 2:

Problem 3:

A 5AµF capacitance is charged so that the p.d. between its home bases is 800V. Calculate how long the capacitance can supply an mean discharge current of 2 ma.

Solution 3:

SELF ASSESSMENT ( 3 )

Find the charge on a 10 AµF capacitance when the applied electromotive force is 250 V. ( Answer: 2.5 megahertz )

Determine the electromotive force across a 1000I?F capacitance to bear down it with 2 AµC. ( Answer: 2 kilovolt )

The charge on the home bases of a capacitance is 6 megahertz when the potency between them is 2.4 kilovolt. Determine the electrical capacity of the capacitance. ( Answer: 2.5 AµF )

For how long must a bear downing current of 2 A be fed to a 5 AµF capacitance to raise the p.d. between its home bases by 500V. ( Answer: 1.25 MS )

A direct current of 10 A flows into a antecedently uncharged 5 AµF capacitance for 1 MS. Determine the p.d. between the home bases. ( Answer: 2 kilovolt )

A 16 AµF capacitance is charged at a changeless current of 4 AµA for 2 proceedingss. Determine the concluding p.d. across the capacitance and the corresponding charge in C. ( Answer: 30V, 480 AµC )

A steady current of 10 A flows into a antecedently uncharged capacitance for 1.5 MS when the p.d. between the home bases is 2 kilovolt. Find the electrical capacity of the capacitance. ( Answer: 7.5AµF )

CAPACITANCE AND PHYSICAL DIMENSIONS ( Conventional Parallel Plate Capacitor )

The electrical capacity of a capacitance depends upon the physical dimensions of the capacitance ( i.e. the size of the home bases and the separation between them ) and the dielectric stuff between the home bases. The electrical capacity of a conventional parallel home base capacitance is given by:

Where, C = Capacitance, unit of step Fs ( F )

Iµ0 = Permittivity of free infinite or the free infinite invariable ( 8.85 x 10-12 F/m )

Iµr = Relative permittivity of the dielectric medium between the home bases

( Iµr has no units as it is a ratio of denseness material/vacuum )

A = Area of one of the home bases ( M2 )

vitamin D = Thickness of the insulator or separation between the home bases ( m )

In order to increase the electrical capacity of a capacitance, many practical constituents employ multiple home bases as shown.

Ten home bases are shown, organizing nine capacitances with a electrical capacity nine times that of one brace of home bases.

Such an agreement has n home bases so capacitance C a?? ( n -1 ) . Thus electrical capacity is so given by:

Problem 1:

A ceramic capacitance has an effectual home base country of 4cm2 and separated by 0.1 millimeters of ceramic of comparative permittivity 100. Calculate the electrical capacity of the capacitance in picofarads ( I?F ) .

If the capacitance in portion ( a ) is given a charge of 1.2AµC what will be the p.d. between the home bases?

Solution 1:

Problem 2:

A waxed paper capacitance has two analogue home bases, each of effectual country 800 cm2. If the electrical capacity of the capacitance is 4425 pF find the effectual thickness of the paper if its comparative permittivity is 2.5.

Solution 2:

Problem 3:

A parallel home base capacitance has nineteen interleaved home bases each 75 millimeter by 75 millimeters and separated by mica sheets 0.2 millimeters thick. Assuming that the comparative permittivity of the isinglass is 5, calculate the electrical capacity of the capacitance.

Solution 3:

n = 19, therefore ( n – 1 ) = 18

A = 75 ten 75 = 5625mm2

Iµr = 5, Iµ0 = 8.85 ten 10-12 F/m

vitamin D = 0.2mm = 0.2 x 10-3m

SELF ASSESSMENT ( 4 )

** Where appropriate take Iµ0 as 8.85 ten 10-12 F/m.

A capacitance consists of two parallel home bases each of country 0.01 M2, spaced 0.1 millimeter in air. Calculate the electrical capacity in picofarads ( pF ) . [ Answer: 885 pF ]

