Alternating Current Circuits

"·'. ~"

i , . ;:,,, .

_'!.. _ _ _ , ~ " ~ ·

-Alternating Current Circuits

,..,.,.._

,-_

:. Alt~rn.ating Current and Voltage

Circuit'Elements in ac Circuits

; ·-Tefmiilology and Analysis of Series J -ii't'circuits

\· ReSonanc~ in Series LCR Circuit .< ·P,ower,in'ac Circuits and'Choke Coil

Short' AriSWer Questions Multiple Choice Questions

. (with Hints and Solutions)

~ ·~ertion-Reason Type Questions

' 0

10.1 Alternating Current and Voltage

A time varying current or voltage according to its law of variation may be periodic or non-periodic. In case of periodic current or voltage, the current or voltage is said to be alternating if:

(a) Its amplitude is constant, and

(b) Alternate half cycle is positive and half negative. This all is illustrated in Fig. 10.1.

Damped

oscillatory Oscillatory

1,,-..._--~'----,,..,.,.,----+

(A) (B)

Fig.10.1 Periodic but not alternating

(A) Amplitude Is not constant,

(B) Direction ls not changing

If the current or voltage varies periodically as 'sin' or 'cos' function of time, the current or voltage is said to be sinusoidal and is what we usually mean by it [Fig. 10.2].

+ /orV I= 1₀ sin rot 0 f'--"=--"1---1'--'---\-V= V₀sin rot (A) Fig. 10.2 - V= V₀ cos rot (B)

Superposition of a number of sinusoidal alternating currents or voltages of different frequencies gives rise to the so called complex alternating current or voltage as shown in Fig. 10.3.

for V ·

8-J·

Of--+--+-~-i;:i Square type (A) Triangular type (B)

Fig. 10.3 Complex alternating current or voltage

Regarding alternating current or voltage, it is worth noting that:

(I) According to convention in physics alternating current is abbreviated as ac* (Physics by Halliday and Resnick) and not as a.c. ( Oxford Dictionary of Physics) or A.C. (Electricity and Magnetism by W.J. Duffin) or AC which is usually done by most of the authors.

(2) It is produced by dynamo or electronic oscillator and is represented by the symbol

-0 .

(3) The frequency of ac in India is 50 Hz, i.e., f=50Hz

so ro= 21tf= 1001t rad/s

(4) The ac can be converted into de with the help of rectifier while de into ac with the help of inverter. (5) It cannot produce 'chemical effects of current' such as

electroplating or electrolysis as due to large inertia ions cannot follow the frequency of ac.

(6) It can be stepped up or down with the help of transformer (while de cannot be).

(7) Alternating current is measured by 'hot wire instruments'.

(8) The 'maximum value' of alternating current or voltage is called 'peak value' and is represented by IO and V0 respectively. So that alternating current or voltage at any time I in a circuit will be given by,

J=J₀sinrol or

V= V

0 sin rot

(9) If the 'average' or 'mean value' of alternating current or voltage is defined for full cycle, it will be zero as

J;

sin ro I dt or { cos ro I dt = 0 so it is defmed for positive ( or negative) half cycle as,

rT/2 r1t/ro

Jo

Idt

Jo

IO sin rot dt

/av

or

I mean= T/2

Io

dt f1t/OO

Jo

dt

'!:..1

'It 0

lorv2

Ql,LJ.1---~,L._--"-L---?----T/2 T (A) Fig. 10.4 - 1, lrms -

Jz

(8)

(10) Effective, virtual or rms value of alternating current is defined as the square root of the average of 12 during a complete cycle, i.e.,

(11) (12) 1 = [{ J 2

dtl

=

[J:''"'

15 sin 2 rot dtl 112 nus

iT

i21t I

ro dt dt 0 0

Irrru =

Ji

[asJ;sin2rotdt=~] All ac instruments read this value, e.g., if we speak about 220 V alternating voltage, we mean V rrns = 200 V.

2

Vo

As V,. = - V0 and V rrns = r;;

'It -v2

V₀

>

V rrns

>

Vav and V₀ = 1.41V nus So, if V nus = 220 V,

V₀

=Ji

x220=311 V and

v ••

=0.9x220=198V

I

Form factor= __!!!'e. 1 ••

. 'd 1 Io 1t 1t So, for smus01 a ac = r;; x - = r;;

-v2 21 o 2-v2

Question I. State whether the following statements are true or false giving reason in brief;

(a) de is more dangerous than ac of same value. (b) The average value of ac is defined over half cycle.

Answer: (a) False; It is a common misconception that de is more dangerous than ac of same value as due to it an electrocuted person sticks to the line while ac repels him away from the line.

Actually in case of 220 V ac the peak value

V₀ =220x-J2=311 V and as 311 V will cause more hann to the human body than 220 V de, ac is more dangerous than de of the same value.

(b) False; No doubt the average value of ac is defined over a half cycle ( as over a full cycle its value will be zero), the half cycle must be either positive or negative.

lfm

a half cycle, one quarter is positive and the other quarter is negative, the average value for the half cycle will be zero as shown in Fig. 10.5.

Average over +ve half cycle

Average over full cycle El T lav= 0 (A) l:il Average over quarter +ve cycle and quarter -ve cycle

(C) Flg.10.5

(8)

• Also the term 'an·ac current'"in use is.n;tpr~per as the term_ ac itself stands for alt;;;ti~-;-current. This is like the.·term·'SI system'. ·

7

J

GRB

Physics for Medical Entrance Exams (2nd Year Programme)

Question II. In what situations is it preferable to use an ac source rather than a de source? When is de preferable to ac?

Answer: Though ac is more dangerous than de of same value, it is preferred over de as:

(I) ac is cheaper than de ( as battery has limited life and is more costly).

(2) It can be easily converted into de with the help of a rectifier.

(3) It can be easily stepped up or down with the help of a transformer.

(4) It can be transmitted over long distances at high voltage without much power loss.

( 5) It can be regulated by using choke-coil, without any significant waste of energy.

de is preferred over ac to study chemical effects of current such as electrolysis and electroplating as ac cannot cause chemical effects of current.

Question lll. Show that average heat produced during a cycle of ac is same as produced by de with I= lrms.

Answer : As for ac, I= 1₀ sin 001, the instantaneous value of heat produced per sec in a resistance R,

H = 12 _{R =Ii sin}2 _ootxR

And so the average value of heat produced during a cycle,

r

Hdt

r

(IJsin2ootxR)dt H

=

_o _ _

=

' ' o '

-av

J;

dt

S:

dt

i.e., H,v=1IJR [as( sin2ootdt=1T]

or H,v

=

1-:.,.R ... (I) [ as I nm=

Ji]

However, in case of de,

H =I_av 2R ... (2)

So from Eqs. (I) and (2), I=Irms

i.e., ac produces same heating effects as de of value I =Inns· This is also why ac instruments which are based on heating effect of current give rms value.

Problem 1.

If

the voltage in an ac circuit is represented by the equation.

V = 22W2sin (3141- <!>) calculate (a) peak and rms value of the voltage, voltage, ( c) frequency of ac.

Solution: (a) As in case ofac,

V=V₀sin(oot-<!>) The peak value

V0 = 220;'2 = 311 V

and as in case of ac,

V. V = -0 ·V =220V nm

.Jz •

nns (b) average (b) In case of ac

2

2

V,v =-V0 =-x311=198.17V 1t 1t (c) As oo = 21tf, 21t/ = 314

.

