1 Basic Concepts of Networks - GATE ACADEMY · QCV CV CV where, 12 12 eq CC C CC By using above...

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1.1 A voltage drop across a non-linear

element is given 22v i i . The power loss across it for a current of sin t is

(A) 1.0 W (B) 2.0 W

(C) 3.0 W (D) 0.5 W

1.2 The bridge circuit in given figure is

balanced. The magnitude of current I is

(A) 2 mA (B) 4 mA

(C) 5 mA (D) 6 mA

1.3 For the waveform ( ) 2 cosv t

6t

the ratio rms

average

V

V

(A) 3

2 2 (B)

3

2

(C) (D) 2

1.4 In the circuit shown in below figure,

current through the 5 resistor is

(A) 0 A (B) 2 A

(C) 3 A (D) 7 A

1.5 For the circuit shown in the figure, the

diode D is ideal. The power dissipated by the 300 resistor is

(A) 0.25 W (B) 0.50 W

(C) 0.75 W (D) 1.00 W

1.6 The root-mean-square value of a voltage

waveform consisting of a Super-imposition of 2 V dc and a 4 V peak-to-peak square wave is

(A) 2 V (B) 6 V

(C) 8 V (D) 12

1.7 A metal wire has a uniform cross-section A, length l , and a resistance R between its two end points. It is

1Basic Concepts

of Networks

1 k�

2 k�4 k�

2 V

sV0V

I

10 �

2 � 5 �5 A 2 A

100 ,0.5 W�

60 cos 314 t

200 �

0.5 W

300 �

1 W

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1.4 Topic Wise GATE Solutions [IN] GATE ACADEMY ®



uniformly stretched so that its length

becomes l . The new resistance is

(A) R (B) 2 R

(C) R (D) e R

1.8 For the circuit shown below, the voltage across the capacitor is

(A) (10 0) Vj (B) (100 0) Vj

(C) (0 100) Vj (D) (0 100) Vj

1.9 The current I supplied by the dc voltage source in the circuit shown below is

(A) 0 A (B) 0.5 A

(C) 1 A (D) 2 A

1.10 The power supplied by the dc voltage source in the circuit shown below

(A) 0 W (B) 1.0 W

(C) 2.5 W (D) 3.0 W

1.11 Which one of the following equation is valid for the circuit shown below?

(A) 2 5 6 7 0I I I I

(B) 3 5 6 7 0I I I I

(C) 3 5 6 7 0I I I I

(D) 3 5 6 7 0I I I I

1.12 The root mean squared value of

( ) 3 2sin( )cos(2 )x t t t is

(A) 3 (B) 8

(C) 10 (D) 11

1.13 A 100 , 1 W resistor and a 800 , 2

W resistor are connected in series. The maximum dc voltage that can be applied continuously to the series circuit without exceeding the power limit of any of the resistors is

(A) 90 V (B) 50 V

(C) 45 V (D) 40V

1.14 The current I shown in the circuit given

below is equal to

(A) 3 A (B) 3.67 A

(C) 6 A (D) 9 A

100 �j

100� �j

(10 0) V� j

10 �

1 �1 V 1 A

I

3 �

6 � 1 �3 V

1 �

1 �1 �

1 � 1 �

1 �

1 �

3I

6I

1I

2I

5I

7I

4I

+–

5V

I

10 �10 A10 �

10 �

10 V

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1.5GATE ACADEMY ® Network Theory : Basic Concepts of Networks



1.15 If 6 VA BV V , then C DV V is

(A) – 5 V (B) 2 V

(C) 3 V (D) 6 V

1.16 Consider a delta connection of resistors and its equivalent star connection is shown below. If all elements of the delta connection are scaled by a factor k, k > 0, the elements of the corresponding star equivalent will be scaled by a factor of

(A) 2k (B) k

(C) 1

k (D) k

1.17 Three capacitors 1 2,C C and 3C whose

values are 10 F, 5 F and 2 F

respectively, have breakdown voltages of 10 V, 5 V and 2 V respectively. For the interconnection shown below, the maximum safe voltage in Volts that can be applied across the combination, and the corresponding total charge in C

stored in the effective capacitance across the terminals are respectively,

(A) 2.8 and 36 (B) 7 and 119

(C) 2.8 and 32 (D) 7 and 80

1.18 The linear I-V characteristic of 2-terminal non-ideal ac sources X and Y are shown in the figure. If the sources are connected to a 1 resistor as

shown, the current through the resistor in amperes is ______A.

