1 )€¦ · x dx dv 1 2v 4v 4 1 2 log 1 2v logx c 4 1 or 2 or 4 log x + log(1 –2v2) = – 4c or logx4(1-2v2) = – 4c [put v =y/x] or x4( 1–2y2/x2) = e-3c= c' Hence the required

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V R2

R1 R3RS

Ib

IS Ib

Vo

Fig.1

01. (a)

Sol: The given network is shown in Fig.1

Apply KCL at node V

2

0

1Ss R

VV

R

V

R

VI

2

0

21s R

V

R

V

R

V

R

V

2

0

21SS R

V

R

1

R

1

R

1VI

0S VV3I …….. (1)

)1RRRR( s321

Apply KCL at node V0

0IR

V

R

VVb

3

0

2

0

1

V

R

VIWhere

1b

0 0b

2 2 3

V V VI 0

R R R

0V3VVV 00

0V2V2 0.

V = 0V ……. (2)

Substituting V in equation (1)

00S V)V(3I

0S V4I

: 3 : Test-9 (Solutions)


4

1

I

V

S

0

01. (b) (i)

Sol: Putting x + y = t so that dy/dx = dt/dx –1

The given equation becomes 1dx

dt =sin t + cost

Or

Integrating both sides, we get

ctcostsin1

dtdx

Or

c

2cos2sin1d2

x [t = 2]

c

cossin2cos2

d2x

2

cdtan1

sec2

= log(1+tan)+c

Hence the solution is cyx2

1tan1logx

01. (b) (ii)

Sol: Given equation isxy

yx

dx

dy 22 which is homogeneous in x and y

Put y = vx, then ,dx

dvxv

dx

dy

(i) becomesv

v1

dx

dvxv

2

Orv

v21v

v

v1

dx

dvx

22

Separating the variables,x

dxdv

v21

v2

Integrating both sides,

cx

dx

v21

dvv2



or

cx

dxdv

v21

v4

4

12

cxlogv21log4

1or 2

or 4 log x + log(1 –2v2) = – 4c or logx4(1-2v2) = – 4c [put v =y/x]

or x4( 1–2y2/x2) = e-3c= 'c

Hence the required solution is

x2( x2–2y2 ) = 'c

01. (c)

Sol: (i) Force acting on the electron

Ee = – 1.60210–19 (–2.5106za )

= 4.00510–13za N

Now F = ma = mdtdv

or ordtmF

dv

CdtmF

v Where C is a constant of integraton

Ct1039626.4Cdt1011.910005.4 17

31

13

Since electrons are leaving the cathode with zero initial velocity, we have at t=0, v=0 so that C=0

v(t) = 4.396261017t m/sec

(ii) Now v(t) = t1039626.4dt

)t(dz 17

z(t) = 117 Ctdt1039626.4 , Where C1 is the integration constant.

12

17

Ct2

1039626.4)t(z

Now at t=0, z(0) =0

C1 =0

z(t) = 2.19811017 t2

(iii) zz23

6

z a7747.795a102.0

10100a

AreaI

)z(J

A/m2



01. (d)

Sol: (a)

The following is the band diagram based on the given data

Assume no scattering, to find the value of KE at x = 2m

Slope of band in this region m/eV105.0

m14

eV05.1

x

Em 6

Find the energy at 2m

E2m = – m(2m –1m) = – (0.5 106 eV/m) (110–6 m) = –0.5eV

Consider that the total energy due to KE and PE is constant

Ek(0) + E0 = Ek (2m) + E2m

KE at x = 0 is 0.5eV 0.5eV + 0 = Ek (2m) – 0.5eV

Ek (2m) = 1eV or Ek (2m) = 1.6 10–19J

(b) Similarly for energy at 4m

E4m = – m(4m –1m) = – 1.5eV

Total energy due to KE and PE is constant

Ek (0) + E0 = Ek (4m) + E4m 0.5eV + 0 = Ek (4m) – 1.5eV

Ek (4m) = 2eV or 3.2 10–19J

Mass of e– = m0 = 9.1 10–31 kg

Speed of e– at 4m =

mm4E2

v k , m = 0.5 m0 (given)

1.5eV

0.5eV

Ax=0 B

x=1m Cx=4m

Dx=5m

2eV



60 30 60 40

Rth

Fig.2

60

60V 60V20V+

+

+

50V

30

40

60 100V

Fig.1

Vth

31

19

101.95.0102.32

v

s/m10186.1 6

Time taken for e– to travel from 4m to 5m

610186.1m4m5

vx

t

= 0.843ps

01. (e)

Sol: Vth is found from Fig.1

100

60100V

90

3060V 6030

= 20 V = 60 V

Apply KVL in Fig.1

50 + 20 = Vth + 60

Vth = 10 V

Rth is found from Fig.2

Rth = (60 // 30) + (60 // 40)

100

4060

90

3060

= 44



+

Rth=44

Vth=10VRL

Fig.3

Thevenin equivalent network

For maximum power transfer

RL = Rth

The value of RL to get maximum power consumed in it is,

RL = Rth = 44

Maximum power consumed in RL = W176

100

444

1010

=0.568W

02. (a)

Sol:

(i)

00tsinz5sin)/100(z

aa

a

E

z

tsinz5cos5100

a

=

atsinz5cos500

NowtB

E . Hence

atsinz5cos500

tB

atcosz5cos500

B

0r0

BBH



0tcosz5cos500

0

z

aa

a

H

0

z

tcosz5sin5500a

0

atcosz5sin2500

0

2

tE

JJEH r0DD

= 6.250

100sin(5z) cos(t) a

atcosz5sin2500

0

2

atcosz5sin625 0

or 00

2

6252500

= 1.884109 rad/sec

= 1.884 Grad/sec

(ii) ED r0Ns

= tsinz5sin100

25.6 0

= tsinz5sin625 0

To the inner conductor =2, a is perpendicular .