I=

314 =

so

Hz i.e., 2X1t

Problem 2.

If

a direct current of value a ampere is superimposed on an alternating current/= b sin 001 flowing through a wire, what is the effective value of the resulting current in the circuit?

I ~ = ?

\__/1-Fig. 10.6

Solution : As current at any instant in the circuit

will

be, /=Ide+ I,0=a

+

b sin 001

So, l,tr=

[r

_o_

l

=

_r

(a+bsinoot)2dt

2 dt] 112 [ I

r

]

112

J:

dt

rJo

. J T 2 · 2 • 2 [ ] m

i.e., l,tr

=

rfo

(a +2absmoot+b sm oot)dt but as and So, I

j' .

- smootdt=O To I

rT . ,

I

rJo

sm-001dt=2 [ I ] 112 l,tr= a2+2b2

Problem 3. Find (a) the average and (b) rms value for the saw-tooth voltage of peak value V0 as_shown in Fig. 10.7.

+V,

t

V

l-,-,---,&...:...=_-J-_ _

-,,,,;:;::..c...::...L,--+

1

-Fig. 10.7

Solution : As the equation of the saw-tooth wave shown in Fig. 10.7

will

be, So, (a) 2Vo (21 ) V=Tt-V0 =V0

T-1

v,v

J

T/2

f'

V dt 0 Vdt T/2 '-"'-=---rT/2 dt

J'

dt

Jo

T/2

=2

rrt2vo(21 - 1)d1

=

Vo

r!o

T 2 www.bathlabooks.com

www.puucho.com

and (b)

ie.,

ie.,

10.2 Circuit Elements in ac Circuits

In de circuits if transient effects are excluded, only resistance R can be used to control the current in the circuit, as inductor acts as short circuit while capacitor as open-circuit.

However, in case of ac circuits all the three, resistance R, inductance L and capacitance C, or any combination of these can be used to control the current in the circuit. Resistance

R,

inductance L and capacitance Care called circuit elements and are shown in Fig. 10.8.

---A,/1/\,-- ~

R

L

-11--

_c

Fig. 10.8 The ac circuit-elements

To understand how R, L and C control the current in an ac circuit we consider the following cases:

(A) A Resistor in an ac Circuit

If an alternating voltage E = E₀ sin

rot

is applied across a resistance as shown in Fig. 10.9 (A), by Kirchhoff's loop-rule at anytime

t,

or R E = IR, i.e,,

I= Eo

sin

rot

R E

I=-R

[as E = E₀ sin rot]

CJ

E o r / ~ lL../ V 0 n h &

r

R rot-E= E0sin rot Circuit (A) Wave-diagram (8) Fig. 10.9

or 1=1₀ sin

rot

with 1₀ =E₀ IR From this it is clear that:

Phaser-diagram*

(C)

( 1) The frequency of current in the circuit is ro and is same as that of the applied voltage.

(2) In a resistance, applied voltage is in phase with the resulting current.

(3) Apart from instantaneous value, current in the circuit is independent of frequency and decreases with inct~:a:~n~~ R (similar to that

in[SL_d~;~~!~~)-

.

J.

J.

1-~ 2

roorf- R

-(A) (8)

Fig. 10.10

NoitE! : At high ·trequency, current is limited only on the surface of a wire; I

so due .to_ decrease in cross-section, resistance of the wire_ is

!

considerably increased. This effect.is cal_led 'Skin effect'. In order to

I

minimise skin effect the conductors carrying ac are made up-of a I

large num~er of strands of fine wire connected in _parallel at their l ends and'insulated throughouttheir length from·each,other. This 1

increa~es the surface area and thus, _decreases the resistance. \

(B) An Inductor in an ac Circuit

If a sinusoidal emf

E

=

E

0 sin

rot

is applied across an

inductorof inductance Las shown in Fig. 10.11, by Kirchhoff s loop-rule we have, , or or

dl

E-L-=0 dt

L

- =

dl

E .

₀ smrot dt di E .

- =

__Q_ sm

rot

dt L

[asE = £₀ sin rot]

L

which on integration gives,

Eo

I

=

Loo

cos

rot

E = E, sin rot

i.e.,

I= 1

₀ sin ₍

rot-

1t ₎ withJ₀ = - 0

E Flg.10.11

2

roL

From this expression it is clear that:

(I) The frequency of current in the circuit is same as that of applied emf but current in an inductor 'lags' the applied voltage by (it/2) [ or voltage leads the cun-ent by (it/2)] as shown in Fig. 10.12 (A) and (B).

rot-Wave-cliagram (A) Fig. 10.12 E

L

Ii

Phaser-diagram (8)

(2) As here /₀ =

(E,jroL),

the quantity

roL

has the

dimensions of resistance as,

[roL] = [rad x HJ= [rad x ohm x sec]= [ohm] sec _ _ _ sec

GRB

Physics for Medical Entrance Exams (2nd Year Programme)

(3)

This quantity is called 'inductive-reactance' and is represented by XL and represents the opposition of a

coil to ac,

i.e.,

(asro=21tf)

With increase in frequency

'inductive reactance', i.e.,

opposition of a given coil to ac increases linearly with frequency. So ifro-+ 0,XL-+ 0 and ifro-+ =,XL-+=. This is why an inductor is called 'low pass-jilter' and as for de ro = 0, i.e., XL = 0, the opposition of an inductance to de is zero, i.e., an inductance acts like a simple conducting wire in steady-state in a de circuit.

L = constt.

t

t

coorf-(A) (B) Flg.10.13

(C) A Capacitor in an ac Circuit

If an alternating voltage E = E₀ sin rot is applied across a capacitor as shown in Fig. 10.14, the Kirchhoff's loop-rule

gives,

C

E;:; E₀ sin rot Flg.10.14

E-'L=o

e

or q = eE₀ sin rot (as E = E₀ sin rot)

I

=

dq = eroE

0 cos rot

dt

or I =10sin( rot+~ )withJ0 = E0ero From this expression it is clear that:

(I) Current in the circuit has same frequency as the applied voltage but leads it by (7t/2) [or voltage across a capacitor lags the current by (it/2)] as shown in Fig.

10.15 (A) and (B).

E , J ~ 3 •

,,I

0 ~ rot- E Wave-diagram (A) Fig.10.15 ~ l e

l

Ve Phasor-diagram (B)

(2) AshereJ₀=eroE₀=

Eo

withXe = -1-,(1/roC)has

Xe roe

(3)

dimensions of resistance as,

[ro~]=Lad

1

s-1

_x

_{c:w]=[:]=ohrn}

and so it represents the opposition of a capacitor to the flow of ac through it and is called 'capacitive

reactance'.

I

l

'h·

.

fr X

As Xe = - =--,wit mcrease m equency e roe 21tfe

decreases non-linearly, i.e., the opposition of a capacitor to ac decreases with increase in frequency. So if ro -+ 0, Xe -+

=

and if ro -+ =,Xe-+ 0. This is why a capacitor is called 'high pass-jilter' and as for de ro-+

0,

Xe-+=, the opposition of a capacitor to de is infinite, i.e., a capacitor acts as open circuit in de circuits in steady-state.

t

~

t

~

Xe Xe roorf-

c ~

(A) (B) Fig.10.16

10.3 Terminology and Analysis of Series

ac Circuits

From the theory of circuit elements, it is clear that in an ac circuit, inductance L, capacitance

e

and resistance R all produce opposition to the flow of current through them due to which a phase difference may be created between applied voltage and the current in the circuit. So the current in a given ac circuit, when voltage,

E= E₀

sin

rot is applied, will be given by,

I=

J₀sin (rot-<!>)

where 1₀ is the 'peak value' of current and <I> is the 'phase difference' between voltage and current in the circuit.

