ALTERNATING CURRENT

2 .

Ex. 1

Sol.

Ex. 2

Sol.

AC AND DC CURRENT :A current that changes its direction periodically is called alternating current (AC). If a current maintains its direction constant it is called direct current (DC).

i

t tconstant dc periodic dc

I

r "I

i

— : :— :p /\ / V t i _ j :___t

variable dc ac ac

If a function suppose current, varies with time as i = Imsin (cot+tf)), it is called sinusoidally varying function.Here Im is the peak current or maximum current and i is the instantaneous current. The factor (cot+<|>) is called phase, co is called the angular frequency, its unit rad/s. Also co=27d where f is called the frequency, its unit S'1 or Hz. Also frequency f = 1/T where T is called the time period.

AVERAGE VALUE :

Average value of a function, from t1 to t2, is defined as <f> =

fdt

t2 tjWe can find the value of fdt

graphically if the graph is simple. It is the area of f-t graph from t1 to t2.

Find the average va lue of current shown graph ically , from t = 0 to t = 2 sec.

From the i - 1 graph, area from t = 0 to t = 2 sec

= 7- x 2 x 10 = 10 Amp. sec.

10Average Current = — = 5 Amp.

2%Find the average value of current from t = 0 to t = — if the current varies as i = Imsin cot.

2jc0)j" lm sincotdt

coIm Chm 1 1 - COS CO—

CO< l > = 2%

CO

2k

CO

= 0

It can be seen graphically that the area of i - 1 graph of one cycle is zero. < i > in one cycle = 0.

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AALLTTEERRNNAATTIINNGG CCUURRRREENNTT

Some questions (Assertion–Reason type) are given below. Each question contains STATEMENT – 1 (Assertion) and

STATEMENT – 2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. So

select the correct choice :

Choices are :

(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1.

(B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.

(C) Statement – 1 is True, Statement – 2 is False.

(D) Statement – 1 is False, Statement – 2 is True.

535. STATEMENT – 1

The alternating current cannot be used to conduct electrolysis.

STATEMENT – 2

The ions due to their inertia, cannot follow the changing E�

.

536. STATEMENT – 1

In a series LCR circuit at resonance, the voltage across the capacitor or inductor may be more than the applied

voltage.

STATEMENT – 2 At resonance in a series LCR circuit, the voltages across inductor and capacitor are out of phase.

537. STATEMENT – 1 By only knowing the power factor for a given LCR circuit it is not possible to tell whether the applied alternating

emf leads or lags the current.

STATEMENT – 2

cos θ = cos (–θ)

538. STATEMENT – 1

In the purely resistive element of a series LCR, AC circuit the maximum value of rms current increases with

increase in the angular frequency of the applied emf.

STATEMENT – 2

2

2maxmax

1I , z R L

z C

ε = = + ω −

ω , where Imax is the peak current in a cycle.

539. STATEMENT – 1 AC source is connected across a circuit. Power dissipated in circuit is P. The power is dissipated only across

resistance.

STATEMENT – 2 Inductor and capacitor will not consume any power in AC circuit.

540. STATEMENT – 1 : In series RLC circuit potential drop across inductive reactance will be same as capacitive

reactance at resonance.

STATEMENT – 2 : At frequency less than resonance frequency for series RLC nature of circuit will be

capacitive, frequency more than resonance nature of overall circuit will be inductive.

541. STATEMENT – 1 : For series RLC network, power factor of circuit in region (1) is positive and in region (2) is

negative.

STATEMENT – 2 : Overall nature of circuit in region (1) is inductive while in region (2) is capacitive.

(1)(2)

I

ffr

542. STATEMENT – 1 : In a series LCR circuit, at resonance condition power consumed by circuit is maximum.

STATEMENT – 2 : At resonance condition effective resistance of circuit is maximum.

543. STATEMENT – 1 : In series L–R circuit voltage leads the current.

STATEMENT – 2 : In series L–C circuit current leads the voltage.

544. STATEMENT – 1 : Average value of a.c. over a complete cycle is always zero.

STATEMENT – 2 : Average value of a.c. is always defined over half cycle.

