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Capacitor Hence^ capacitors are^ always attached
in parallel to^ circuit^ so that
An electronic component that^ is^ used the circuit keeps on working even
to (^) store electrical (^) energy and^ supply when (^) capacitor is fully charged.
current to^ the circuit^ when^ need
be. →
metal plates Explain why capacitor stores energy ,
not (^) charge.
Symbol : • For a
parallel plate (^) capacitor,^ both
plates have^ equal^ and^ opposite
charge,^ hence^ net^ charge^ is^ zero^.
I ←^ Parallel^ plate capacitor • Work is^ done in separating those
insulator charges
therefore energy^ 1 is dielectric material (^) ) (^) stored. Uses (^) of capacitor 1 Functions (^) ) (^) Capacitance
- (^) Stores (^) energy 2. (^) Time (^) delay circuits It is the (^) charge stored on one
- (^) smoothing in (^) rectifier plate (^) of a (^) capacitor per unit
- (^) Power (^) supplies 5.^ Tuning circuits (^) potential (^) difference.
Charging a^ capacitor C^ =^9 v^ or^ q=CV→^ pod^.
Positive terminal^ pulls charge capacitance
electrons (^) from one ¥ (^) terminal plate and negative SI Unit :^ Farad 11= deposits them ⑦ on^ the^ other^.^ Define^
the (^) Farad 111=1 :^ It is the > (^) + - • Both capacitance^ of^ a^ capacitor^ when achieveplates begin^ to^ 1C^ of^ charge^ is^ stored^ across^ a
equal &^ opp.^ p.d.^ Of IV.
charge.
- Charging (^) process (^) stops when (^) the Factors (^) affecting capacitance
p.cl. across^ the^ capacitor is^ equal^ to
EMF of the^ battery. 1.* surface Area^ of plates
I =^ +^ -^ Bigger^ SA^ of^ plate^ >^ more^ capacitance
"
I =^ 2.^ Insulation
- I - i^ ' II better^ insulators^ →^ more^ capacitance
- (^3). Distance between (^) plates lamp (^) off after Lamp^ still^ on^!^ more
capacitor charges^ less^ distance^ →^ stronger^ field^ →^ capacitance
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Energy stored^ in^ a^ capacitor capacitors^ in^ series
g-^ -^ CV Tc=qgt (^) ±, = I = ++ =
✓μ ^^ Cz^ + - +
c. (^) ,
° For the same V (^) , C (^) , stores more > V '^ '^ '^ '='^ '^ '^ '^ '^ '^ ': (^) charge than Cz (^). + -
= = VT 19T^ )
o =^ I^ > ° As^ q ✗^ C^ , C^ , > Cz .
(^092 9) , (^) qlc Charge :^ Each (^) capacitor gains the
same magnitude of charge
on one of its^ plates. VIV (^) As V= (^) Energy go (^9) , = 92=93 = (^) 9T ^ (^) R (^) charge i (^) E-- Vxq Voltage : (^) Each (^) capacitor achieves " (^) a different p.cl. based
to
' under^ = Energy on^ their^ capacitance^. "^ graph^ stored^ in ° 0 : (^) capacitor 9=95-9^ q=cv^ Text
>qk 44=44 Tt
Aof A-- lzbh (^) or smallermore capacitors^ achieve pd and^ vice^ versa^. =tzqV so^ Ecap^
: Izvq Lil, --^ 4-^ Vt
sum of all^ P.d. is^ equal
v v^ to^ EMF^ of source^.
if g--^ CV^ so^ if V^ - - Eso (^) Vi 1- (^) V2 -1 (^) V3 = (^) Vt Ecap (^) :{V44^ Ecap - -1-21% **E cap : Izu Ecap -.^ 1- 2C^ Ñ Capacitance :^ V1^ +^ V2^ +^ V3^ =^ Vt
* * as^ g--^ CV^ so^ V=^9
BE (^) cap z -1244?^
- (^) V (^) ;) T II. +^ %: %: %. VIV ^ As (^9) , =^ 92=93^ =^ Ut V - = -1 Area =^ W^ -^ D^ by battery
÷ iv.^ ☐^ by^ battery^
: (^) Va ✗ (^) ( E. + IgE,^ )^
- -41¥ ) =. During (^) charging (^) process o = > (^) way of the (^) workdone (^) = I (^0) or 91C (^) by the (^) source is , +^ {at^ É, cell (^) / (^) Battery stored in (^) capacitor and the^ other (^) half (^) for identical (^) capacitors , G-^ = En
lost as heat^ in capacitance <
connecting wires^!^ of each^ no. -
-9 cap.
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Discharging of a^ capacitor Time^ Constant^ :^ The^ time^ taken
C forits^ the^ capacitor^ to^ decrease
±,^ I^ • the rate charge^ /voltage/^ current^ in
of discharge^ circuit^ ,^ to^ § of its^ initial
+^
of (^) directlya^ capacitor is
proportional value^.
to the^ chaise on q =^ g.^ e-^ that Éc
capacitor.
if t^
- RC (^) so II. ✗^ x^ > (^) x= Koe
- " t g = g. e-^7 ¥ such relations^ follow exponential functions!^ q=%e^
q = 9oz or^9 = Lego 9 ¥ ✗^9 >^ q=^ g.^ e-^
Éc
e e' so 1g =^00368 (^ 36.8%^ ) 1 ¥.^ ✗^
V >^ V^ =^ V.^ e-^ Éc
After t=T^ ,^ remaining^ charge^ is 1 ¥,^ ✗^
I >^ I^ =^ Ioe
- ¥' 36.8% (^) of its^ initial^ charge.