To calculate I and <I> without dealing with the theory we adopt the following readymade procedure and terminology:

(!) Specify the total resistance R of the given circuit which is independent of frequency and so is same for both de and ac.

(2) Compute inductive and capacitive reactances, ifL and C

are given, using,

XL

=WL

and Xe = -1- with ro = 21tf roe

(3) Calculate the reactance

'X': The combined opposition of inductive and capacitive parts to the flow of ac through a circuit is called 'reactance' X and is given by,

X=XL-Xe

The unit of reactance is ohm and it can be positive, negative or zero as XL>, < or= Xe.

(4) Calculate the impedance

'Z': The total opposition of

an ac circuit to the flow of ac through it is called 'impedance' and is given by,

Z=~R2 +X2

The unit of impedance is also ohm and it is always positive.

(5) The peak value of current

1

₀ is calculated by applying Ohm's law to ac circuits by replacing resistance

R

by impedance Z, i.e.,

V Vo

V=IZso that]= -or1₀ =

-Z

Z

(6) The 'phase difference' $ between I and Vis given by,

"' X . "' _1

(X)

tan.,=

R,

i.e., ., = tan

R

Note:. In ac Circuits expressions tOr Z.and $ can

be

easily kept in mind with the-help.of so called_ 'ac triangle~ as·shown_ in Fig.10.17.

t

z

ab X Triangle R -Flg.10.17

z

= [RZ + X'Jl/2 $ = tan-1_(X/R) (7)

In

case of ac circuits as V = IZ V2 = J 2 (R 2 + X 2) =

(IR)2

+ (IX)2 [as

z

=

(R2

+

X2)1/2] i.e., V2 =Vf +V} [asVR=JRandVx=lX]

with, Vx=IX=J(XL -Xe)= VL - Ve i.e., in case of ac

V," VR + Vx but V2 =Vi +V} and Vx* VL + Ve but Vx= VL - Ve

Now we shall illustrate this theory by considering the given cases:

(A) ac Applied to LR Circuit

If an ac, E = E0 sin OJI, is applied to an L-R circuit as shown

in Fig. 10. I 8 (A), the reactance, X -XL =OJL

Soimpedance,Z=~R2 +X2 =~R2 +OJ2L2

and hence, ... (I)

and

So current in the circuit at any time will be given by, J=J₀sin(OJt-$)

L R

E;::E₀sin rot

(A) Fig.10.18

~~

--1"> k:'...J..!..$ _ __,__/ VR-(B) ... (2)

where

1

0 and$ are given by Eqs. (I) and (2) respectively.

From this it is clear that:

(I)

In L-R circuit, current lags the applied voltage by an angle of tan-1 (

OJL!R)

rad.

(2) Across the resistance, voltage and current are in same phase while across the inductance voltage leads the current by (it/2) rad; so VL leads VR and Jby (it/2) rad in phase. This all is shown in Fig.

I 0.18

(B).

(3) If instead ofac, de is applied, OJ=O soX=O, Z =Rand hence

1

₀ =

(EofR)

and$=

0°,

i.e., inductance becomes ineffective.

(4) Voltage across R andL will be given by, Vn =JR and VL = IX=

IOJL

with V2 = Vf +V; (not V= Vn + VL)

(B) ac Applied to CR

Circuit

If an ac

E

=

E

₀ sin OJI is applied to a

CR

circuit as shown in Fig. 10.19 (A), the reactance,

X=XL -Xe =-(OJ~) So, impedance,

Z=~R2 +X2 +~R2 +(l/OJC)2

Eo

Eo

and hence,] 0 = - = -oc=======

Z R2+(0J!Cr

and _$=tan-'(!)

-1 ( I )

= tan

OJCR

So current in the circuit at any time will be given by, J:=J0sin(OJI-$)

... (!)

... (2)

GRB

Physics for Medical Entrance Exams (2nd Year Pr"ogramme)

E= E0 sin rot

(A) (B)

Fig.10.19

where

1

0 and cl> are given by Eqs. (1) and (2) respectively.

From this it is clear that:

(I) Current leads the applied voltage by an angle tau-1 (1/weR).

(2) Reactance is negative, which implies voltage across capacitor ewill lag the current through itby

(1t/2)

rad. (3) If instead of ac, de is applied,

w

= 0, so

X

= =,

Z

= =

andhenceJ0= (EIZ)=O, i.e., there will beno current in the circuit.

(4) Voltage across Rand ewill be given by, I

VR =JR and Vc=IX =

-we

with V2 =Vi +VJ

(C) ac Applied to

LC

Circuit

In this situation as,

(andnotV = VR +Ve) and E = E₀ sin rot Fig.10.20 R=O,XL =WL 1 X c = -we So, _{X=XL -Xe =(wL- w~)} Z=~R2+X2 =X

E

E

and hence, I_O = _Q_ = 0 z [wL-(1/we)] and cj>=tan-1

(!)

= tan _1 [wL - (1/wC)] 0 (asR

=

0) -I ( )

1t

tan oo =-2 So current in the circuit at any time will be given by,

J=J0sin (wt- cj>)

where

I

O and cl> are given by Eqs. (!) and (2) respectively.

From this it is clear that:

(!) Current in the circuit lags the applied voltage by (7t/2) rad.

(2) If

)

I

However, as

b

is the natural frequency of L-C vLC

circuit. So in case of L-e circuit

if

the frequency of applied ac becomes equal to the natural frequency of the circuit, the amplitude of current in the circuit becomes infinite due to zero impedance. This situation is called 'resonance' and is depicted in Fig. 10.21.

(0 = COo r o -(A)

t

z

(B) Fig.10.21

Problem 4. An alternating voltage E = 200

..fi.

sin (100 t) Vis connected to a lµF capacitor through an ac ammeter. What will be the reading of the ammeter?

(a) 20 mA (b) 20..fi. mA

(c) 10..fi.mA (d) 40mA

Ans. (a)

Solution : Comparing E = 20o..fi.sin (I00t)with E = E₀ sin wt we find that, and So, E0 = 2oo..fi.

v

w

= I00(rad/s) 1 I X e = - = - - - - 104 Q we I

oo

x

10-6

And as ac instruments read nns value, the reading of ammeter will be,

i.e.,

[asErrns=iandZ=Xc] 20mA

Problem 5. A 0.21 H inductor and a 12 ohm resistance are connected in series to a 220

v;

50 Hz ac source. Calculate the current in the circuit and the phase angle between the current and the source voltage.

Solution: Here XL = wL= 2rcjL= 2rc x 50x 0.21= 2JrcQ So, Z=~R2 +X2 =J122 +(2lrc)2

= .J144 + 4348

i.e., So, (a) and (b) Z=.J4492 = 67.02Q V 220 I= - = - - = 3.28 A Z 67.02

~

= tan-1 ( ; ) = tan-1 ( 2 1~) = tan -l (5.5) = 79.7°

i.e., the current will lag the applied voltage by 79.7° in phase.

Problem 6. When 100 volt de is applied across a coil, a current of! amp flows through it; when 100 V ac of50 Hz is applied to the same coil, only 0. 5 amp flows. Calculate the resistance and inductance of the coil.