1.19 The current in amperes through the resistor R in the circuit shown in the figure is _______ A.

RR

R R RR R

R

BV

DVCV

AV

5 V

10 V

2 �

1 �

2 A

aR

cRbR

CR BR

AR

2C 3C

1C

0

1

2

3

1 2 3 4 5 6

Curr

ent

(A)

Voltage (V)

Source Y

Source X

Source Y

Source X 1 �

1 A1 R� �

1�1�

1�

1 Volt

I

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1.20 The current xI in the circuit given

below in milli-ampere is _______.

1.21 A circuit consisting of dependent and

independent source is shown in the figure. If the voltage at Node-1 is –1 V, then the voltage at Node-2 is _______V.

1.22 In the given circuit, the mesh currents

1 2,I I and 3I are

(A) 1 2 31 A, 2 A and 3 AI I I

(B) 1 2 32 A, 3 A and 4 AI I I

(C) 1 2 33 A, 4 A and 5 AI I I

(D) 1 2 34 A, 5 A and 6 AI I I

+–

100� 100�

100� 10 mA1V

xI

1 A

0.5 �I1 I3

I21 21

4 RV

1RV1

322I1 � �

+–

2I

6� 1�

2� 1�1I 3I

5V

2 A

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1.1 D 1.2 D 1.3 A 1.4 B 1.5 C

1.6 C 1.7 B 1.8 D 1.9 A 1.10 D

1.11 D 1.12 C 1.13 C 1.14 A 1.15 A

1.16 B 1.17 C 1.18 1.75 1.19 1 1.20 10

1.21 2 1.22 A

ConceptsofAbsorbedandDeliveredPower

ConceptsofVoltmeterandAmmeter

Given : ( ) sini t t …(i)

22v i i …(ii)

According to question,

From equation (i) and (ii),

2( ) sin 2sinv t t t

Instantaneous power consumed by the element is given by,

( ) ( )P v t i t

2(sin 2sin )sinP t t t

2 3sin 2sinP t t

31 cos 22sin

2

tP t

1 cos 2 3sin sin 3

22 4

t t tP

1 3 cos 2 sin 3

sin2 2 2 2

t tP t

The power loss across the element is given by,

1

DC term 0.5 W2avgP

Hence, the correct option is (D).

Given : The bridge is balanced. Given circuit is shown below,

Since, the bridge circuit is balanced the product of opposite arm impedances are equal i.e., AD BC AB CDZ Z Z Z

1 k 2 k 4 kABZ

0.5 kABZ

The current through branch AB is given by,

3

24 mA

0.5 10AB

ABAB

VI

Z

…(i)

Scan for Video

Explanation

Scan for Video

Explanation

Non-linear

element

v

i

Scan for

Video Solution

1 k�

2 k�4 k�

2 V

sV0V

I

D B

C

A

Answers BasicConceptsofNetworks

Explanations BasicConceptsofNetworks

1.1 (D)

1.2 (D)




From figure,

1 k 4 k

1 k 4 k 0.5 k 2 kABI I

[By CDR] From equation (i),

5 k

4 mA7.5 k

I

6 mAI


Key Point 1. Voltage division rule for resistor :

2. Voltage division rule for inductor :

3. Voltage division rule for capacitor :

Concept of Voltage division rule for capacitor : Capacitors 1C and 2C are connected in series.

So, same current will flow through them and hence, same charge will be stored in both the capacitors. Let, Q be the charge stored in capacitor 1C

and 2C then,

1 21 2C C eqQ C V C V C V

where, 1 2

1 2eq

C CC

C C

By using above equations,

11 C eqC V C V

1

1 21

1 2C

C CC V V

C C

1

2

1 2C

CV V

C C

Similarly,

22 C eqC V C V

2

1 22

1 2C

C CC V V

C C

2

1

1 2C

CV V

C C

1 k�

2 k�4 k�

2 V

sV

I

D B

C

A4mAABI �

0.5k

ABZ

� �

Scan for

Video Solution

1R

V

1RV

2R2RV

1

1

1 2

R

RV

V

R R

��

�1R

VR1

1

1

1 2

R

RV

V

R R

��

�2R

VR2

1L

V

1LV

2L2LV

1

1

1 2

R

RV

V

L L

��

�1L

VL1

1

1

1 2

R

RV

V

L L

��

�2L

VL2

1C

V

1CV

2C2CV

1

1

1 2

R

RV

V

C C

��

�1C

VC2

1

1

1 2

R

RV

V

C C

��

�2C

VC1




Given : ( ) 2 cos6

v t t

The average value of ( )v t is given by,

avgV DC term = Constant term = 2 V

…(i)

The rms value of ( )v t is given by,

2

2 1 1 92 4

2 22rmsV

…(ii)

From equations (i) and (ii),

932

2 2 2rms

avg

V

V

Hence, the correct option is (A).