Hence dzddSoradzddS

Q = dSdQ s

2

0

2.0

0

0 dzdtsinz5sin625

20

2.0

00 5

z5cos625 sint

= 4.427 sint nC

At

2

t , Q = 4.427 nC427.42

sin

02. (b)

Sol: The transfer characteristics of a p channel enhancement type MOSFET can be well explained by

considering the working mechanism of the MOSFET.

Working of a P - channel EMOSFET:

In a MOSFET the gate to source terminal is reverse biased and the drain to source terminal is

forward biased. For a P-Channel E–MOSFET VGS is negative voltage applied and VSD, is positive

voltage applied and the substrate is connected to the VDD. The circuit diagram of a P-Channel

EMOSFET is represented as below.

Initially channel will not be formed in the MOSFET till |VGS| < |VT| (Threshold voltage) and current

through the FET is zero and this region of operation is called cutoff region. when the applied

voltage VGS exceeds the threshold voltage minority carriers i.e., the holes from the n –

substrate and holes from the p+ wells are acquired to form the channel. As soon as the channel is

formed when source to drain voltage is applied flow of charge carriers takes place and hence the

current ID. As the source to drain voltage is increased current increases linearly and hence this

region of operation is known as ohmic or linear region. As the source to drain voltage is still

increased the more number of holes are accumulated towards drain than at the source which results

in reduction of channel passage which is nothing but channel length modulation and hence

current ID reduces and becomes constant and this region of operation is called as saturation region

and the point at which the current ID is constant is called pinchoff voltage.

The characteristics of a P-Channel E–MOSFET are represented as below:



P-channel EMOSFET

P+

P+

N-substrate

DGS

VGS

VDS

ID

(mA) VGS = -20V

-4V

-8V

-12V

-16V

VDS0 -5 -10 -15 -20

-8

-6

-4

-2

Fig: Drain characteristics ofP – channel enhancement

MOSFET

-8 -4-6 -2 VGS,

V

ID

mA

10 0

IDS

S

Fig: Transfer characteristic ofP-channel enhancement typeMOSFET

THRESHOLD VOLTAGE

Threshold voltage is defined as the minimum voltage required to let the MOSFET into conduction

state in P- channel E-MOSFET the conduction takes place when VGS < VT or |VGS| > |VT|. As VGS



is made negative, the current |ID| increases slowly at first, and then much more rapidly with an

increase in |VGS|. Typically the value of VT for the P-channel standard MOSFET is -4V, and the

power supply voltage of –12V for the drain.

Three methods by which VT can be reduced are as following:

1. Polycrystalline silicon doped with boron is used as the gate electrode, This reduces

contact potential difference and hence reduces VT as whole.

2. If silicon crystal uses [100] direction orientation the value of VT reduces one-half that

obtained with [111] orientation.

3. The silicon nitride approach makes use of a layer of Si3N4 and SiO2, whose dielectric

constant is about twice that of SiO2 and as a result decreases VT

02. (c) (i)

Sol: Here = y – sin x and = cos x.

By Green’s theorem ]xdycosdx)xsiny[C

dydxyxR

2/xx

0x

x/x2y

0ydydx1xsin

2/

0

/x20 dx/)1x(sin

2/

0

2/

0

2/

0dxxxcos.1xxcosx

2dx)1x(sinx

2

2/

0

22

2

xxsin

4

2

2

481

2

2

2

0 A

Y B

Fig.8.13

y = 0

2x

2

y





02. (c) (ii)

Sol: Here K3J3I)yz3(K)zxy2(J)y(If 232 at the point (2, –1, 1)

Directional derivative of f in the direction I + 2J +2K

3

2

221

K2J2I,K3J3I

222

3

233/2.32.31.1

02. (c) (iii)

Sol: The sum 6 can be obtained as follows: (1,5), (2,4), (3,3), (4,2), (5,1), i.e., in 5 ways.

The probability of A’s throwing 6 with 2 dice is36

5

The probability of A’s not throwing 6 is 31/36.

Similarly the probability of B’s throwing 7 is 6/36, i.e.,6

1.

The probability of B’s not throwing 7 is 5/6.

Now A can win if he throws 6 in the first, third, fifth, seventh etc. throws,

The chance of A’s wining

......36

5

6

5

36

31

6

5

36

31

36

5

6

5

36

31

36

5

......

6

5

36

31

6

5

36

31

6

5

35

311

36

532

61

30

61

636

36

5

6/5)36/31(1

1.