A waxed paper capacitance has two analogue home bases, each of effectual country 0.2m2. If the electrical capacity is 4000 pF find the effectual thickness of the paper if its comparative permittivity is 2. [ Answer: 0.885 millimeter ]

Calculate the electrical capacity of a parallel home base capacitance holding 5 home bases, each 30 millimeter by 20 millimeters and separated by a dielectric 0.75 millimeter midst holding a comparative permittivity of 2.3. [ Answer: 65.14 pF ]

How many home bases does a parallel home base capacitance have if its electrical capacity is 5nF, each home base is 40mm by 40mm and each insulator is 0.102mm midst with a comparative permittivity of 6? [ Answer: 7 ]

A parallel home base capacitance is made from 25 home bases, each 70mm by 120mm interleaved with isinglass of comparative permittivity 5. If the electrical capacity of the capacitance is 3000pF determine the thickness of the mica sheet. [ Answer: 2.97mm ]

The electrical capacity of a parallel home base capacitance is 1000pF. It has 19 home bases, each 50mm by 30mm separated by a insulator of thickness 0.40mm. Determine the comparative permittivity of the insulator. [ Answer: 1.67 ]

CAPACITORS CONNECTED IN PARALLEL AND SERIES

CAPACITORS CONNECTED IN PARALLEL

The figure above shows three capacitances, C1, C2 and C3 connected in analogue with a supply electromotive force V applied across the agreement. ( Note: merely like resistances in analogue, the supply electromotive force V is the same across each parallel capacitance )

V = V1 = V2 = V3

When the bear downing current I reaches point Angstrom it divides, some fluxing into C1, some fluxing into C2 and some into C3. Hence the entire charge QT ( i.e. QT= I x T ) is divided between the three capacitances. The capacitances each shop a charge and these are shown as Q1, Q2 and Q3 severally. Hence,

But, QT=CV ( where C is the entire tantamount circuit electrical capacity )

And, Q1=C1V

Q2=C2V

Q3=C3V

Therefore, CV = C1V + C2V + C3V ( where C is the entire tantamount circuit electrical capacity )

Dividing throughout by the common V giving,

C = C1 + C2 + C3 aˆ¦aˆ¦ . + Cn

The tantamount electrical capacity of a group of analogue connected capacitances is the amount of the electrical capacities of the single capacitances.

CAPACITORS CONNECTED IN SERIES

The figure above shows three capacitances, C1, C2 and C3 connected in series across a supply electromotive force V. Let the p.d. across the single capacitances be V1, V2 and V3 severally as shown.

Let the charge on the home base ‘a ‘ of the capacitance C1 be +Q C. This induces and equal but opposite charge of -Q Cs on home base ‘b ‘ . The music director between home bases ‘b ‘ and ‘c ‘ is electrically isolated from the remainder of the circuit so that an equal but opposite charge of +Q Cs must look on home base ‘c ‘ , which, in bend, induces an equal and opposite charge of -Q Cs on home base ‘d ‘ , and so on.

Hence when capacitances are connected in series the charge on each is the same.

QT = Q1 = Q2 = Q3

In a series circuit: V = V1 + V2 + V3 ( Similar to resistances in series )

Since, so ( where C is the entire tantamount circuit electrical capacity )

Dividing throughout by the common Q giving,

( Where C is the entire tantamount circuit electrical capacity )

For series connected capacitances, the reciprocal of the tantamount electrical capacity is equal to the amount of the reciprocals of the single electrical capacity.

For particular instance of two capacitances in series,

Hence,

i.e.

Problem 1:

Calculate the tantamount electrical capacity of two capacitances of 6I?F and 4I?F connected in

( a ) Analogue

( B ) Series.

Solution 1:

Problem 2:

What electrical capacity must be connected in series with a 30I?F capacitance for the tantamount electrical capacity to be 12I?F?

Solution 2:

Problem 3:

Capacitances of 1I?F, 3I?F, 5I?F and 6I?F are connected in analogue to a direct electromotive force supply of 100V. Determine ( a ) the tantamount circuit electrical capacity, ( B ) the entire charge and ( degree Celsius ) the charge on each capacitance.