Key Point

If any function is in the form of

0 1 2

3

( ) sin sin 2

sin 3 ...

x t A A t A t

A t

0 1 2

3

( ) cos cos 2

cos3 ...

x t A A t A t

A t

0 1 2( ) cos sin 2 ...x t A A t A t

Then,

avgX Constant term DC value 0A

2 2

2 1 20 ....

2 2rms

A AX A

Given circuit is shown below,

. Method 1 :Nodal Analysis .

Applying KCL at node A,

52 10A A BV V V

5 50A A BV V V

6 50A BV V …(i)

Applying KCL at node B,

25 10B B AV V V

2 20B B AV V V

3 20A BV V …(ii)


10 VAV and 10 VBV

The current through 5 resistance is given by,

10

2 A5 5BV

I

Hence, the correct option is (B).

. Method 2 : Mesh Analysis .

Applying KVL in loop shown in figure, 2 10( 5) 5( 7) 0I I I

17 85 0I

5AI

From figure, The current through 5 resistor is,

7 5 7 2 AI


10 �

2 � 5 �5 A 2 A

10 �

2 � 5 �5 A 2 A

I

A B

10 �

2 � 5 �5 A 2 A

A B

CD

I 5I � 7I �

1.3 (A)

1.4 (B)





. Method 1 :

For positive half cycle, diode is ON and for negative half cycle, diode is OFF. Hence the arrangement of given circuit is half wave rectifier.

Therefore, output rms voltage is given by,

2m

rms

VV

where, mV maximum value of supply voltage,

60

30 V2rmsV

For a resistive load, rms value of current is given by,

Total resistance

rmsrms

VI

30 1

100 200 300 20rmsI

Power dissipated in 300 resistor is given by,

2(300 ) 300rmsP I

2

1 1(300 ) 300 300

20 400P

3

(300 ) 0.75 Watt4

P

Hence, the correct option is (C).

Key Point

1. If any sinusoidal signal ( )v t is passed

through half wave rectifier then,

mavg

VV

,

2m

rms

VV

2. If any sinusoidal signal ( )v t is passed

through full wave rectifier then,

2 m

avg

VV

,

2m

rms

VV

. Method 2 :

From figure,

60cos(314 )

( )100 200 300

ti t

cos(314 )

( )10

ti t

For a resistive load,

Power dissipated is given by,

2

0

1( )

T

avgP i t R dtT

Since, given circuit is a half wave rectifier and the load is resistive,

/2

2

0

1( )

T

avgP i t R dtT

where, 2 2

314T

Scan for

Video Solution

100 ,0.5 W�

60 cos 314 t

200 �

0.5 W

300 �

1 W

I

100 ,0.5 W�

60 cos 314 t

200 �

0.5 W

300 �

1 W

I

Scan for

Video Solution

100 ,0.5 W�

60 cos 314 t

200 �

0.5 W

300 �

1 W( )i t

1.5 (C)




2/

0

1 cos300

2 10avg

tP dt

3

0.75 W2 2avgP


Key Point For any signal ( )v t ,

1. 0

1( )

T

avgV v t dtT

where, T Time period of signal

2. 2

0

1( )

T

rmsV v t dtT

According to question, 1 2( ) ( ) ( )v t v t v t

where,

. Method 1 :

Calculation of rmsV :

22

02

1(4) 0

TT

TrmsV dt dtT

2

0

16 T

rmsV dtT

/2

0

16 Tt

T

1

16 02rms

TV

T

8 V


. Method 2 :

1 2( ) ( ) ( )v t v t v t … (i) From figure (a) and (b),

1 2 VrmsV

2 2 VrmsV

From equation (i),

2 2

1 2rms rms rmsV V V

2 22 2rmsV

8 VrmsV Hence, the correct option is (C).

Key Point If the waveform is in the form of, 1.

3

rms

AV ,

2avg

AV

2.