545. STATEMENT – 1 : In series LCR circuit resonance can take place.

STATEMENT – 2 : Resonance takes if inductance and capacitive reactance are equal.

546. STATEMENT – 1 : KVL rule is also being applied in AC circuit shown below.

STATEMENT – 2 :

~

8V

V=10V VC in the circuit = 2V.

547. STATEMENT – 1 : AC generators are based upon EMI principle.

STATEMENT – 2 : Resistance offered by capacitor for alternating current is zero.

548. STATEMENT – 1 : For sinusoidal a.c. ( I = I0 sin ωt ) 0rms

II

2= .

STATEMENT – 2 : The r.m.s. value of alternating current is defined as the square root of the average of I2 during

a complete cycle.

549. STATEMENT – 1

Rate of heat generated when resistance is connected with AC source depends on time.

STATEMENT – 2 RMS voltage may be greater than maximum AC voltage.

550. An inductor, capacitor and resistance connected in series. The combination is connected across AC source.

STATEMENT – 1 : Peak current through each remains same.

STATEMENT – 2 : Average power delivered by source is equal to average power developed across resistance.

551. STATEMENT – 1 : In alternating current direction of motion of free electrons changes periodically.

STATEMENT – 2 : Alternating current changes its direction after a certain time interval.

552. STATEMENT – 1 : When frequency is greater than resonance frequency in a series LCR circuit, it will be an

inductive circuit.

STATEMENT – 2 : Resultant voltage will lead the current.

553. STATEMENT – 1 : When capacitive reactance is smaller than the inductive reactance in LCR circuit, e.m.f. leads

the current.

STATEMENT – 2 : The phase angle is the angle between the alternating e.m.f. and alternating current of the

circuit.

554. STATEMENT – 1 : An alternating current shows magnetic effect.

STATEMENT – 2 : Alternating current varies with time.

Hint & Solution

535. (A) 536. (B) 537. (A) 538. (D) 539. (A) 540. (B)

541. (C) 542. (C) 543. (B) 544. (B)

545. (A) 546. (C) 547. (C) 548. (A)

549. (C) 550. (B) 551. (B) 552. (A)

553. (C) 554. (B)

537. For a certain values of cos θ (power factor) two values of θ are possible. One is positive the other is much negative. Accordingly the applied emf may lead or lag.

538. The maximum value of rms current rms rms

z R

ε ε= = . It does not depend upon ω.

542. av

VIcosP

2

φ=

At resonance condition cos φ = 1 But Z = R

Which is minimum.

543. L–R circuit

V

H

N

RWL

C–R circuit.

VR

Y Cw

I

544. For half cycle Imean = 0.636I0

or Emean = 0.636 E0

Average value is always defined over a half cycle cause in next half cycle it will be opposite in direction. Hence

for one complete cycle, average value will be zero.

545. At resonant frequency

XL = XC ∴ Z = R (minimum) Therefore current in the circuit is maximum.

546. Voltage will be added vectorially.

547. C

1X

C=

ω

ω ≠ O area for AC.

1/ 2 1/ 2T 2 /

2 2 2

0

0 0 0rms T 2 /

0 0

I dt I sin t dtI

I2

dt dt

π ω

π ω

ω

= =

∫ ∫

∫ ∫.

549. Rate of heat generated depends on time.

550. Average power consumed by capacitor or inductor is zero.

551. Motion of electron is random with drift velocity opposite to the direction of current.

552.

VL

VR

V –VL C

VC

φ

V

I

553. L C

1L

X X CtanR R

ω −− ωφ = =

When XL > XC then tan φ is positive i.e. φ is positive (between 0 and π/2). Hence e.m.f. leads the current.

554. Like direct current, alternating current also produces magnetic field. But the magnitude and

direction of the field goes on changing continuously with time.

SHORT REVESION

1. Capacitance Of A n Isolated Spherical C onductor :C = 4 k £„£ R in a m edium C = 4 k g „R in airO r 0

* This sphere is at infinite distance from all the conductors .* The capacitance C = 4 k e nR exists be tw een the surface o f the sphere & earth .