36

5

03. (a) (i)

Sol: We have fx = 4x3 – 4x + 4y ; fy = 4y3 + 4x – 4y

And r =fxx = 12x2 – 4, s = fxy = 4, t =fyy = 12y2– 4…… (i)

Now fx = 0,fy = 0 given x3 –x+y = 0, ……. (i)

y3 +x –y = 0 ……. (ii)

Adding these, we get 4(x3+y3) = 0 or y = –x,

Putting y = – x in (i), we obtain x3 – 2x = 0, i.e. .0,2,2x



Corresponding values of y are .0,2,2

.vealsoisrandve42020srt),2,2(At 22 Hence 2,2f is a minimum value.

)2,2(At also both rt – s2 and r are +ve.

Hence )2,2(At , is also a minimum value.

At (0,0) ,rt – s2 = 0 and, therefore, further investingation is needed.

Now f(0,0) = 0 and for points along the x-axis, where y =0, f(x,y) =x4–2x2= x2(x2 –2), which is

negative for points in the neighbourhood of the origin.

Again for points along the line y = x, f(x,y) 2x4 which is positive.

Thus in the neighbourhood of (0,0) there are points where f(x,y) < f(0,0) and there are points where

f(x,y) > f(0,0). Hence f(0,0) is not an extreme value i.e., it is a saddle point.

03. (a) (ii)

Sol: Let

2/

0dx

xcosxsin

xsinI

Then

2/

0dx

x2

1cosx

2

1sin

x2

1sin

I

=

2/

0dx

xsinxcos

xcos

Adding

2/

0

2/0

2/

0 2|x|dxdx

xcosxsin

xcosxsinI2

Hence I = /4

03. (b)

Sol: Figure below shows the location of charges 5nC and –8nC along with their images. The point A

and A lies on a line which is perpendicular to the plane z=0 and the distances of these points from

the z=0 plane is same. This is also true for the points B and B



Let C be the midpoint of A and B then

214

,2

35,

242

C

= (3,–1,2.5)

V at C(3,–1,2.5) =)CB(d4

108)CA(d4

105)BC(d4

108)AC(d4

105

0

9

0

9

0

9

0

9

2220

9

45.2)51(234

105

2220

9

15.2)31(434

108

2220

9

45.2)51(234

105

2220

9

15.2)31(434

108

= 1.311V

03. (c) (i)

Sol: The probability that A can solve the problem is ½.

The probability that A cannot solve the problem is2

11 .

Similarly the probabilities that B and C cannot solve the problem are4

11and

3

11

The probability that A,B and C cannot solve the problem is

4

11

3

11

2

11

Hence the probability that the problem will be solved, i.e., at least one student will solve it

4

3

4

11

3

11

2

111

5nC

–5nc

A(2,–5,4)

A(2,–5,–4)

B(4,3,1)

B(4,3,–1)

–8nc

8nc

Z=0Perfectly conductingplane



03. (c) (ii)

Sol: Since the total probability is unity

6

01dx)x(f

i.e., 6

4

4

2

2

01dxk6hxkdx2dxkx

or 1kx62/kxxk22/xk6

424

2

2

0

2

or 2k +4k + (–10k+12k) = 1 i.e., k = 1/8

mean of 6

0dx)x(fxX

2

0

4

2

6

4

2 dx)k6hxx4dxhx2dxhx

6

4

26

4

34

2

22

0

3 2/xk63/xk2/xk23/xk

3248

120k33/152k12k3/8k

03. (d)

Sol: k1005.0

50

I

Vr

0

A0

V2VV,VV

I2g DSGS

tGS

Dm

V/mA91.09.02

5.02gm

L0mi

0 R//rgV

V =–0.91(100k//10k) = –8.3

For I = 1mA or twice the current

tGSt2GS2

tGS

2tGS

D

D VV2VVVV

VV

I

I1

2

1

2

1

V5.29.0229.0V 2GS

–

C

Vi

RG

+VDD

C

+

V0

I=500A

RL

10k

10M



V/mA3.1g2gI

I

g

g12

2

1

2

1

mmD

D

m

m

k502

100r

I

I

r

r02

D

D

02

01

1

2

AV = –1.3 (50k//10k) = –10.8

04. (a) (i)

Sol: 2/113

19

2

16

6

13XE

2/933

181

2

136

6

19XE 2

E(2X + 1)2 = E (4X2 + 4X + 1) = 4E (X2) + 4E(X) + 1

= 4(93/2) + 4(11/2) + 1= 209

04. (a) (ii)

Sol: Let f(x) = 3x –cosx – 1

f(0) = – 2 = – ve, f(1) = 3 – 0.5403 – 1

= 1.4597 = +ve.