2V

2V

4V

2 ( )v t

– 2V

( )v t

1( )v t

2

T

T

2

T T

t

t

t

Fig. (a)

Fig. (b)

Fig. (c)

( )v t

T 2T 3T

A

t

A

2

T Tt

( )v t

1.6 (C)




2

rms

AV ,

2avg

AV

3.

rmsV A , 0avgV

4.

2

3rmsV A ,4avg

AV


Before stretched :

Area 1A A

Length 1l l

Resistance 1R R

Volume 1V V = Area Length

1V V A l …(i)

After stretched :

New length 2l l

New Volume 2( )V = New length New area

2 2V l A …(ii)

Since, resistivity and volume of the material remains same.


2A l l A

New area 2( )A

A

Since, Resistance Length

Area

[ Resistivity]

11

1

l lR R

A A

…(iii)

New resistance,

2

22

2 /

l l lR

A A A

From equation (iii),

2 22 1R R R


Key Point

A metal wire has a uniform cross-section A, length l , and a resistance R between its two end points. It is uniformly stretched so that its length becomes l then the resistance

between two end points will be 2R .


. Method 1 : Voltage division rule :

A

2

T T

A�

t

( )v t

A

2

T T

A�

( )v t

t

100j� �

100j �

(10 0) V� j

10 �

100j� �

100j �

(10 0) V� j

10 �

cV

1.7 (B)

1.8 (D)




From figure,

100

(10 0)10 100 100C

jV j

j j

[By VDR]

0 100 VCV j


. Method 2 : Mesh Analysis :

Applying KVL,

(10 0) 10 ( 100 ) 100 0j I j I j I

1 AI

The voltage across capacitance is given by,

100 100 1 (0 100) VcV j I j j



. Method 1 : Nodal analysis :


1 01

AVI [From fig. 1 VAV ]

1

1 01

I

0 AI


. Method 2 : Mesh analysis :

Applying KVL in loop-I 1 1 0I

0 AI


. Method 3 : Source Transformation :

Applying source transformation in the above figure,

Applying KVL in above loop, 1 1 0I

0I Hence, the correct option is (A).

100j� �(10 0) V� j

(10 100)j� �

cV

Z

100 �j

100� �j

(10 0) V� j

10 �

I

cV

Scan for

Video Solution

1 �1 V 1 A

I

1 �1 V 1 A

I

A

B

1 �1 V 1 A

I

I

1 �1 V 1 A

I

1 �

1 V

I

1V

1.9 (A)





3

3 6 1eq

VI

Z

3

1 A2 1

The power supplied by the 3 V DC source is,

3 1P VI 3 W



From figure, Potential at point A, B, and C are same i.e. A B CV V V V Hence, resistance between A and C can be neglected. The modified figure can be represented as,


3 5 6 7 0I I I I Hence, the correct option is (D).

Given : ( ) 3 2sin cos 2x t t t

( ) 3 sin 3 sinx t t t

2sin cos sin( ) sin( – )A B A B A B

The rms value of ( )x t is given by,

2 2

2 (1) ( 1)(3) 10

2 2rmsx



Maximum power dissipated in 100 resistor

is,

2max max1W 100P I

max

10.1 A

100I

Maximum power dissipated in 800 resistor

is,

2max max 800 2P I

max

20.05 A

800I

As both resistors are connected in series therefore, 2 W resistor limits the current to maximum value of 0.05 A in the circuit.

The maximum current that can flow in this circuit is, max 0.05 AI

Scan for

Video Solution

3 �

6 � 1 �3 V

eqZ

I

1 �

1 �1 �

1 � 1 �

1 �

1 �

3I

6I

1I

2I

5I

7I

4I

+–

5V

A

C

B

1 �

1 �

1 �

5 V

1 �1I

2I

3I

6I

1 �

1 �

5I7I

A

1 100R � � 2 800R � �

+ _

1W 2 W

1.10 (D)

1.11 (D)

1.12 (C)

1.13 (C)




Then maximum voltage that can be applied without exceeding power limit is,

max max 1 2( )V I R R

max 0.05 (800 100)V

max 900 0.05 45 VV



. Method 1 : Nodal analysis :


10

10 010 10 10

A A AV V V

10 100 0A A AV V V

3 90AV

30AV V

From figure,

10

AVI

( 30)

3 A10

I


. Method 2 : Mesh analysis :

Applying KVL at loop 1, 1 110 10 10( 10 ) 0I I I

120 10 110I I …(i)