Solution : In case ofa coil, i.e., LR circuit,

I=~withZ=~R2 +X2 =~R2 +(ro£)2

z

So when de is applied, ro = 0, so Z = R V . V 100

andhence l=-,1.e., R=-=-=lOOQ

R I 1

and when ac of 50 Hz is applied, V . V 100

I = - ,.e. Z=-=-=200Q

z' '

I 0.5

but Z=~R2 +ro2£2 ,i.e.,ro2£2 =Z2 ""R2

i.e., (21tfLJ2. =2002-1002 =3xl04 (asro=21t/) So, L .J3xl02 =

,fj,

H=0.55H

21t X 50 1t

Problem 7. ,An ac source of angular frequency ro is fed across a resistor Rand a capacitor C in series. The current registered isl

If

now the frequency of source is changed to ro/3 (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original

frequency ro. [AIIMS 2008]

(a)

~

(b)

t

(c)

1

(d)

H

Ans. (b)

Solution : According to given problem,

I=~= VI [R 2 + (1/Cro

)2 ]

112 ... (1)

z

and I=V/[R2 _{+ (3/Cro) 2]"2 ... (2)}

z

Substituting the value off from

Eq.

(1) in (2), 4(R 2 +_I_)= R 2 +

---2....,-'

C2ro2 C2ro

i.e., _l_=~R2

c2ro2 5

So that,

Problem 8. A 50

w;

I 00 V lamp is to be connected to an ac mains of200 V, 50 Hz. What capacitance is essential to be put in series with the lamp?

(a) 9.2 mF (b) 4.6 µF (c) IR4 µF (d) 9.2µF

Ans. (d)

V2 1002

Solution: As resistance of the lampR = _§__ = - - = 200 Q and W 50

h / V lOO 1 A h th I . .

t e max. current = - = - = - ; sow en e amp 1s put m

R 200 2

series with a capacitance and run at 200 V ac, from V = JZ we have, Z=~= 200=400Q

I 1/ 2 Now as in case of CR circuit,

Z= R2+(ro~r,i.e.,R2+(

00

~r

=160000 or

CJ)~

_{r = ]6X 10}4 - (200)2 = 12x 104

So, -1-=.Ji2x102 ore 1 F roe · 1001t x .Ji2 x 102

i.e., C=~µF=9.2µF

1t.Ji2

Problem 9. A capacitor of 10 µF and an inductor of 1 Hare joined in series. An ac of 50 Hz is applied to this combination.

What is the impedance of the combination? (a)

_!±

Q (h) 14 Q (c) 15 Q

31t 1t 1t

Ans. (b)

(d) 16 Q

1t

Solution : Here XL= roL = 21tjL

=

21t x 50 x I = 100 1tQ

1 I X c = = -roC 21tfC I 103 = - - - = -Q 21tx 50x !Ox 10-6 1t So, X=XL-Xc =1001t- 1~ 3 =102[ 1t 2 : 10

]n

So, Z=~R2 +x2 = IXI= 102 [l0-1t9.86] Q i.e.,

Problem 10. A 200 km long telegraph wire has capacity of0.014 µF /km. If it carries an alternating current of frequency 5 kHz, what should be the value of an inductance required to be connected in series so that the impedance is minimum?

(a) 0.72 mH (b) 0.18 mH (c) 0.96 mH (d) 0.36 mH

Ans. (d)

GRB

Physics for Medical Entrance Exams (2nd Year Programme)

Solution : As in _case of L-e circuit impedance,

Z=IXL-Xel

So, it wHI be minimum when,

.

.

I

I

XL =Xe ,.e, roL=- orL=

2

roe ro

e

(21tf)2e

Here as the wire is 200 km and has capacitance 0.014 µF/km

e

= 0.014 x 200µF= 2.SµF and hence, L I (21t x 5x 103

)2

x 2.8x 10-6 I 1001t2 X 2.8 i.e.' L IO

mH=

0.36

mH

1t2 X 2.8

10.4 Resonance in Series

LCR

Circuit

If an ac

E

=

E

0 sin

cot

is applied to a circuit containing

L,

e

and R in series as showu in Fig. 10.22, then as,

So, and C E = E0 sin rot Fig. 10.22 X = XL - Xe = ( col -

~e)

Z=JR2+X2 =JR2 +[col-(VcoC)]2

Eo

Eo

Io

= - = --.=========

z

JR2 +[coL-(l/coe)J2 <!>=tan _, X -=tan _, [col - (I/ roe)]

R R

... (1)

... (2) So current in the circuit at any time t will be given by,

I=

J₀sin (

cot-

<j>)

where /0 and <I> are given by Eqs. (!) and (2) respectively. From this it is clear that in case of series LCR circuit:

(I) Current in the circuit may lag, lead or be in phase with the applied voltage depending on the fact that XL > Xe, XL <Xe or XL =Xe respectively.

1 .

(2) If XL

=

Xe, i.e., col= roe

1

I

i.e., co=

.fie

or

f

=

2

1t.fic

= fo

i.e., the frequency f of applied ac is equal to the natural frequency of the circuit / 0

k•

the circuit is

27t

said to be in

'resonance'.

(3) In case of series resonance, i.e.,!=

Jo:

(i) XL

=

Xe, i.e., 'inductive reactance' is equal to capacitive reactance.

(ii) X=XL -Xe=

0,

i.e., 'reactance' of the circuit is

zero.

(iii) Z

=

~~R-2_+_X_2

=

_{R, i.e., 'impedance' of the}

circuit is minimum and is equal to resistance [Fig. 10.23 (A)].

( . ) I 1v _O

= -

Vo

= -

Vo max. i.e., current m . .

t

h e czrcuz . ·1 . zs

Z

R

maximum [Fig. 10.23 (B)].

'

'

1

'

_'

1

'

'

'

z

_'

_'

I

'

'

'

R

---'

min. (A) (B) Fig. 10.23

(v) Before resonance, current in the circuit leads the applied voltage (as XL <Xe) and after resonance it lags the applied voltage (as Xe

<

XL) and at resonance <j>

=

1an-1 _(X/R)

₌

_1an-1 _(0/R)

₌

_{0, i.e.,}

current is in 'phase' with applied voltage.

( vi) Currents through L and C are same but

I

80° out of phase with respect to each other so that net PD

across reactance is zero,

i.e., Vx= VL-Ve=O

with V= VR

(vii) The 'power factor' of the circuit, R

PF= cos<j> = -=!=max. (asZ= R = min.)

z

and hence power consumed by the circuit, Pav= V rms I rms cos<j>

I

=

-V

₂ 0J0 = max.

(viii) The series resonant circuit is called 'acceptor circuit'

as at resonance its impedance is minimum

and it most readily accepts that current out of many currents whose frequency is equal to its natural frequency. In radio or TV tuning we receive the desired station by making the frequency of the circuit equal to that of the desired station.

(ix) At resonance as

1

0

=

EofR,

roL

So, VL =1₀ XL= - E0

R

i.e., VL =QE0 withQ=

0:,

and is called 'quality factor' of the circuit. Thus, at resonance the voltage drop across inductance ( or capacitance) is Q times the applied voltage. Hence, the chief characteristic of series resonant circuit is 'voltage magnification'.