2. Spherical Capacitor :It consists o f tw o concentric spherical shells as shown in figure. Here capacitance o f region betw een the tw o shells is and that outside the shell is C2. We have

Ci =4 k e 0 ab

b - aand C2 = 4k e n b

Depending on connection, it may have different combinations o f and C2.

3.

4.

Parallel P late Capacitor :(i) U niform D i-electric M edium :

I f tw o parallel p lates each o f area A & separated by a d istance d are charged w ith equal & opposite charge Q, then the system is called a parallel plate capacitor & its capacitance is given by,

C e0erAd

in a m ediume- A

C = —h— w ith air as m edium

This result is only valid when the electric field between plates o f capacitor is constant.

(ii) M edium Partly A ir :

W hen a d i-electric slab o f th ickness t & relative perm ittiv ity e r is introduced betw een the plates o f an air capacitor, then the distance betw een

the plates is effectively reduced by

the d i-electric slab .

irrespective o f the position o f

(iii) C omposite M edium : C = t , U

- r 2 - r 3

e r3H

1

Cylindrical Capacitor :It consist o f tw o co-axial cylinders o f radii a & b, the outer conductor is earthed . The di-electric constant o f the medium filled in the space between the cylinder is

e r . The capacitance per unit length is C =2kg Farad0 r

M f) m

5. C oncept of variation of parameters:

As capacitance o f a parallel plate capacitor isC =kA

-, if either o f k, A or d varies in the region between

, I f all dC's are in parallel CT = J dC

the plates, w e choose a small dc in between the plates and for total capacitance o f system.

1 rI f all dC's are in series — = -C T J e 0 k (x )A (x )

C ombination O f Capacitors :(i) Capacitors In Series :In this arrangem ent all the capacitors w hen uncharged get the same charge Q but the potential difference across each will differ ( if the capacitance are unequal).

1 1 1 1 1+C c r

(ii)Qi cl5v

eq. Cj C2 C3

Capacitors In Parallel :W hen one plate o f each capac ito r is connected to the positive term inal o f the battery & the o ther plate o f each capac ito r is connected to the negative term inals o f the battery , then the capacitors are said to be in parallel connection.The capacitors have the same potential difference, V but the charge on each one is different ( if the capacitors are unequal).c = c + c + c + ..... + c .eq. 1 2 3 n

Energy Stored In A Charged Capacitor :C apacitance C, charge Q & potential difference V ; then energy stored is

1 1 1 Q2U = — CV 2 = — Q V = — — . This energy is stored in the electrostatic field set up in the di-electric

m edium betw een the conducting plates o f the capacitor .

\+i rC2,v

q 3+! | c 3,v11

Q +v_-

8. H eat produced in switching in capacitive circuit

D ue to charge flow always some amount o f heat is produced when a switch is closed in a circuit whichcan be obtained by energy conservation as -H eat = W ork done by battery - Energy absorbed by capacitor.

9. Sharing O f Charges :W hen tw o charged conducto rs o f capacitance & C2 at po ten tia l V) & V 2 respectively are connected by a conducting wire, the charge flows from higher potential conductor to lower potential conductor, until the poten tia l o f the tw o condensers becom es equal. The com m on poten tia l (V) after sharing o f charges;y = net charge = qt + q2 = + C2V2

net capacitance Cj + C2 C j+C 2charges after sharing q t = C(V & q2 = C 2V. In this p rocess energy is lost in the connecting w ire

as h e a t . This loss o f energy is U . .. - U ,C>J initial real

c t c 22(C1 + C2) ( v , - V 2)2 .