Solution 3:

Problem 4:

Capacitances of 3I?F, 6I?F and 12I?F are connected in series across a 350V supply. Calculate ( a ) the tantamount circuit electrical capacity, ( B ) the charge on each capacitance, and ( degree Celsius ) the p.d. across each capacitance.

Solution 4:

Problem 5:

For the agreement shown, find ( a ) the tantamount electrical capacity of the circuit, ( B ) the electromotive force across QR and ( degree Celsius ) The charge on each capacitance.

Solution 5:

SELF ASSESSMENT ( 5 )

Capacitors of 2AµF and 6AµF are connected ( a ) in analogue and ( B ) in series. Determine the tantamount electrical capacity in each instance. [ Answers: ( a ) 8I?F ( B ) 1.5I?F ]

Find the electrical capacity to be connected in series with a 10AµF capacitance for the tantamount electrical capacity to be 6AµF. [ Answer: 15I?F ]

What value of electrical capacity would be obtained if capacitances of 0.15AµF and 0.10AµF are connected in ( a ) series and ( B ) analogue? [ Answers: ( a ) 0.06I?F ( B ) 0.25I?F ]

Two 6AµF capacitances are connected in series with one holding a electrical capacity of 12AµF. Find the entire tantamount circuit electrical capacity. What electrical capacity must be added in series to obtain a electrical capacity of 1.2AµF? [ Answers: ( a ) 2.4I?F ( B ) 2.4I?F ]

For the agreement shown below, happen ( a ) the tantamount circuit electrical capacity and ( B ) the electromotive force across a 4.5I?F capacitance. [ Answers: ( a ) 1.2I?F ( B ) 100V ]

Three 12AµF capacitances are connected in series across a 750V supply. Calculate ( a ) the tantamount electrical capacity, ( B ) the charge on each capacitance and ( degree Celsius ) the p.d. across each capacitance. [ Answers: ( a ) 4AµF ( B ) 3mC ( degree Celsius ) 250V ]

If two capacitances holding electrical capacities of 3AµF and 5AµF severally are connected in series across a 240V supply, determine ( a ) the p.d. across each capacitance and ( B ) the charge on each capacitance. [ Answers: ( a ) 150V, 90V ( B ) 0.45 megahertz on each ]

Capacitances of 4AµF, 8AµF and 16AµF are connected in parallel across a 200V supply. Determine ( a ) the tantamount electrical capacity, ( B ) the entire charge and ( degree Celsius ) the charge on each capacitance. [ Answers: ( a ) 28 AµF ( B ) 5.6mC ( degree Celsius ) 0.8mC, 1.6mC, 3.2mC ]

DIELECTRIC STRENGTH

The maximal safe working electromotive force is the maximal electromotive force that can be applied to the terminuss of a capacitance without doing harm to the capacitance.

The maker specifies this electromotive force. The bound is necessary so that the field strength in the insulator does non transcend a value that would do the insulator to breakdown and free its insulating belongingss. The figure quoted by the maker for a capacitance is besides known as the dielectric strength and will be in Vs per meter.

Tocopherol is the dielectric strength ( V/m ) , V is the applied possible difference across the parallel home bases ( V ) and vitamin D is the distance ( m ) .

**Note: Equation identical to Electric Field Strength ( Potential/Voltage Gradient ) .

Problem1:

A capacitance is to be constructed so that its electrical capacity is 0.2AµF and to take a p.d. of 1.25kV across its terminuss. The insulator is to be mica and has a dielectric strength of 50MV/m. Find ( a ) the thickness of the isinglass needed, and ( B ) the country of a home base presuming a two-plate building. ( Assume Iµr for isinglass to be 6 ) .

Solution 1:

ENERGY STORED IN CAPACITORS

The energy, W, stored by a capacitance is given by,

Where,

W is the energy ( in Joules ) ,

C is the electrical capacity ( in Farads ) , and

V is the possible difference ( in Volts ) .

Problem 1:

( a ) Determine the energy stored in a 3AµF capacitance when charged to 400V.