Note : The ·theory of ac circuits con_taining:circuit elements in _parallel is

beyond the scope of this book, eventheri irl case of circuitshown:in

Fig .. 10.24. This theory reveals that the condition of resonance iS,

w=Jl~-~

and-impeda.nce at resonance is max. while current is min., i.e.,'

l Z=-=max. le l RC

L

R

C

E=E₀sin rot Fig.10.24

and

i

₀=EoCR=min l And'if R->0

oo=

~

with Z =·co and 1₀ =O. this.is shown'in Fig.

lo.is.'

vlC

f-i\

R= 0 I \ . L

t _______

J \. ___ RC z Ol

=.

0\, c o - ro=00o ro--(A) (B) Flg.10.25

So in case.of 'parallel resonance', at resonance:

{i) The current.in the·circuit is-minimum and that is why a_ parallel resonan_t circuit is callE!d a 'rejet;!or circuit' and due to its large impedance it is widely used in radio.

'(ii) If the resistance in the circuit is zero,

IL= Vo sin (mt

'-2:)

x,

2

and le= Vo. sin.(rot+2:)

Xe 2

So the net·current in the circuit, l=JL+lc=O

10.5 Power in ac Circuits and Choke Coil

(A) Power in ac Circuits

If an ac voltage E

=

E₀ sin

cot

is applied to a circuit and the resulting current is,

with,

and

the instantaneous power consumed by the circuit,

P= VJ=IoE

0 sincotsin(cot-q,)

So the average* power consumed,

or

s:

Pdt Pav = ~ T

-fo

dt = Eolo

rr

sincotsin(cot-q,)dt

T

Jo

EI

rT

P,. = ; /

Jo [

cos q, - cos(2cot - q,) ]dt which on simplification with

T=

(27t/co) gives,

I

Pav =

2

Eol o COS q> = E rrnsl nm COS q> ... (!) cos q, is called 'power factor' of the circuit and expression (!) is the desired result and from this it is clear that:

(I) In case ofa 'pure resistance' asX= 0,

(2) (3)

R R

PF

=cosq,=-=-=l

Z R

So,

In case of a pure inductance or capacitance as

R

=

0,

R

0

PF

=cosq>=-=-=0

Z X

So, Pav=ErmsJrmsx0=0

i.e., average power consumed per cycle in a pure inductance or capacitance is zero. So in an ac circuit if the phase difference between V and J is 7t/2, i.e., if,

V

=

V₀sin

cot

and 1=10 cos

cot

the power consumed by the circuit is zero (min.). Such current is called 'wattless'.

In actual practice an inductor or capacitor has large reactance and a finite small resistance; so the phase difference between E and J is large but less than 90° and so power consumed by inductor or capacitor is not zero but very small.

* Herej!Y~~~ge is-defined over a

cy,cclee•-~---J

GRB

Physics for Medical Entrance Exams {2nd Year Programme)

(B) Choke Coil

In a direct current circuit, current is reduced by means of a rheostat (resistance). This results iu a loss of electrical ·energy J2R per sec as heat in the resistance.

L, r Choke coil Flg.10.26

The current in an ac circuit can however be reduced by means of a device which involves very small amount ofloss of energy. This device is called 'choke coil' or ballast and consists of a copper coil wound over a soft iron laminated core. This coil is put in series .with the circuit in which current is to be reduced.

As this circuit is a LR circuit, the current in the circuit,

I=~ withZ=~(R +r)2 +(roL)2

So due to large inductance L of the coil, the current in the circuit is decreased appreciably. However, due to small resistance of the coil

r,

the power loss in the choke,

P,v

=

V rms J rms COS «p ~ 0

r r r

cos«p=-=-;=====-~0 Z ~r2+co2L2 roL as

Problem 11. An alternating emf of frequency 50 Hz is applied to a series circuit of resistance 20 ohm, an inductance ofl 00 mH and a capacitor of30 µF. Does the current lag or lead the applied emf and by what angle?

Solution : As here, So, and hence, ~-r l. e.' and so, XL

=

OlL

=

2irjL = 271 X 50X 0.1= !On= 31.4 ohm I I X e = = -roC 271jC 21t X SOX 30x 10-6 = 106.16ohm X=X, -Xe =31.4-106.16 =- 74.76 , «p=tan-1 (1)=tan-l (-::76) «I>= - tan -I (3.73) = - 75° 1=1₀ sin[rot-(-75°)] =10 sin(rot + 75°) i.e., current will lead the applied voltage by 75°.

Problem 12. A I 00 V ac source of frequency 500 hertz is connected to anLCRcircuitwithL= 8.1 mH, C= 12.5 µFandR= 10 Q all connected in series. Find the potential difference (PD)

across the resistance.

Solution : As here, So, and hence, So,

and so,

Xi

=

roL

=

21t/L

=

2 X 1t X 500 X 8.1 X !0-3 = 8.11t = 25.434

n

I 1 X e = = -roC 271jC 1

= . ,

-21t X 500 X 12.5 X 10-6 '

103

=--=25.478Q 12.571 X=X, -Xe = 25.434-25.478= - U044 Q Z=~R2 +X2 =~(10)2 +(U044)2 "'lOQ V 100 1=-=-=lOA Z 10

VR =IR= !Ox 10= IOOV

Problem 13. A 750 hertz, 20 V source is connected to a resistance ofl 00 ohm, an inductance of0.1803 henry and a capacitance of I 0 microfarad all in series. Calculate the time in which the resistance (thermal capacity 2 Jf'C) will get heated by I O"C.

(a) 12.4 min (b) 5.8 min (c) 2.8 miu (ti) 1.2 min

Ans. (b)

Solution : As in this problem,

Xi= ml= 271/L = 271 x 750 x 0.1803 = 849.2

n

I I I

and, X e = - = - - = - - - - ~ 21.2Q roe 271fC 271 x 750x 10-5

So, X =Xi -Xe= 849.2-21.2= 828Q

andhence,Z=~R2 +X2 =~(100)2 +(828)2 =834Q

but as in case of ac,

i.e., P,v

=

V nn, I nm COS «p

=

V nm X V""' X

!!_

z

z

P

=(v=)2

xR ~v Z = ( 20 ) 2 x 100= U0575W 834 and

as,E=Px

t=mc80= (TC)80 t (TC)x80 2x!O P U0575 = 348s == 5.8 min. www.bathlabooks.com

_,,

www.puucho.com

Problem 14. An LCR series circuit with 100

n

resistance is connected to an ac source of200 Vand angular frequency 300 radls. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. Calculate the current and the power dissipated in the LCR circuit

Solution : When capacitance is removed, the circuit becomes LR

with,

tanq>=XL ,i.e., XL=Rtanq>=I00../3Q R

and when inductance is removed the circuit becomes CR with, tanq>=Xe, i.e., Xe =Rtanq>=IOO,J3Q

R

as here XL

=

Xe so the circuit is in series resonance and hence

asX=Xi -Xe =0,i.e.,Z=~R2 +X2 =R, So, I =Vans= Vans= 200 = 2 A

nns Z R 100

and Pav =V nns / nns COS q>= 200X 2X 1=400W

Problem 15. An inductor 20 x 10-3 _{henry, a capacitor 100 µF}

and a resistor 50 .Q are connected in series across a source of emf

V = 10 sin 3141. Find the energy dissipated in the circuit in 20 minutes. lfresistance is removed from the circuit and the value of inductance is doubled, thenfind the variation of current with time

in the new circuit.

Solution : Here, time of I cycle T = 1/50 s. So, we have to calculate the average energy as time >> T.