10. Remember:(i) The energy o f a charged conductor resides outside the conductor in its EF, w here as in a condenser

it is stored w ithin the condenser in its EF.(ii) The energy o f an uncharged condenser = 0 .(iii) The capacitance o f a capacitor depends only on its size & geom etry & the di-electric betw een the

conducting surface. (i. e. independent o f the conductor, like, whether it is copper, silver, gold etc)

EXERCISE # IQ l

Q 2

Q 3

Q.4

Q 5

Q 6

Q 7

Q 8

Q 9

(a)(b)

A solid conducting sphere o f radius 10 cm is enclosed by a thin metallic shell o f radius 20 cm. A charge q = 20pC is given to the inner sphere. Find the heat generated in the process, the inner sphere is connected to the shell by a conducting wire

The capacitor each having capacitance C = 2pF are connected with a battery o f em f 30 V as shown in figure. W hen the switch S is closed. Find(a) the amount o f charge flown through the battery(b) the heat generated in the circuit(c) the energy supplied by the battery(d) the amount o f charge flown through the switch S

The plates o f a parallel plate capacitor are given charges +4Q and -2Q . The capacitor is then connected across an uncharged capacitor o f same capacitance as first one (= C). Find the final potential difference between the plates o f the first capacitor. ____ +1 ^ ____

In the given netw ork if potential difference betw een p and q is 2 V and C2 = 3Cj. Then find the potential difference betw een a & b. a c, C2

li-l qCj c2

Find the equivalent capacitance o f the circuit betw een point A and B .

2C 4C 8C/—II— — 11---- ----1 1---- --- 1 1----- ------^

:C : - C : - C : - CInfiniteV__II__ __1 1___ __ 1 1___ __1 1___ section>

2C 4C 8C+3q+q

The tw o identical parallel plates are given charges as shown in figure. I f the plate area o f either face o f each plate is A and separation betw een plates is d, then find the amount o f heat liberate after closing the switch.

S

Find heat produced in the circuit shown in figure on closing the switch S.

In the following circuit, the resultant capacitance between A and B is 1 pF. Find the value o f C.

2pF

+20pC' -20|iC

+50pC ̂ -50pCr5pF

lq p

X8 |iF _^

2|,tF T ~|~2|jF izm-f

6qF - j- -,-4gF

J h h

Three capacitors o f 2pF, 3pF and 5pF are independently charged w ith batteries o f em f’s 5V, 20V and 10V respectively. A fter disconnecting from the voltage sources. These capacitors are connected as shown in figure with their positive polarity plates are connected to A and negative polarity is earthed. N ow a battery o f 20V and an uncharged capacitor o f 4pF capacitance are connected to the junction A as shown with a switch S. W hen switch is closed, find : the potential o f the junction A. final charges on all four capacitors.

Q.10 Find the charge on the capacitor C = 1 pF in the circuit shown in the figure.

lpF IjiF 10V

7 %lpF lpF lpF lpF

lpFC=luF lpF

: :ipF ^ipF:UpF ::ipF z z

Q .l l

Q .12

Find the capacitance o f the system shown in figure.

I Plate area = A

k = 1 k = 2

k = 3 II

The figure shows a circuit consisting o f four capacitors. Find the effective capacitance betw een X and Y.

— 1 1--------- ------ 1 1--------lpF IjiF

L inF2jiF

------ 1 1--------

Y

Q. 13 Five identical capacitor plates, each o f area A, are arranged such that adjacent plates are at a distance'd ' apart, the plates are connected to a source o f em f V as shown in figure. The charge on plate 1 is________________ and that on plate 4 i s ___________ .

Q . 14 In the circuit shown in the figure, intially SW is open. W hen the switch is closed, the charge passing throughthe sw itch______________in the direction

to

XA Er

I

60 V

60 V SW2|jF3nF

13 b

T

Q . 15 In the circuit shown in figure, find the amount o f heat generated when switch s is closed.

Q . 16 Two parallel plate capacitors o f capacitance C and 2C are connected in parallel then following steps are performed.(i) A battery o f voltage V is connected across points A and B .(ii) A dielectric slab o f relative permittivity k is slowly inserted in capacitor C.(iii) Battery is disconnected.(iv) Dielectric slab is slowly rem oved from capacitor.Find the heat produced in (i) and w ork done by external agent in step (ii) & (iv).