( B ) Find besides the mean power developed if this energy is dissipated in a clip of 10Aµs.

Solution 1:

Problem 2:

A 12AµF capacitance is required to hive away 4J of energy. Find the p.d. to which the capacitance must be charged.

Solution 2:

Problem 3:

A capacitance is charged with 10mC. If the energy stored is 1.2J, determine ( a ) the electromotive force and ( B ) the electrical capacity.

Solution 3:

SELF ASSESSMENT ( 6 )

** Where appropriate take Iµ0 as 8.85 ten 10-12 F/m.

When a capacitance is connected across a 200V supply the charge is 4AµC. Find ( a ) the electrical capacity and ( B ) the energy stored. [ Answer: ( a ) 0.02AµF ( B ) 0.4mJ ]

Find the energy stored in a 10AµF capacitance when charged to 2kV. [ Answer: 20 J ]

A 3300pF capacitance is required to hive away 0.5mJ of energy. Find the p.d. to which the capacitance must be charged. [ Answer: 550 V ]

A capacitance is charged with 8mC. If the energy stored is 0.4J, determine ( a ) the electromotive force and ( B ) the electrical capacity. [ Answer: ( a ) 100V ( B ) 80 AµF ]

A capacitance, dwelling of two metal home bases each of country 50 cm2 and spaced 0.2mm apart in air, is connected across a 120V supply. Calculate ( a ) the energy stored ( B ) the electric i¬‚ux denseness and ( degree Celsius ) the possible gradient ( i.e. electric field strength ) . [ Answer: ( a ) 1.593AµJ ( B ) 5.31AµC/m2 ( degree Celsius ) 600kV/m ]

D.C TRANSIENTS

Networks of capacitances and resistances ( known as C-R circuits ) form the footing of many timing and pulse determining circuits and are therefore frequently found in practical electronic circuits.

When a District of Columbia electromotive force is applied to a capacitance C and resistance R connected in series, there is a short period of clip instantly after when the electromotive force is connected that the current flowing in the circuit and electromotive forces across C and R are altering.

These altering values are called transients.

Charging A Capacitor

The figure above shows a series connected C-R circuit.

When the switch S is closed, so by Kirchhoff ‘s valotage jurisprudence: V = Vc + VR

The battery electromotive force V is changeless.

The capacitance electromotive force Vc is given by,

The electromotive force bead across R ( i.e. VR ) is given by,

Hence at all times:

At the blink of an eye of shutting S ( i.e. initial circuit status ) , presuming there is no initial charge on the capacitance, Q is zero ( i.e. Q0 ) , therefore Vc is zero ( i.e. VC0 ) . ( Note: From equation Vc = Q / C ) .

Therefore from equation V = Vc + VR,

V = 0 + VR ( i.e. V = VR = IR )

A short clip subsequently at clip T1 seconds after shuting S, the capacitance is partially charged to, state, Q1 C because current has been fluxing. The electromotive force VC1 is now,

If the current flowing is I1 amperes, so the electromotive force bead across R has fallen to

VR1 = I1R Vs. Therefore from equation V = Vc + VR

A short clip later still, say at clip T2 seconds after shuting S, the charge has increased to Q2 C and VC has increased to,

Since V = VC + VR and V is a changeless, so VR decreases to I2R. Thus VC is increasing and ‘I ‘ and VR are diminishing as clip additions.

Ultimately, a few seconds after shuting S ( i.e. at the concluding or steady province status ) , the capacitance is to the full charged to, state Q Cs, current no thirster flows, i.e. I = 0, and therefore VR = IR = 0. It follows from equation V = Vc + VR that V = VC.

Curves demoing the alterations in VC, VR and ‘I ‘ with clip are shown below.

The curve demoing the fluctuation of VC with clip is called an exponential growing curve and the graph is called the ‘capacitor electromotive force / clip ‘ feature.

The curves demoing fluctuations of VR and ‘I ‘ with clip are called exponential decay curves, and the graphs are called ‘resistor electromotive force / clip ‘ and ‘current / clip ‘ features severally.