So a root of f(x) = 0 lies between 0 and 1. It is nearer to 1. Let us take x0 = 0.6

Also f(x) = 3+sinx

Newton’s iteration formula gives

n

nnn

n'

nn1n xsin3

1xcosx3x

xf

xfxx

n

nnn

xsin3

1xcosxsinx

Putting n = 0, the first approximation x1 is given by

0

0001 xsin3

1xcosxsinxx

6.0sin3

16.0cos6.0sin6.0

6071.05729.03

183533.05729.06.0

Putting n = 1 in (i), the second approximation is



1

1112 xsin3

1xcosxsinxx

= 6071.057049.03

18213.057049.06071.0

Clearly, x1 = x2

Hence the desired root is 0.6071 correct to four decimal places

04. (b)

Sol: To find the transfer characteristics vary the input voltage

Case (i):- for Vin 5V, D2 is in reverse bias

So V0 = Vin

For Vin –5V, D1 is in reverse bias

So V0 = Vin

V0 = Vin –5V Vin 5V

Case (ii):- for Vin > 5V, D1 is in reverse bias

D2 is in forward bias

k20

5VI in

V0= 5 + 10k I

+ +

––

Vin V0

D2D1

–

+5V

10k

+

–5V

10k

10k

+ +

––

Vin V0

10k 10k

10k

I +

–5V



k20

5V5V in

0 10k

2

1V0 Vin + 2.5V for Vin > 5V

Case (iii):- for Vin –5V, D1 is in forward bias

D2 is in reverse bias

k20

5VI in

V0 = –5 +k20

5Vin 10k

V0 = –5 +2

Vin + 2.5

2

1V0 Vin – 2.5 for Vin –5V

Transfer characteristics

V0 = Vin for –5 Vin + 5

+ +

––

Vin V0

10k 10k

10k

I–

+5V

–5V

5 6

5

V0Slope=1/2

–5Vin

5.5



5 10I1 4

1

20V12A12V

I1

+

+

+

Fig.1

1V

10I1I1 5

1

Fig.3

+

+

I

5 10I1

1

20V12A12V

I1

+

+

+

Fig.2

+ Vth+ 5I1

V0 =2

1Vin + 2.5 for Vin + 5V

V0 =2

1Vin – 2.5 for Vin –5V

04. (c)

Sol: The given circuit is shown in Fig.1

Vth is found from the circuit shown in Fig.2

I1 = 12 A

Write the mesh equation

12 5I1 10I1 Vth 20 = 0

815I1Vth = 0

Vth = 815(12) = 172 V (∵ I1 = 12A)

Rth is found from the circuit shown in figure3



Write the mesh equation

1+10I15II = 0

110I5II = 0 (∵ I1 = I)

16I = 1

I = A16

1

Rth = 16

16

11

I

V

The thevenin’s equivalent circuit with load 4 is shown in figure4

i =172

16 4 = 8.6 A

Rth = 16

Fig.4

4 VTh=172V+

–

i





04. (d)

Sol: (i) Let the transistor be in saturation

VCE = 0.2V , VBE = 0.8V

mA36.5k2.2

2.012IC

By superposition Theorem

IB = I1 – I2

= mA618.0100

8.012

15

8.012

hFEIB = 300.618 = 18.56mA which is greater than Ic.

Since Ic < hFE IB transistor is an saturation

V0 = 0.2V

(ii) Assume the transistor in active region then

VBE act = 0.7 V & I1 =1R

30.11, I2 = 0.127 mA

IB = I1 – I2 =1R

30.11 – 0.127 mA

Neglecting ICO . IC = IB = mA)8.3R339

(1

VCE = 12–2.2IC = 20.35 –1R

746

But for the transistor to be in active region

VBC = VBN –VCN 0.5 or

1R746

35.207.0 0.5

Or R1 36.9k or R1(min) = 37k

(iii) By superposition theorem

V696.012.15100

151.

15100

100VBN

Hence transistor is in cut-off

V0 = 12V



a

b

c

Fig.1

c b

a a

N

c b

z z

z

z

z z

(iv) For transistor to remain in cut-off VBE = 0V

ICBO = I2 –I1

= mA053.015

1

100

12

ICBO(T) = ICBO(T0) 10/TT 02 ; T0= 250

0.053 = 10 10–6 2(T–25)/10

T–25 = 124.10C

Maximum temperature T = 149.10C

05. (a)

Sol:

Line voltage, VL = 415 V

Phase impedance = z = (5+j8.66)

= 1060

Assume ‘abc’ sequence

Vab = 4150,

Vbc = 415120,

Vca = 415120

With -connected load,

605.41

6010415

Iab

0V BN

Iz

I1

–12V

ICBO



Ibc = 41.5180

Ica = 41.560

Ia = Iab Ica =9035.41 = 71.990

With Y – connected load,

306.2393

30415VaN

VbN = 239.6150

VcN = 239.690

9096.236010

306.239IaN

IbN = 23.96150

IcN = 23.9630

Total line current

Ia = 71.990 + 23.9690

= 95.8690 = j 95.86

Ib = 95.86150 (or) 95.86-210o

Ic = 95.8630

05. (b) (i)

Sol: We divide the interval (0.01) into five steps i.e. we take n = 5 and h = 0.02. The various

calculations are arranged as follows:

x y (y – x)/(y + x) = dy/dx Old y + 0.02 ( dy / dx ) = new y

0.00 1,0000 10000 1,0000 + 0.02(1,0000) = 1.0200

0.02 1.0200 0.9615 1.0200 + 0.02(9615) =1.0392

0.04 1.0392 0.926 1.0392 + 0.02(926) = 1.0577

0.06 1.0577 0.893 1.0577 + 0.02(893) =1.0756

0.08 1.0756 0.862 1.0756 + 0.02(862) = 1.0928

0.10 1.0928

Hence the required approximately value of y = 1.0928



05. (b) (ii)