Q . 17 The plates o f a parallel plate capacitor are separated by a distance d = 1 cm. Two parallel sided dielectricslabs o f thickness 0.7 cm and 0.3 cm fill the space between the plates. I f the dielectric constants o f the two slabs are 3 and 5 respectively and a potential difference o f440V is applied across the plates. F ind :

(i) the electric field intensities in each o f the slabs.(ii) the ratio o f electric energies stored in the first to that in the second dielectric slab.

Q. 18 A 10 pF and 20 pF capacitor are connected to a 10 V cell in parallel for some tim e after which the capacitors are disconnected from the cell and reconnected at t = 0 with each o th e r , in series, through wires o f finite resistance. The +ve plate o f the first capacitor is connected to the -v e plate o f the second capacitor. D raw the graph which best describes the charge on the +ve plate o f the 20 pF capacitor with increasing time.

List of recommended questions from I.E. Irodov.3.101, 3.102, 3.103, 3.113, 3.117, 3.121, 3.122, 3.123, 3.124, 3.132, 3.133, 3.141, 3.142, 3.177, 3.184,

3.188, 3.199, 3.200, 3.201, 3.203, 3.204, 3.205

EXERCISE # II

Q l

Q 2

Q 3

(000

Q.4

(0000*0

Q 5

Q 6

Q 7

Q 8

(a) For the given circuit. Find the potential dilFerence across all the capacitors.(b) H ow should 5 capacitors, each o f capacities, lp F be connected so

as to produce a to tal capacitance o f 3/7 pF.

6uF,II— p H fiY ’

I7nF

9|TF

-±c25V

II— M l 8|xF

The gap betw een the plates o f a plane capacitor is filled w ith an isotropic insulator w hose di-electric

constant varies in the direction perpendicular to the plates according to the law K = Kj 1 + sin — X d

where d is the separation, between the plates & Kj is a constant. The area o f the plates is S. Determine the capacitance o f the capacitor.

Five identical conducting plates 1 ,2 ,3 ,4 & 5 are fixed parallel to and equdistant from each other (see figure). Plates 2 & 5 are connected by a conductor while 1 & 3 are joined by another conductor. The junction o f 1 & 3 and the plate 4 are connected to a source o f constant e.m.f. V0. F in d ; the effective capacity o f the system betw een the terminals o f the source, the charges on plates 3 & 5.Given d = distance betw een any 2 successive plates & A = area o f either face o f each plate .

A potential difference o f 300 V is applied betw een the plates o f a plane capacitor spaced 1 cm apart. A plane parallel glass plate with a thickness o f 0.5 cm and a plane parallel paraffin plate with a thickness of 0.5 cm are placed in the space betw een the capacitor plates find :Intensity o f electric field in each layer.The drop o f potential in each layer.The surface charge density o f the charge on capacitor the plates. Given th a t : kglass = 6, kparaffin= 2

A charge 200pC is imparted to each o f the tw o identical parallel plate capacitors connected in parallel. A t t =0, the plates o f both the capacitors are 0.1m apart. The plates o f first capacitor move tow ards each other w ith relative velocity O.OOlm/s and plates o f second capacitor m ove apart w ith the same velocity. Find the current in the circuit at the moment.

A parallel plate capacitor has plates w ith area A & separation d . A battery charges the plates to a potential difference o f V0. The battery is then disconnected & a di-electric slab o f constant K & thickness d is introduced. Calculate the positive w ork done by the system (capacitor + slab) on the m an who introduces the slab.

A capacitor o f capacitance C0 is charged to a potential V0 and then isolated. A small capacitor C is then charged from C0, discharged & charged again, the process being repeated n times. The potential o f the large capacitor has now fallen to V. Find the capacitance o f the small capacitor. I f V0 = 1 0 0 volt, V=35volt, find the value o f n for C0 = 0.2 pF & C = 0.01075 pF . Is it possible to rem ove charge on C0 this way?