~'7

10 sin 314 t Fig.10.27

Energy consumed in time

t=V=Jrrns cos q>t 1₀ V₀ R = - · - · - /

..fj, ..fj,

z

E=VtRt

2Z2

Now,Z=~R2 +(

ooL-cicr

(·:1

0

=~)

= (50)2 + (314 X 20 X 10-3 - I ) 2 314x IOOx 10-• = .J3153.6

=

56 ohm :. Energy consumed 102 x 50x 20x 60 2x 3153.6 95IJ

When resistance is removed,

p 1t cos q>=- = Oor q> =-Z' 2 Z' =-

_roe

1- -roE 314 x40x 10-3 _{= 19.3} 314 X 10-4 I= Eo sin (rot+ q,)

Z'

=

.IQ_

sin (314 t + 1t! 2)= 0.52cos 3!4t

19.3

Problem 16. A box P and a coil Q are connected in series with an ac source of variable frequency. The emf of source is constant at IOV. Box P contains a capacitance of l µF in 4.9 mH and a resistance of68Q in series. The frequency is adjusted so that the maximum current flows in P and

Q.

Find the impedance of P and

Q

at this frequency. Also find the voltage across P and Q respectively. 1µF C 32Q 4.9mH 68Q R, L R₂ BoxP -coilQ Fig.10.28

Solution : As this circuit is a series LCR circuit, current will be

maximum at resonance, i. e. , I

ro=--

_..fic

I 105 rad = ~(4.9x 10-3 )(10-6) 7 s

with I =Eo IOV

_!_A

max. R _{(32+ 68) 10} So the impedance Zp =[Rf

+(ro~rr

2

=[(

32₎2 _+Co,

:10-6

rJ/2

= -.J5924 =

n

n

and ZQ =[Ri +

(roL)

2]"2

=[(68)2 +( 4.9xl0; 3

x 1 o ' r f 2

=

.J

9524

=

97.6

n

GRB

Physics for Medical Entrance Exams {2nd Year Programme) and hence and 1 Vp =IZp =-X(77)=7.7V 10 1 VQ =IZQ =-X (97.6)=9.76V 10

Problem 17. A series LCR circuit containing a resistance of 120 .Q has angular resonance frequency 4 x 105 rad s-1. At

resonance the voltages across resistance and inductance are 60 V

and 40 V respectively. Find the values of L and C. At what frequency the current in the circuit lags the voltage by 45°?

. V 60 1

Solution: At resonance as X = 0, I= - = - = -A

R 120 2

V

and as VL = IXL = Irol, L = __!,__

roI So, L= 40 -0.2mH

G)

x

4

x 105 and as ro₀ = - -I C = - -1

..fLC'

Lro~ I i.e., 0.2xl0-3 X (4 X !05

)2

Now in case of series LCR circuit,

C

tan,P

XL-Xe

R

So current will lag the applied voltage by 45° if,

1 m l -tan45° roe R I 32µF i.e., lxl20=rox2xl0--4 - 1 ..,; ro(l/ 32) X 10 i.e., ro 2 -6xl05 ro-16xl010 =0 i.e., 6X 10 5

_±

_{J(6X 10}5 )2 + 64 X 1010 r o = ~ -2 i.e., ro= 6x 10 5 + !Ox 105 _{8x 105 rad} 2 s

Problem 18. A currentof4Aj/ows in a coil when connected to a 12 V dcsource. Jfthe same coilis connected to a 12 V, 50 rad/sac source a current of 2.4 A flows in the circuit. Determine the inductance of the coil. Also find the power developed in the circuit if a 2500 µF capacitor is connected in series with the coil.

Solution: IncaseofacoilasZ=JR2 +ro2L2

V V

i.e., I=z= JR2 +ro2 L2 So when de is applied as ro = 0,

I=V i.e. R=12=3.Q

R' ' 4 ... (!)

and when ac is applied,

or So,

i.e.,

but as

I=~ i.e. Z=(~)=(E.)=5.Q

z' '

I 2.4 X}. = 52 -R2 = 52 -32 =42 XL =4.Q XL 4 XL =mL L=-=-=0.08.Q ' ro 50

Now when the capacitor is connected to the above circuit in

series, as So, and hence So, i.e., 1 103 - - - = - = 8 . Q roe 50 x 2500 x 10..,; 125 Z=JR2 +(XL-Xc)2 =J32 +(4-8)2 =5.Q V 12 I=-=-=2.4A Z 5 Pav = V ""'/ nns cos qi

=(Inns xZ)I nns x(i)

Pav =I~R=(2.4)2 x3=17.28W

Problem 19. A choke coil is needed to operate an arc lamp at 160 V (rms) and 50 Hz. The arc lamp has an effective resistance of

5 .Q when running at 10 A (rms). Calculate the inductance of the choke coil. Jfthe same arc lamp is to be operated on 160 V(dc), what additional resistance is required? Compare the power losses in both cases.

Solution: As for lamp VR =IR= 10

x

5 = 50 V, so when it is connected to 160 V ac source through a choke in series,

V2 =Vi +Vf ,VL =J1602-502 =152V and as, VL =IXL = ImL = 21tjLI

So, L = VL 152 4.84 x 10-2 H 21tjl 2x 1t x 50x 10

Now the lamp is to be operated at 160 V de; instead of choke if additional resistance r is put in series with it,

i.e., Arc ramp .._\I,,. - -R V= V0 sin rot Fig.10.29 V = I (R + r), i.e., 160= 10(5+ r) r= 11.Q

In case of ac, as choke has no resistance, power loss in the

choke will be zero while the bulb will consume,

P =12 R = 102 X5= 500W

However, in case of de as resistance r is to be used instead of choke, the power loss. in the resistance

r

will be,

PL= 102 X 11= 1100W

While the bulb will still consume 500 W, i.e., when the lamp is run on resistance r instead of choke more than double the power consumed by the lamp is wasted by the resistance r.

Problem 20. A LC circuit (inductance 0.0 I H, capacity I µF) is connected to a variable frequency ac source as shown. Draw a rough sketch of the current variation as the frequency is changed from

I

kHz to

2

kHz.

Solution: As shown in Fig. I 0.30 (A) here Land Care connected in parallel to the ac source, so resonance frequency,

L (A)

t

I 1.6 kHz (B) Flg.10.30

f=

I 21tM 21t~O_OJx 10-6

=

104 = 1.6 kHZ 21t

,_

and as in case of parallel resonance, current in LC circuit at resonance is zero, so theJ:f curve will be as shown in Fig. 10.30 (B).

Problem 21. An ac source is connected to two circuits as shown in Fig. I 0.31 (A) and (B). Obtain current through the resistance R

at resonance in both the circuits.

L C R

v,

v.

I V (A)

v '

-R I I C V (B) Fig.10.31

Solution: (A) Circuit (A) is the case of 'Series resonance' and in case of series resonance as VL and Ve are equal in magnitude but have phase difference of 180° relative to each other,

Vx=VL-Ve=0

andas V2=V;+V;,

So

So in this situation current through R,

V V

I =

Z

=

Ji'

[

as at resonance Z = R]

(B) Circuit (B) is the case of "Parallel resonance". IfV' is the voltage across

Lor

C, we have

V' . (

,c)

IL = - s m wt--XL 2

V'

=--cosrot XL V' . (

,c)

le =-sm OJI+-Xe 2 V' =--cosoit Xe l=h +le =V'(-1- - -1-)cosoit Xe XL

but as in case of resonance as XL= Xe, in this situation current through resistance R is zero, i.e., I= 0.