Sol: Let f(z) = u + iv, where u = sin 2x/(cosh 2y – cos 2x)

dy

ui

x

u

x

vi

x

u)z(f '

22 x2cosy2cosh

y2sinh2x2sini

)x2cosy2(cosh

)x2sin2(x2sinx2cos2x2cosy2cosh

= 22 x2cosy2cosh

y2sinhx2sin2i

x2cosy2cosh

2y2coshx2cos2

By Milne-Thomson’s method, we express f (z) in terms of z by putting x = z and y = 0

)0(i

z2cos1

2z2cos2)z(f 2

= zeccos

zsin2

2

z2cos1

2 22

Integrating w.r.t.z, we get f(z) = cot z + ic, taking the constant of integration as imaginary since u

does not contain any constant.

05. (c) (i)

Sol: (a) Here f(z) =z2 –z+1 and = 1

Since f(z) is analytic within and on circle C:z| = 1 and = 1 lies on C

by Cauchy’s integral formula

i2dz1z

1zzc

.,e.i1)(fz

zfci2

1 2

(b) In this case, = 1 lies outside the circle

C: |z| =1/2. so (z2 –z+1)/(z – 1) is analytic everywhere within C.

by Cauchy’s theorem 0dz1z

1zzc

2

05. (c) (ii)

Sol: f(z) has simple poles at z = 0, /2, 3/2,–– ….

Only the poles z = 0 and z = /2 lies inside

|z| =2

zcosz

zsin2/zLtzf

2zLt0fsRe

2/x2/x



2

1 1

10.50.5

2V

2V+

+

i1

Fig.1

1A

a b c

=

zsinzzcos

zsinzcos2/zLt

2/z

2

2/

1

form

0

0Being

And

zcosz

zsin2/zLt2/fsRe

2/z=

2

2/

1

zsinzzcos

zsinzcos2/zLt

2/x

Hence sum of residues 022

0

05. (d)

Sol: The given circuit is shown in Fig.1 where a,b,c are marked as nodes.

V2VC

Apply KCL at node ‘a’

01

VV

2

2V

5.0

2V baaa

0V2V22V8V4 baaa

10V2V7 ba -------- (1)

Apply KCL at node ‘b’

15.0

V

1

2V

1

VV bbba

1V22VVV bbba

3V4V ba -------- (2)



By solving equation (1) and (2)

Va = 1.769 V

Vb = 1.192 V

1

VVi ba

1

1

192.1769.1

A577.0i1

05. (e)

Sol: (i)

Method: I

By redrawing the circuit

I

VR 0 Apply KCL output node.

0IR

RR

1VV

RV

2

4

3

1

24

3

1 RR

RV

R

VI

0

24

3

1

R

RR

R

R

11

I

V

+

–

ZL

ILR1 R2

R3R4

+

–

I

R1 R2

R3R4

+–

4

30 R

R1VV

V R0

V



24

3

1 RRR

R1

R4R2 = R1 R3

Method: II

R0 The output current is independent of the load

0R

R

R1VV

R

VVI

2

4

3LL

1

inLL

24

3

1 RR

R

R

1

R1R3 = R4R2

Comment:

The given circuit is a voltage to current converter i.e., output current depends only on the input

voltage and is independent of output voltage

1

inL R

VI

+

–

IL

R1 R2

R3R4

4

3L R

R1V

Vin

ZL

+

–VL

24

3

1L

1

inL RR

R

R

1V

R

VI

0 [ I2 is independent of Vc]





(ii). If Vin = 5V

mA5R

5

R

VI

11

inL

R1 = 1k

We know that R1R3 = R4R2

3

4

1

2

R

R

R

R

Choose R1 = 1k, R2 = 2k, R3 = 1k, R4 = 2k,

ZL can be any value ( IL is independent of ZL)

06. (a)

Sol: (i) zyx1 a4a10a12D nC/m2

r1 = 2.5

r2 = 1

From Boundary condition

n1n2 DD

xzyxn1n1 a.a4a10a12a.DD

= 12

xn1 a12D

So, xn2 a12D

We have zyt1 a4a10D

0

z

0

yt1 5.2

a45.2

a10E

From boundary condition

t2t1 EE

z0

y0

t2 a5.24

a5.210

E

t20t2 ED



zy a5.2

4a

5.210

= zy a6.1a4

So n2t22 DDD

= zyx a6.1a4a12 nC/m2

12

6.1)4(tan

DD

tan22

1

n2

t212

= 19.750

(ii)2

n22 E

Ecos

2

n20

EE

60cos

m/V621

12E n2

So, E2n = 6 xa

x0n22n2 a6ED

x0n1 a6D

x0

x0n1 a4.2

5.2a6

E

2

t20

EE

60sin

2D

1Dn1D1

2t1D

X

t2D

(1)r1

(2)r2 n2D

1En1E1

n1E

X

t1E



3623

12E t2

So, 36E t1

n1

t11 E

Etan

011 77

4.236

tan

1

n11 E

Ecos

)77cos(4.2

cosE

E 01

n11

= 10.67V/m

06. (b) (i)