W hen the switch S in the figure is thrown to the left, the plates o f capacitors Cj acquire a potential difference V. Initially the capacitors C2C3 are uncharged. Thw switch is now thrown to the right. W hat are the final charges q l3 q2 & q3 on the corresponding capacitors.

v\ s □ _ C 2

T

Q 9

(i)

©(iii)

Q .10

Q .l l

Q .12

(i)

(u)

Q.13

Q. 14

Q.15

Q.16

A parallel plate capacitor with air as a dielectric is arranged horizontally. The lower plate is fixed and the other connected with a vertical spring. The area o f each plate is A. In the steady position, the distance between the plates is cIq. W hen the capacitor is connected with an electric source with the voltage V, a new equilibrium appears, with the distance betw een the plates as d j . M ass o f the upper plates is m.Find the spring constant K.W hat is the maximum voltage for a given K in which an equilibrium is possible ?W hat is the angular frequency o f the oscillating system around the equilibrium value d ,. (take amplitude o f oscillation « d,)

An insolated conductor initially free from charge is charged by repeated contacts with a plate whichafter each contact has a charge Q due to some m echanism . I f q is the charge on the conductor after the first

operation, prove that the maximum charge which can be given to the conductor in this way isQq

Q - q ‘

A parallel plate capacitor is filled by a di-electric w hose relative perm ittivity varies w ith the applied voltage according to the law = aV , w here a = 1 per volt. The same (but containing no di-electric) capacitor charged to a voltage V = 156 volt is connected in parallel to the first "non-linear" uncharged capacitor. Determ ine the final voltage V f across the capacitors.

A capacitor consists o f two air spaced concentric cylinders. The outer o f radius b is fixed, and the inner is of radius a. If breakdown of air occurs at field strengths greater than E^, showthat the inner cylinder should have radius a = b/e if the potential o f the inner cylinder is to be maximum

radius a = b / Ve if the energy per unit length o f the system is to be maximum.

S^ - i - 3 |iF i lO V ^p ^6nF5 V I

Find the charge flown through the switch from A to B when it is closed. 6̂ F _L“5V ip6(iF

Figure show s th ree concen tric conducting spherical shells w ith inner and outer shells earthed and the middle shell is given a charge q. Find the electrostatic energy o f the system stored in the region I and II.

The capacitors shown in figure has been charged to a potential difference o f V volts, so that it carries a charge C V with both the switches S, and S2 remaining open. Switch S, is closed at t=0. At t=RjC switch S, is opened and S2 is closed. Find the charge on the capacitor at t=2R jC + R2C.

In the figure show n initially switch is open for a long time. N ow the switch is closed at t = 0. Find the charge on the rightmost capacitor as a function o f time given that it was intially unchanged.

Q.17

Q.18

Q 1

(i)

©

(iii)

Q 3

Q.4

Q 5

(i)

©

In the given circuit, the switch is closed in the position 1 at t = 0 and then moved |-----LVto 2 after 250 ps. Derive an expression for current as a function o f time for -L2ov yt > 0. Also plot the variation o f current with time. I t

Find the charge which flows from point A to B, when switch is closed.

EXERCISE # III

■40V

A a B

II II ii M IIII II5\xF 5^lF

II5^lF

___11___

II5\iF

II5 n F

120V

rsoon:0.5 (iF

Two parallel plate capacitors A & B have the same separation d = 8.85 x 10-4 mbetweenthe plates. Theplate areas o f A & B are 0.04 m2 & 0.02 m2 ----------------- 1 ̂ | ^ ^respectively. A slab o f di-electric constant (relative "1 Tpermittivity) K=9 has dimensions such that it can exactly Ifill the space between the plates o f capacitor B . I liovthe di-electric slab is placed inside A as shown in the figure (i) A is then charged to a potential difference o f 110 volt. Calculate the capacitance o f A and the energy stored in it.the battery is disconnected & then the di-electric slab is rem oved from A . Find the w ork done by the external agency in removing the slab from A .the same di-electric slab is now placed inside B, filling it completely. The tw o capacitors A & B are then connected as shown in figure (iii). Calculate the energy stored in the system. , j E E . , ,

two square m e.a.Mc p,a,.s , in side are kept 0.01 in apart, like a parallel plat e capacitor, in air in such a way that one o f their edges is perpendicular, to an oil surface in a tank filled with an insulating oil. The plates are connected to a battery o f e.m.f. 500 v o lt . The plates are then lowered vertically into the oil at a speed o f 0.001 m/s. Calculate the current draw n from the battery during the process.[di-electric constant o f oil = 11, e 0 = 8.85 x 10-12 C2/N 2m2] [ JEE '94, 6 ]

