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CAPACITORSCAPACITORS

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BASIC CONSTRUCTION

INSULATOR

CONDUCTOR

CONDUCTOR

+-

TWO OPPOSITELY

CHARGED CONDUCTORS

SEPARATED BY AN

INSULATOR - WHICH MAY

BE AIR

The Parallel Plate Capacitor

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CAPACITORS STORE

• CHARGE&

• ENERGY

USES

1. Storing energy as in flash photography

2. Time delays in electronic circuits

3. As filters in electronic circuits

4. In tuning circuits

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Charge stored [Q] depends on p.d. [Volts] applied [V]

Q

V

Gradient = C = V

Q

The capacitance, C, is defined as

the charge required to raise the potential by one volt.the charge required to raise the potential by one volt.

Hence Q = C V

C is measured in FARADS, [F], - more often : F, nF or pF

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Q

Q

Q

C

---

+++

+Q -Q

Q = CV

Charging a Capacitor

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WORK DONE BY THE CELL W = Q V

As VOLTS = Joules per Coulomb

The pd across the capacitor builds up as more charge is added

Voltage

Charge0

Q

V

Work done in charging = average volts x charge

QVQV

W2

1]

2

0[

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Volts V

Charge Qq

v

The work done in adding such an infinitesimally small amount of charge, q, that Vremains constant is given by:

w = V.q

This is the area of the strip. Hence the total work done = the energy stored in the capacitor is the area under

the graph

QVE21 Or if we combine with Q =

CV,

2

2

1CVE

CQ

E2

21or

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What is the charge and the energy stored when a 50 F capacitor is charged to (a) 100 V, (b) 200 V?

(a) Q = CV = 50 x 10-6 x 100 = 0.005 C

E = 0.5CV2 = 0.5 x 50 x 10-6 x 1002 = 0.25 J

(b) Q = CV = 50 x 10-6 x 200 = 0.01 C

E = 0.5CV2 = 0.5 x 50 x 10-6 x 2002 = 1.00 J

Why does doubling the p.d. in (b) quadruple the energy stored??

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At all times after the switch is closed V = VR + VC

Initially Vc = 0, so V = Vr and so initial current = ?

Finally I =0 when capacitor is charged and VC = V

time

Volts

VC

RV

I 0

I0

time0

VC

CR

V

VR

Charging a Capacitor

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CAPACITOR DISCHARGE

R

+ -

I

V

Initially V = V0

Hence initial current, I0 = ? R

VI 0

Final Current = ?

Volts, V

Time, t0

V0t

RCeVV1

0

i.e. the decay is exponential

What if R and / or C is larger?

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CAPACITOR DISCHARGE

The current also falls exponentially and is given by :

tRCeII1

0

tRCeVV1

0

Current I

Time, t

I0Note the area under this graph is the initial charge stored.

tRCeQQ1

0

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CAPACITOR DISCHARGE tRCeVV1

0

R times C has units of seconds and is called the

time constant of the circuit

If we put t=RC into the equation above, it becomes V=V0 e-1, which works out to:

i.e. when time is equal to R x C the p.d. across

the capacitor has dropped to 37% of its original p.d.

The capacitor is almost discharged in 5 time constants.

Try putting t = 5 x R x C

V=0.37V0

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CAPACITOR DISCHARGE tRCeVV1

0

Taking logs to the base “e”

tRC

VV1

lnln 0

y = c + m x

In the form

Hence Plot

ln V

t

Gradient is negative and equal to

RC

1

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The Parallel Plate Capacitor

Area of Plate

overlap = A

d d = plate separation

Medium relative permittivity = r

d

AC r0

0 = the permittivity of free space = 8.86. X 10-12 F m-1

For air or a vacuum, r = 1

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Question 1 A 1000 F capacitor is charged to 100 V and then discharged through a 2000 resistor.

(a) What will be the initial current?

(b) What will be the current after 4 s?

AR

VI 050.0

2000

100 (a)

(b)VeeeVV RC

t

5.13100100 261010002000

4

0

mAARV

I 75.600675.020005.13

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Question 2 A capacitor is charged to 100 V and then discharged through a resistor whose value

is gradually reduced in order to maintain a constant current of 200 mA though it. The capacitor becomes discharged after 10 s.

(a) What is the initial value of the resistor?

(b) What is the capacitance of the capacitor?

500200.0100

IV

R(a) Initially the p.d. is 100 V, so

(b) Q = I t = 0.200 x 10 = 2.0 C

Q = C V, so 2.0 = C x 100 Hence C = 2 x 10-2 F or 20 000 F

The capacitor was initially charged to 200 V with 2.0 C of charge