Problem 22. For the circuit shown in Fig. 10.32 current in inductance is 0.8 A while in capacitance is 0.6 A. What is the

current drawn from the source?

L C

I _V

Fig.10.32

Solution: If an ac source E = E0 sin OJI is applied across an

inductance and capacitance in parallel, current in inductance will lag the applied voltage while across the capacitor will lead,

and so,

h

= ~sin (wt - ~)

XL 2

= - 0.8 cos 0)/

le =~sin(oii+~)

Xe 2

= + 0.6 cos 0)/

So the current drawn from the source,

l=IL +le =-0.2cosoit i.e.,

II

0l=0.2A

GRB

Physics for Medical Entrance Exams (2nd Year Programme)

Problem 23. For a resistance Rand capacitance C in series, the impedance is twice that of a parallel combination of the same elements. What is the frequency of applied emf?

Solution: As shown in Fig. 10.33 (A), in case of series

combination, Zs =~R2 +X~ =[R2 +(1/roC)2]112 R C (A) (B) Flg.10.33 In case of parallel combination,

JR = V sin rot and le = _!:'._sin (rot+~)

R Xe 2

So, I= IR + le = V sin rot+__!'.'._ cos rot

R Xe i.e., J=/₀ sin(Olt+q,) with J₀cosq,=-V R and Slll . "' ' t ' = -V Xc

and as according to given problem,

Zs= 2Zp, i.e.,

zJ

=

4Z;

(R2ro2C2 + 1) R2

i.e., ro2C2 4 (l+R2ro2C2)

i.e., (l+R2ro2C2

)2

=4R2ro2C2 or or 1 + R2 ro2C2 = 2RroC (RroC -1)2

=

0 I ro=-RC l f= 21tRC

Problem 24. A RC series circuit of R = 15

n

and C = IOµF is connected to 20 volt DC supply for very long time. Then capacitor is

disconnected from circuit and connected to inductor of! 0 mH Find

amplitude of current. [AIIMS 2016 (Evening)J

(a) 0.2v'!O A (b) 2v'!O A

(c) 0.2 A (d) v'!O A

Ans. (a)

Solution : After infinite time potential difference across capacitor is 20 volt so charge will be maximum

Q= CV= !Ox!0-6 x20= 2x 10-4 C

When charged capacitor is connected to inductor then there

will be LC oscillations I 2 Q2 -L/₀ = -2 2C lo =_g__= 2xl0-4

..JLG

~!Ox 10-3 X !Ox 10-6 2xl0-4 = 0.2v'!O A ~10-7

Problem 25. In an LC oscillatorcircuitL= lOmH,C =40µF . .lf

initially at t

=

0 the capacitor is fully charged with 4 µC, then find

the current in the circuit when the capacitor and inductor share

equal energies. [AIIMS 2016 (Mornlng)J

(a) 0.2 mA (b) 4.4 mA (c) 0.3 mA (d) 2 mA Ans. (b) L = 10 mH Solution: Uc = Q 2 = 0.2µJ (total) 2C I I J , 2 UL =Uc =D.lµJ=-L1 2 i = 0.44xl0-2 A= 4.4 mA C=40µF Fig. 10.33 (C)

Problem 26. Alternating voltage is generated by rotating a coil in a normal magnetic field with 50 Hz frequency. 1fV nns

=

220 V, then the maximum flux passing through the coil is :

(a) 0.99 Wb (c)

0.6

Wb Ans. (a) [AIIMS 2016 (Morning)J (b) 0.4 Wb (d) 0.7Wb

Solution : q, = NAB cos 8

fmax=NAB emf=- dq, =(NABsinB)ro dt l.414x220 = ₀_₉₉Wb 2x3.14x50 (·: 8

=

rot) www.bathlabooks.com

www.puucho.com

1. Which out of A.C., AC, ac and a.c. do you think to be correct abbreviation for 'alternating current' in physics?

[Ans. ac]

2. For an ac can ever:

(a) rrns value be equal to peak value (b) average value be equal to peak value (c) rrns value be equal to average value (d) all the three values be equal

[Ans. Yes; if the ac is square wave.]

3. What is the average and rrns value of voltage for square wave

having peak value V0 [Fig. 10.34] ?

r

V 0 I

r,

T 2

t

T iT 1' Fig.10.34 [Ans. V.,

=

V""'

=

V0]

I

2T

[Hint: Think carefully; you would not have to calculate.]

4. Explain why a current carrying wire for de contains a single

thick wire while for ac a number of thin wires insulated from each other along the length are used?

5. Define reactanceX and impedance Z. Can these be negative? If yes, when and what does it imply?

[Ans. Reactance can be negative when Xe> XL. It implies current

leads the applied voltage in the circuit.]

6. Can an ac source be connected to a circuit and yet deliver no power to it? If so under what circumstances?

[Ans. Yes, when the phase difference between voltage and current

is 90'.]

7. Calculate the average and rrns value of voltage for triangular wave having peak value Vo as shown in Fig. 10.35.

t

V

Fig. 10.35

[Ans. V,v = (Vof2) and Vnns = Vof../3]

2T

[Hint: The equation of voltage will be V = 4 V₀t!T]

i

8. The potential difference V across and the current I flowing through an instrument in an ac circuit is given by,

V

=

Scosro

t;

I= 2sinro

t

What is the power dissipated? [Ans. Zero, as$= 90°]

9. A 10 ohm electric iron is connected to 120 volt 60 Hz wall outlet. What is the peak potential difference on the electric iron?

[Ans. 170 VJ

10. What is the inductive reactance of a coil if the current

through it is 20 mA and voltage across it is

l

00 V? [Ans. 5 kn]

11. The electric current in an ac circuit is given by i

=

io

sin

rot.

What is the time taken by the current to change from its maximum value to the rrns value?

"

[Ans. T/8 or-]

4co

12. A voltage of IO V and frequency 1000 Hz is applied to a 0.1 µF capacitor in series with a resistor of 500 Q. Find the power factor (PF) of the circuit and the power dissipated. [Ans. PF= 0.3, P,v = O.QJ8 W]

13. Calculate the rrns voltage across the inductor and capacitor. Explain the reason that how it is possible that voltage on these element(s) is higher than supply voltage.

[AIPMT (Mains) 20041 XL =25Q Xc=75Q R=O 250V Fig.10.36 [A ns. urrent m c1rcmt : C . . . I

= -

V = - - -250

= -

250

=

SA X (75 - 25) 50 VL =IXL= 5

x

25=125 V and Vc=IXc=Sx 75=375V

Voltage on capacitor is more than that of supply voltage because the phase difference between VL and Ve is 180° (i. e, out of phase).] 14.

For

the given circuit write down the (AIPMT (Mains) 2004]

(i) phase difference between 11 and /R₁ and

r

•

GRB

Physics for Medical Entrance Exams (2nd Year Programme)

Flg.10.37

[ Ans. (i)

!:

rad or 90°, (ii)

!:

rad or 90°]

2 2

15. In a series LCR ac circuit at off-resonance, for what values of the angular frequency will the some voltage lead the current in the circuit? JAIPMT (Mains) 20051

[Ans. Voltage will lead the current if i.,XL > i.Xc =>XL> Xe => (ll >

n-l

vu

16. An alternating current of 4.0 A flows through a silver voltameter for 15 minutes. The electrochemical equivalent of silver is I. I 18 x I 0--6 kg/C. How much silver is liberated ?