A parallel plate capacitor C is connected to a battery & is charged to a potential difference V. Another capacitor o f capacitance 2C is similarly charged to a potential difference 2 V volt. The charging battery is now disconnected & the capacitors are connected in parallel to each other in such a way that the positive terminal o f one is connected to the negative terminal o f other. The final energy o f the configuration i s :

(A) zero (B) | CV2 (C) ^ CV2 (D) | CV2 [ JEE '95, 1 ]2 6 2

The capacitance o f a parallel plate capacitor with plate area 'A1 & separation d is C . The space between the plates is filled with two wedges o f di-electric constant K ! & K2 respectively. Find the capacitance o f the resulting capacitor.

[ JEE '96, 2 ]

A

t f|2hL

Two capacitors A and B w ith capacities 3 pF and 2 pF are charged to a potential difference o f 100 V and 180 V respectively. The plates o f the capacitors are connected as shown in figure w ith one w ire from each capacitor free. The upper plate o f a is positive and that o f B is negative, an uncharged 2 pF capacitor C with lead wires falls on the free ends to complete the circuit. C alculate: the final charges on the three capacitorsThe amount o f electrostatic energy stored in the system before and after the completion o f the circuit.

[ JEE '97 (cancelled)]

z ^3hf 100V

Z I 2pF B 180 V

Q 6 An electron enters the region between the plates o f a parallel plate capacitor at a point equidistant from eitherplate. The capacitor plates are 2 x 10_2m apart& 10_1 m long. Apotential difference o f 300 volt is kept across the plates. Assuming that the initial velocity o f the electron is parallel to the capacitor plates, calculate the largest value ofthe velocity o f the electron so that they do not fly out o f the capacitor at the other end. [ JEE '97, 5 ]

Q .7 For the circuit shown, which o f the folio wing statements is tru e? v,=30v v2=20vs j closed, V j = 15 V ,V 2 = 20 V

(B) w ith S3 closed, V j = V2 = 25 V(C) w ith Sj & S2 closed, V j = V2 = 0(D) w ith Sj & S2 closed, V j = 30 V, V2 = 20 V [JE E '99, 2 ]

Q . 8 Calculate the capacitance o f a parallel plate condenser, with plate area A and distance between plates d, when filled with a medium whose permittivity varies a s ;

e (x )= e 0 + P x 0 < x < - |

e (x) = g 0 + P (d - x) < x < d . [ R E E 2000, 6]

Q. 9 Two identical capacitors, have the same capacitance C. One o f them is charged to potential V, and theother to V2. The negative ends o f the capacitors are connected together. W hen the positive ends are also connected, the decrease in energy o f the combined system is [ JEE 2002 (Scr), 3 ]

( A ) A ( v ,2 - V | ) ( B j i c ^ + v l ) ( C ) i c ( v , - v J ( D ) i c ( v , + v J

Q. 10 In the given circuit, the switch S is closed at time t = 0. The charge Q on the capacitor at any instant t is given by Q (t) = Q0 ( 1-e ,/l). Find the value o f Q0 and a in terms o f given parameters shown in the circuit.

[JEE 2005]

Q .l l Given : R j = IQ , = 2Q , Cj = 2pF, C2 = 4pFThe time constants (in pS) for the circuits I, II, III are respectively

Ci1 C2 111 11

R1T V —W/v—

—r 2

(I) (II) (III)(B) 18, 4, 8/9 (D) 8/9, 18 ,4

(A) 18, 8 /9 ,4 (C) 4, 8/9, 18 [JEE 2006]

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