Sol: Given circuit is

53 DD II

k1

0V

k1

V10 53

V3 + V5 = 10V ------ (1)

2tGSnD VVL

Wk

2

1I

33

k1

V]1VV[102

2

1I 52

433

D3 ------- (2)

+10V

1k

1k

V3

V4

M4

ID5

M3

ID4

ID3



k1

V]1VV[102

2

1I 52

543

D4 ------- (3)

By equating (2) & (3)

[V3 – V4 – 1]2 = [V4 – V5 – 1]2

V3 – V4 – 1 = V4 – V5 – 1

V3 + V5 = 2V4

V4 = 5V

From equation (3) 35

254

3 10V]1VV[1022

1

[5 – V5 – 1]2 = V5

016V9V 525

V5 = 6.55 or V5 = 2.45

V5 = 6.55 results in ID5 = 6.55 mA, V3 = 4.45 V this is not possible so take V5 = 2.45 V

V3 = 10 – 2.45 = 7.55 V

V3 = 7.55 V, V4 = 5 V, V5 = 2.45 V



06. (b) (ii)

Sol: For AC analysis short circuit the DC sources

Vo = – gm Vi

k10k7.14

k7.4 ………… (1)

M1.0)M10||M47(

M10||M47VV sigi

25.8

24.8Vsig …………(2)

Substitute (2) in (1)

Vo = –110-3sigV

25.8

24.8k10

k7.14

k7.4

AV =sig

0

V

V= –3.19 V/V

To find 1pf frequency, short circuit capacitors C2 & C3.

Find 1C

gmViVsig 47M 4.7k

V0100k

10M

DG

+

–

Vi

10k

Rsig=0.1M

RG = 8.24 M

1ThR

RG = 47M//10M

Vsig

0.1M C1=0.01F

RG=8.24 M gmVi 10 k4.7 k

G D

S



1ThR = Rsig + RG

1C =1THR C1 = 8.35 106 0.0110–6

1C = 0.0834

1pf =0834.014.32

1

2

1

1C

1pf = 1.909 Hz

To find 3pf frequency, short circuit C1 & C2 capacitors

k7.14k10k7.4R 3Th

633Th3C 101.0k7.14CR 3101.07.14

Hz3.108101.07.1414.32

1

2

1f

33C

3p

To find 2pf short circuit capacitors C1 & C3

Ix = )Vg(R

Vxm

s

x

Ix = xms

x VgR

V

0.1M

RG = 8.24 M[ Vi = 0] 10 k4.7 k

3ThR

–gmVxR0

RS

G

RL

–

+

Vi=VxRG

S

+––

+Vx

IVx

Ix

2k

D



k66.03

k2

Rg1

R

gR1

1

I

VR

sm

s

ms

x

xTh 2

04186.0F10.21

k2CR 2ThC 22

Hz88.232

1f

2C2p

Conclusion:

AV = – 3.19 V/V

Hz909.1f1p

Hz88.23f2p

Hz3.108f3p

fL = Max of (321 ppp f,f,f ) =

3pf = 108.3 Hz

06. (c) (i)

Sol: x0 = 0 , y0 = 1, h = 0.2, f(x0, y0) = 1

k1 = hf (x0 , y0) = 0.21 = 0.2000

2400.01.1,1.0f2.0k2

1y,h

2

1xhfk 1002

2440.012.1,1.0f2.0k2

1y,h

2

1xhfk 1003

2888.0244.1,2.0f2.0k2

1y,hxhfk 3004

4321 kk2k2k6

1k

2468.04568.14

12888.04880.04800.02000.0

6

1

Hence the required approximate value of y is 1.2428.



06. (c) (ii)

Sol: To find C.F.

It’s A.E is (D – 2)2 = 0,

D = 2/2

Thus C.F = (c1+c2x) e2x

To find P.I.

2

22x2

2 x2D

1x2sin

2D

1e

2D

18I.P

Now

x22x22 e

12

1xe

2D

1

[∵ by putting D = 2, (D–2)2 = 0, 2(D –2) = 0]

2

ex x22

x2sin4D42

1x2sin

4D4D

1x2sin

2D

1222

x2cos

8

1

2

x2cos

4

1xdx2sin

4

1

And

22

22

22 x...

2

D

2

32

2

D21

4

1x

2

D1

4

1x

2D

1

2

3x2x

4

1x....

4

D3D1

4

1 222

Thus P.I = 4x2 e2x e2x + cos 2x + 2x2 + 4x + 3

Hence the C.S. is y = (c1 + c2 x) e2x + 4x2 e2x + cos2x +2x2 + 4x + 3

07. (a)

Sol: Apply thevenin’s theorem to the resistor 82K–10K and Vcc = 24V circuit

221

ccth R

RR

VV

V61.2101082

24



Rth = 82K || 10K

K91.81082

1082

The equivalent circuit now becomes

Applying KVL to base emitter loop

Vth = (Rth+RB) IB2 + VBE2 +VBE1+IE1Re ----(1)

IE2 = (IB2+IC2) = [IB2 + 2 IB2]

= (1+2) IB2 = 51 IB2

IB1 = IE2 = 51 IB2

Also IE1 = (IB1+ IC1) = (1+1)IB1

= 101 IB1 = 5151 IB2

Substituting this value in equation (1)