[AIPMT (Mains) 2005[

[Ans. No silver will be liberated because with ac, anode and cathode are interchanged in each half cycle.]

17. What is the value of power factor at resonance in a series

LCR circuit? [AIPMT (Mains) 20061

[Ans. For resonance condition in an LCR circuit, the inductive and capacitive reactances are equal and the phase difference is 0° so power factor= cos 0°

=

I.]

18. In the series LCR circuit at resonance the applied ac voltage is 220V. What is the potential drop across R?

[AIPMT (Mains) 2006]

[ Ans. At resonance the net potential drop across L and C' is zero.

Hence, whole of the potential drop of220 Vis across the resistance.]

19. An LCR series circuit having 220 V ac source, inductance

L

= 25 mH and resistance

R

= 100

n.

If voltage across inductor is just double of voltage across resistor then fmd out frequency of source. [AIPMT (Mains) 20081

[Ans. According to question, VL =

zv,

so IXL= 2JR =>coL = 2R

co 2R 8000

Frequency f = - = - = - - = 1273.88 / s.]

2lt 2rtL 27t

20. A periodic voltage V varies with time t as shown in figure. T is the time period. Find the rms value of the voltage.

V [AIPMT (Mains) 2008[

v,1---.

T/4 T Flg.10.38

J

T/4 2

f4Il'(!__)

· v,dt , ₄

[ Ans. Root mean square value < V >

=

0

T =

-f

ft

T

=f!=~]

Muliiple Choi~e Questions" '"'

3- - ' "

=·

. >"

Jl,!QIJ

l

Only One Choice is Correct _{6. If the instantaneous current in a circuit is given by}

I= 2 cos ( rot

+

9), the rms value of the current is:

1. Frequency of ac mains in India is:

(a) 30 cps (b) 50 cps (c) 60 cps (d) 120 cps

2. An ac generator produces an output voltage E

=

170 sin 3 77 t volt, where tis in sec; the frequency ofalternating voltage is: (a) 50 Hz (b) 110 Hz

(c) 60 Hz (d) 230 Hz

3. The peak voltage of220 V ac mains in volt is: (a) 155.6 (b) 220

(c) 311.0 (d) 440

4. If Eo represents the peak value of the voltage in an ac circuit, the rms value of the voltage will be:

(a) Eofrc (b) 2Eofn (c) Eo/2 (d) E₀

/.J2

5. In an ac circuit, therms value of current Inns is related to the peak current Io by the relation:

(a) Inns

=

(Jht/o) (b) Inns= (Iof.fi.) (c) Inns=(.J2)/o (d)Inns=rtlo

(a) 2 amp (b)

Jz

amp (c) 2../z amp (d) zero amp 7. In general, in an alternating current circuit:

(a) the average value of current is zero

(b) the average value of square of current is zero ( c) average power dissipated is zero

( d) the phase difference between voltage and current is zero 8. A direct current of 5 A is superimposed on an alternating

current I

=

10 sin cot flowing through a wire. The effective value of the resulting current will be:

(a) (15/2) A (b)

s./3

A

(c)

5-J5

A (d) 15 A

9. The reactance of a capacitor Xe in an ac circuit varies with frequency f of the source voltage. Which one of the following represents this variation correctly?

10. 11. 12. 13. 14. 15. 16. 17. 18. (a)

t~

(b)

XclL__

t I /

,_

,_

(c)

t~

,-

,-The reactance of a coil when used in the domestic ac power supply (220 V -50 Hz) is 100 Q. The self-inductance of the coil is nearly:

(a) 3.2 henry (b) 0.32 henry(c) 2.2 henry (d) 0.22 henry Ohm's law expressed as E = IR:

(a) may never be applied to ac

(b) applies to ac in the same manner as to de

( c) always applies to ac circuits when Z is substituted for R (d) tells us that, E,rr= 0.707(Emax) for ac

In an ac circuit the current:

(a) is in phase with the voltage (b) leads the voltage

( c) lags the voltage

(d) any of the above depending on the circumstances

An ac is applied to a resistive circuit. Which is true of the following?

(a) Current leads in phase ahead of voltage (b) Current lags behind the voltage in phase (c) Current and voltage are in the same phase

( d) Any of the above may be true depending upon the value of the resistance

In a circuit containing an inductance of zero resistance, the

current lags behind the applied alternating voltage by a phase angle:

(a) 90° (b) 45° (c) 30° (d) 0°

A coil of inductance 5.0 mH and negligible resistance is connected to an alternating voltage V = 10 sin {I OOt), The peak current in the circuit will be:

(a)2A (b)5A (c)IOA (d)20A

An alternating voltage E (in volt) 200..fz sin (1001) is connected to a I µF capacitor through an ac ammeter. The reading of the annneter shall be:

(a) 10 mA (b) 20 mA (c) 40 mA (d) 80 mA

A coil of resistance R and inductance L is connected to a battery of E volt emf. The final current flowing in the coil is:

(a) E!R (b) EIL

(c) El(R2 ₊

_o/£

2)112 _{(d) ELl(R}2 ₊

_£

₂₎₁₁₂

Ao alternating current of frequency fis flowing in a circuit containing a resistance R and a choke L in series. The impedance of the circuit is equal to:

( a ) R _ _ _ _ _ (b)R+21tjL (c)~(R2+4it2

J

2L2) (d)R/21tjL 19. 20. 21. 22.

In a circuit containing R and L as the frequency of the

impressed ac increases, the impedance of the circuit:

(a) decreases (b) increases

( c) remains unchanged

( d) first increases and then decreases

An alternating voltage is connected in series with a resistance Rand an inductance L. If the potential drop across the resistance is 200 V and across the inductance is 150 V, the applied voltage is :

(a) 350 V (c) 500 V

(b) 250 V (d) 300 V

A sinusoidal voltage

Vo

sin

rot

is applied across a series combination of resistance R and inductor L. The amplitude of the current in this circuit is:

(a)

Vo

(b)

Vo

~R2 +ro2L2 ~R2 -L2ro2

(c)

Vo

(d)

Vo

(R +Lro) R

An inductive circuit contains a resistance of 10 Q and an inductance of2 H. If an ac voltage of 120 V and frequency 60 Hz is applied to this circuit, the current would be nearly: (a) 0.72 A (b) 0.16 A (c) 0.48 A (d) 0.80 A 23. A coil having an inductance of..'.. henry is connected in series

7t

with a resistance of300 Q. If20 volt from a 200 cycle source are impressed across the combination, the value of the tangent of the phase angle between the voltage and the

current is : (a) tan-1

~

4 (b) tan-1

,i

5 (c) tan-1

~

4 (d) tan-I± 3

24. A 50 W, JOO V lamp is to be connected to an ac mains of200 V, 50 Hz. The capacitance of the capacitor essential to be put in series with the lamp is:

(a) 4.6 µF (b) 2.9 µF (c) 1.5 µF (d) 9.2 µF

25. The natural frequency of a circuit of negligible resistance, capacity C and inductance L is I 600 Hz. If the values of capacity and inductance each are doubled, the frequency will become:

(a) 800 Hz (b) 400 Hz (c) 1200 Hz (d) 3200 Hz

26. In the circuit shown in Fig. 10.39, the voltage in Land in C

are:

Fig. 10.39 (a) in phase

(b) out of phase by 90° (c) out of phase by 180°

( d) in a phase difference which depends upon the values of L and C