- go^ :^ initial^ charge^ It^ :o) Vo :^ initial^ voltage^ It^ :o)
Io :^ initial^ current^ It^ :o)
- (^9) , V (^) ,^ I^ : quantities at^ time^ t^.
- R :^ resistance •^ C : capacitance 91C ^ - (^) T, > T, as^ graph 2 No • reduces (^) charge at a
faster rate.
Of (^) , -^ l^ l^ l^ l^ l^ l--l^ l^ l^ l^ l^ l-- O ==^ =-^ T,
0 Tz T,
% (^) >
t /s
T =^ RC , where T is^ Time Constant
Larger Time^ constant^ >^ dischargeslower.
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Examiner’s^ For Use
5 A solid metal sphere, of radius r , is insulated from its surroundings. The sphere has charge + Q. This charge is on the surface of the sphere but it may be considered to be a point charge at its centre, as illustrated in Fig. 5.1.
+ Q
r
Fig. 5. (a) (i) Define capacitance. .................................................................................................................................. ............................................................................................................................ [1] (ii) Show that the capacitance C of the sphere is given by the expression C = 4 πε 0 r.
[1]
(b) The sphere has radius 36 cm. Determine, for this sphere, (i) the capacitance,
capacitance = ............................................ F [1]
Charge stored^ per unit^ potential (^) difference.
C--
g- C=q/^ ✗ 4H§ as V=^ 9-^ C=4lTEor 41T for C=q÷u÷Eor
( =^ UH^ C- or C- UH^ (8.85×10-1410.36)
( =^40 × 10 -^ "^ F
4. 0 × 10 - "
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Examiner’s^ For Use
5 (a) State two functions of capacitors in electrical circuits.
- .....................................................................................................................................
- ..................................................................................................................................... [2] (b) Three capacitors, each marked ‘30 μF, 6 V max’, are arranged as shown in Fig. 5.1.
A B
Fig. 5. Determine, for the arrangement shown in Fig. 5.1, (i) the total capacitance,
capacitance = ......................................... μF [2] (ii) the maximum potential difference that can safely be applied between points A and B.
potential difference = ........................................... V [2]
stores (^) energy Used in (^) tuning circuits
I
30 μF^30 μF
30 μF
É=ÉiÉ 30 μF 60 μF^ ÷ :-O -1%
Ct - -20μF
20
q, 92 q=CVe.^ we^ assume^ Max^ 6V^ on^ the
- • q same in^ series^ smallest^ capacitor i.^ e-^ 30mF
9 , =^ 9h^9 ,^ =^ Nt
30 μF 60 μF^ *✗ t.gg (^) qv,=CzVz GV, = CTVT 1 t^ smaller^ " C^ " more (^30) ×6=60 ✗ (^) Vz 30 ×6=20 ✗^ Vt bv?^?^ " v " v. =zv Vt=9V✓✓
maxv =^ 6+3=9^ V
\
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Examiner’s^ For Use
(c) A capacitor of capacitance 4700 μF is charged to a potential difference of 18 V. It is then partially discharged through a resistor. The potential difference is reduced to 12 V. Calculate the energy dissipated in the resistor during the discharge.
energy = ........................................... J [3]
DE = E C (^1) Vi-^ Vi^4
1- I 4 7 00 ✗^10 -^ b)^ (^122 -^184
= - 0.^423 J
O o (^42)
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(b) Switch S is now moved to position Y. State what happens to the potential difference across capacitor P and across capacitor Q. capacitor P: .............................................................................................................................. ................................................................................................................................................... ................................................................................................................................................... capacitor Q: ............................................................................................................................. ................................................................................................................................................... ................................................................................................................................................... [4] [Total: 8]
It (^) discharges (^) from 3V^ to^ ov^ as^ it^ is connected (^) across resistor.
It (^) changes (^) up (^) from bv^ to^ 9V^ and^ battery's
p.cl is^ greater than^ 6N.
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7 (a) Explain what is meant by the capacitance of a parallel plate capacitor. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] (b) A parallel plate capacitor C is connected into the circuit shown in Fig. 7.1.
120 V A
X Y
S
C
Fig. 7. When switch S is at position X, the battery of electromotive force 120 V and negligible internal resistance is connected to capacitor C. When switch S is at position Y, the capacitor C is discharged through the sensitive ammeter. The switch vibrates so that it is first in position X, then moves to position Y and then back to position X fifty times each second. The current recorded on the ammeter is 4.5 μA. Determine (i) the charge, in coulomb, passing through the ammeter in 1.0 s,
charge = ....................................................... C [1]
It is The^ charge stoned^ on^ one^ plate of a^ capacitor
per unit^ potential^ difference across^ the^ plates^. C-- Ig
where (^) or is (^) charge on (^) one (^) plate and^ U^ is the (^) pod. across them.
→.
q =^ It of =^ 4.5×10-6^ ×^1 of =^ 4.5×10-6^ (
4. 5 ×10-
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