The name ‘exponential ‘ shows that the form can be expressed mathematically by an exponential mathematical equation, as shown below.

Growth of capacitance electromotive force,

Decay of resistance electromotive force,

Decay of resistance current,

TIME CONSTANT ( I„ – ‘TAU ‘ ) FOR A C-R CIRCUIT

As shown earlier, if a changeless District of Columbia electromotive force is applied to a series connected C-R circuit, a exponential transient growing curve of capacitance electromotive force VC consequences as shown below.

With mention to the figure below, the changeless electromotive force supply is replaced by a variable electromotive force supply at clip t1 seconds. The electromotive force is varied so that the current flowing in the circuit is changeless.

Since the current flowing is a changeless, the curve will follow a tangent, AB, drawn to the curve at point A.

Let the capacitance electromotive force VC reach its concluding value of V at clip t2 seconds.

The clip matching to ( t2-t1 ) seconds is called the clip invariable of the circuit, denoted by the Grecian missive ‘tau ‘ , I„ . The value of the clip changeless is CR seconds, i.e. for a series connected C-R circuit,

( seconds )

Where C is electrical capacity ( F ) , R is the opposition ( a„¦ ) and I„ is the clip changeless ( s )

DISCHARGING A Capacitor

When a capacitance is charged ( i.e. with the switch in place ‘A ‘ ) , and the switch is so moved to place ‘B ‘ , the negatrons stored in the capacitance maintain the current flowing for a short clip.

Initially, at the blink of an eye of traveling from A to B, the current flow is such that the capacitance electromotive force VC is balanced by equal and opposite electromotive force ( Kirchhoff ‘s 2nd jurisprudence ) , i.e. VC = VR = IR.

Finally the transients decay exponentially as current is reduced to zero, i.e. VC = VR = 0.

The transient curve stand foring the electromotive forces and current are shown below.

The equations stand foring the transeunt curves during discharge period of a series connected C-R circuit are:

Decay of electromotive force,

Decay of current,

When a capacitance has been disconnected from the supply it may still be charged and it may retain this charge for some considerable clip. Thus safeguards must be taken to guarantee that the capacitance is automatically discharged after the supply is switched away. This is done by linking a high value resistance across the capacitance terminuss.

Problem 1:

A capacitance is charged to 100V and so discharged through a 50ka„¦ resistance. If the clip invariable of the circuit is 0.8s, determine:

( a ) The value of the capacitance,

( B ) The clip for the capacitance electromotive force to fall to 20V,

( degree Celsius ) The current flowing when the capacitance has been dispatching for 0.5s, and

( vitamin D ) The electromotive force bead across the resistance when the capacitance has been dispatching for one second.

Solution 1:

Problem 2:

A 0.1AµF capacitance is charged to 200V before being connected across a 4ka„¦ resistance. Determine,

( a ) The initial discharge current,

( B ) The clip invariable of the circuit, and

( degree Celsius ) The minimal clip required for the electromotive force across the capacitance to fall to less than 2V.

Solution 2:

SELF ASSESSMENT ( 8 ) – District of columbia Transient

An uncharged capacitance of 0.2AµF is connected to a 100V District of Columbia supply through a resistance of 100ka„¦ . Determine the capacitance electromotive force 10ms after the electromotive force has been applied.

[ ANSWER: 39.35V ]

A circuit consists of an uncharged capacitance connected in series with a 50ka„¦ resistance and has a clip invariable of 15ms. Determine,

( a ) The electrical capacity of the capacitance and

( B ) The electromotive force bead across the resistance 5ms after linking the circuit to a 20V District of Columbia supply.

[ Answer: ( a ) 0.3I?F ( B ) 14.33V ]

A 10AµF capacitance is charged to 120V and so discharged through a 1.5Ma„¦ resistance. Determine,

The capacitance electromotive force 2s after dispatching has commenced, and

How long it takes for the electromotive force to fall to 25V.

[ Answer: ( a ) 105V ( B ) 23.53s ]

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