Vth = (Rth +RB)IB2 + VBE2 +5151 IB2 Re

= (Rth+RB+ 5151Re) IB2 + VBE2 +VBE2

2.61 = (8.91+100+51510.1) IB2 + 0.7 + 0.7

mA1094.101.624

4.161.2I 3

2B

IC2 = 2 IB2 = 501.94 10–3 = 0.097mA

IE2 = (IC2+ IB2) = (1.9410–3+0.097) = 0.09894mA

IB1 = IE2 = 0.09894 mA

Re = 100

100k

Q1

Q2

IB2

IC2

8.91k

Rth RB

IE2

IC1

IB1

IE1

VTh

1k

24 V



IC1 = 1IB = 100 = 100 0.09894 = 9.894 mA

IE1 = (IC1+IB1) = 9.99mA

To find voltage Vo1 and Vo2

Vo1 = Vcc – (Voltage drop across RC1(1K))

= 24 – (IC1 1K) = 24 – 9.894 = 14.11V

Similarly Vo2 = IE1Re = 9.99 0.1 = 0.999V

To find I1 and I2 consider the original circuit

Apply KVL to base – emitter circuit

VBN–100IB2–VBE2–VBE1 –Vo2 = 0

VBn = 100 IB2 + VBE2 + VBE1 + Vo2

= 100 (1.9410–3)+0.7+0.7+0.999

= 2.593V

mA2593.010

593.2

R

VI

2

BN2

I1 = IB2+I2

= 1.9410–3 + 0.2593

= 0.261mA

0.1k

52k

10k

100k

Vo2

Q1

Q2IB2

N

I2

I1



1 1

1

I1I2

1F 1F

1FV1

V2

Fig.1

1F1F

1

T – Network 1

11

1F

T – Network 2

07. (b)

Sol: The given 2-port NW is shown is fig.1

The given network consist of two ‘T’ NW’s which are there parallel

j

11

1j

1Z

111 z

j

11

1j

1

1)j

1(

1

2

j11

1j

1

j2

11

Y

2

1

j

1j

jj1

Z2



122 Z

j

1j

jj1

j)

j1(

12

2

j

1j

jj1

j21

1Y2

21 YYY

2

2

( j ) j j

1 j2 1 j2 2j 2 j

j j( j )2 j 2 j1 j2 1 j2

j2

j

2j1

)j(

j2

j

2j1

j2

j

2j1j2

j

2j1

)j(

Y 2

2



07. (c)

Sol: Let x,y and z ft. be the edges of the box and S be its surface.

Then S = xy +2yz+2zx ………….. (i)

And xyz = 32 …………… (ii)

Eliminating z from (i) with the help of (ii), we get

y

1

x

164xy

xy

32xy2xyS

S/x = y – 64/x2 = 0 and S/y = y – 64/y2 = 0

Solving these, we get x = y = 4.

Now r =2S/x2 = 128/x3, s=2S/xy = 1, t = 2S/y2 = 128/y3.

At x =y =4, rt –s2 = 22–1 = +ve and r is also +ve.

Hence S is is minimum for x = y = 4. Then from (ii), z = 2,

Otherwise (by Lagrange’s method):

Write F = xy+2yz +2zx+(xyz – 32)

Then 0yzz2yx

F

…….. (iii)

0yxz2xyF

…….. (iv)

0xyx2y2z

F

…….. (v)

Multiplying (iii) by x and (iv) by y and subtracting, we get 2zx –2zy = 0 or x = y

[The value z = 0 is neglected, as it will not satisfy (ii)]

Again multiplying (iv) by y and (v) by z and subtracting. We get y = 2z.

Hence the dimensions of the box are x = y = 2z = 4

Now let us see what happens as z increases from a small value to a large one. When z is small, the

box is flat with a large base showing that S is large. As z increases, the base of the box decreases

rapidly and S also decreases. After a certain stage, S again starts increasing as z increases. Thus S

must be a minimum at some intermediate stage which is given by (vi). Hence S is minimum when

x = y = 4 ft and z = 2ft.



07. (d)

Sol: (i) D.C analysis

mA0135.0K101K100

073IB

= 13.5A

IC = IE = IB = 1.35mA

mS54054.025

35.1

V

Ig

T

Cm

(ii) Small signal analysis

eqmvi

0 RgAV

V

(iii)10

1ratioTurns

V

V

0

L

08. (a)

Sol:

+–0.73V

100K

IC

1K

Vi100k r gmV

10:1

RL

Req = 10030 = 3k

+

–R1

R2

V0

VI

1zV2zV



Assume that the cut-in voltage of zener diodes are V

Case-I: (VI > 0)

If VVV2z0

VVVR

R2zI

1

2

If I1

20z

2

1I V

R

RVVV

R

RV

2

[∵due to negative feed back]

If VVVVVR

RV

22 z0z2

1I [ due to limiter diodes]

Case-II: (VI < 0)

If VVV1z0

VVVR

R1zI

1

2

If I1

20z

2

1I V

R

RVVV

R

RV

1

[∵due to negative feed back]

If VVVVVR

RV

11 z0z2

1I [∵due to limiter diodes]

V0

VI

VV2z

VV1z

VVR

R1z

2

1 VVR

R2z

2

1

Fig: Transfer characteristics

Slope =l

2

R

R



s

3

3

Fig. 4

3

3

2

IL(s)

+

3s

Vs

6

+

+

L i(0) = 6 V

V2s

1

R2

3H

R1

Ce2 tu(t)

3

1

2

K

Fig.1

+

12V

33

(1/3) F

+

R3

08. (b)

Sol: The given circuit is shown in Fig. 1.

With the switch, K in position 1, for t < 0 , the behaviour of the circuit is shown in Fig. 2 after the

steady state is reached at t = 0.

vc(0) = 6 V, iL(0) = 2 A

With the switch in position 2,

i.e., for t > 0 the circuit is shown in Fig. 3.

iL(0+) = iL(0) = 2 A

vc(0+) = vc(0

) = 6 V

The transform circuit of Fig. 3 is shown in Fig. 4. After further simplification, it appears as in Fig.

5.

F3

1

3

Fig. 3

3

3

2

iL(t)

+

3He2t u(t)

R2

R1

C

3

Fig. 2

12V

3

3

vc(0)

R31

+

iL(0)L



s

33

Fig. 5

+

3 +3 s

s

6

+

+

6

V1(s) IL(s)

3

2s

1

Writing the nodal equation in V1(s):

0s33

6V

s

33

s

6V

3

2s

1V

111

0)1s(3

6V

)1s(3

6Vs

)2s(3

1)2s(V 111

[V1(s +2)– 1](s+1)+(sV1–6+ V1+6)(s + 2)=0

V1[(s + 2) (s + 1) + (s + 1) (s + 2)] = (s + 1)

)2s(2

1

)2s()1s(2

1sV1

)1s()2s(6

25s12

)1s(3

64s2

1

)1s(3

6VI 1

L

08. (c) (i)

Sol: The poles of5z2z

3z)z(f

2

are given by z2 +2z+5 = 0

i.e., by

i212

2042z

(a) Both the poles z = –1 + 2i and z = –1–2i. lie outside the circle |z| = 1. Therefore, f(z) is analystic

everywhere within C.

Hence by Cauchy’s theorem

c 2

0dz5z2z

3z

(b) Here only one pole z = –1 +2i lies inside the circle C:|z+1–i| = 2. Therefore, f(z) is analystic

within C except at this pole.

i21fsRe 5z2z

3zi21zLt]zfi21z(Lt

2i21zi21z



2/1ii4

i24

i21z

3zLt

i21z

Hence by residue theorem c

2i2/1ii2i21fsRei2dzzf

(iii) Here only the pole z = – 1 – 2i lies inside the circle C: |z+1+i|=2. Therefore, f(z) is analytic

within C except at this pole.

5z2z

3zi21zLti21fsRe

2i21z

i

2

1

i4

i24

i21z

3zLt

i21z

Hence by residue theorem,

ci2i

2

1i2i21sfRei2dzzf

08. (c) (ii)

Sol:

1n 1nnn

0 nxsinbnxcos2

xfLet ………….. (i)

Then

2

0

x200 2|xsin1xcosx|

1dxxsinx

1

2

0

2

0n dx)xsinnxcos2(x2

1nxdxcosxsinx

1

2

0dxx1nsinx1nsinx

2

1

2

0

22 1n

x1nsin

1n

x1nsin1

1n

x1ncos

1n

x1ncosx

2

1

1n.1n

2

1n

1n2cos

1n

1n2cos2

2

12

When

2

0

2

01 xdx2sinx2

1xdxcosxsinx

1,1n

2

1

4

x2sin1

2

x2cosx

2

12

0

Finally,

2

0

2

0n dxx1ncosx1ncosx2

1nxdxsinxsinx

1b



=

2

0

22 1n

x1ncos

1n

x1ncos1

1n

1nsin

1n

x1nsinx

2

1

When n = 1,

2

0

2

01 dxx2cos1x2

1xdxsinxsinx

1b

2

0

2

4

x2cos

2

x1

2

x2sinxx

2

1

Substituting the values of ’s and b’s, in (i) , we get

x sin x = .....x3cos13

2x2cos

12

2xcos

2

1xsin1

22

08. (d)

Sol: Let f(x) represent an odd function in ( –1,1) so that

1n

n xnsinbxf

Where 1

0n dxxnsinxf1

2b

2

1

0

1

2

1 dxxnsin4

3xdxxnsinx

4

12

1

2

122

2/1022 n

xnsin

n

xncos

4

3x2

n

xnsin

n

xncosx

4

12

2222 n

2/nsin

2

ncos

n4

1ncos

n4

12

n

2/nsin

n4

1

2

ncos

n4

12

Thus 0b;41

b 221

0b;3

4

3

1b 4223

etc0b;5

4

5

1b 6225

Hence .....x5sin5

4

5

1x3sin

3

4

3

1xsin

41xf

22222



Documents

1 )€¦ · x dx dv 1 2v 4v 4 1 2 log 1 2v logx c 4 1 or 2 or 4 log x + log(1 –2v2) = – 4c or logx4(1-2v2) = – 4c [put v =y/x] or x4( 1–2y2/x2) = e-3c= c' Hence the required