Applying KVL in loop 2, 110 10( 10 ) 0I I I

110 20 100I I …(ii)

From equations (i) and (ii), 1 4 A, 3 AI I

Hence, the correct option is (A). . Method 3 : Source transformation :

Applying source transformation, modified circuit is shown below,

Simplified circuit is shown below,

From figure, 9 5

5 10I

[By CDR]

3 AI



I

10 �10 A10 �

10 �

10 V

10 �

10 �10 V 10 A 10 �

AV

A I

10 �

10 � 10 �10 V 10 A+–

1I

I

10I �

1

2

10 �

10 �10 V 10 A 10 �

I

10 �10 � 10 A1A 10 �

I

(10 10) 5� �� 9 A 10 �

I

RR

R R RR R

R

BV

DVCV

AV

5 V

10 V

2 �

1 �

2 A

i

i

1.14 (A)

1.15 (A)




Given : 6 VA BV V

From figure, 6

A 3 A2 2

A BV Vi

The same current will flow through branch between node C and D. Using source transformation :

From above figure, 3 1 2 5 VDCV

5 VCD C DV V V

Without using source transformation :

5 1 5 VDCV

5 VCDV


Key Point (i)

For a particular one port network, Incoming current = Outgoing current

Hence,

(ii) The branch in which ,I V or P is to be

calculated, source transformation is not applied for that branch.

Given delta to star conversion is shown below,

Before scaling :

b cA

a b c

R RR

R R R

… (i)

If ' , ' , 'a a b b c cR kR R kR R kR

After scaling :

' '

'' ' '

b cA

a b c

R RR

R R R

( ) ( )

' b cA

a b c

kR kRR

kR kR kR

2

'( )

b cA

a b c

k R RR

k R R R

… (ii)


'A AR kR


Given : 1 110 F, 10 VC V

2 25 F, 5 VC V

3 32 F, 2 VC V

RR

R R RR R

R

BV

DVCV

AV

5 V

10 V

2 �

i

2 V 1 �

3A

3A

RR

R R RR R

R

BV

DVCV

AV

5 V

10 V

2 �

1 �

2 A

3 Ai �

3 A

5 A

Scan for

Video Solution

1N 2N

2 �BVAV

DVCV

1N

AV

CV

3 A

3 A BV

DV

3 A

3 A

2N

aR

cRbR

CR BR

AR

1.16 (B)

1.17 (C)




. Method 1 :

Charge in a capacitor is given by,

Q CV

Capacitor

Breakdown voltage or maximum operating

voltage

Maximum charge stored in capacitor

1 10 FC 10 V 1(max) 100 CQ

2 5 FC 5 V 2(max) 25 CQ

3 2 FC 2 V 3(max) 4 CQ

Since, 2C and 3C are in series.

So, Charge on both capacitor will be same and

equal to the min ( 2 3,Q Q )

23 4 CQ

Equivalent capacitance of 2C and 3C is,

2 323

2 3

5 2 10F

5 2 7

C CC

C C

Equivalent voltage is,

2323

23

4 C2.8 V

(10 / 7) Feq

QV V

C

In parallel, voltage will be same, hence 2.8 V will appear across 1C also.

Charge stored in 1C is given by,

1 1 eqQ C V

1 10 F 2.8 V 28 CQ

In parallel, total charge is given by,

1 23TQ Q Q

(28 4) C 32 CTQ


. Method 2 :

Capacitor 1C 2C 3C

Value (in F ) 10 5 2

Breakdown Voltage (Volts)

10 5 2

From above figure, 0 2 3V V V

3 02 0 0

3 2

22

2 5 7

C VV V V

C C

023 0 0

2 3

55

5 2 7

VCV V V

C C

Now, 0 210 V, 5 VV V 3and 2 VV

(i) If 0 10 VV ,

Then 2

2 10 202.85 V

7 7V

So, 2 5 VV [No Breakdown]

3

5 10 507.13 V

7 7V

So, 3 2 VV [Breakdown]

Hence, 0V must be less than 10 V.

(ii) If 2 5 VV

0

7 355 =17.5 V

2 2V

0 10 VV [Breakdown]

Hence, 2V must be less than 5 V.

(iii) If 3 2 VV

0

72 2.8 10 V

5V

[No Breakdown]

2C 3C

1C

2C 3C

1C2V 3V

0V




Also, 02

20.8 V 5 V

7

VV

[No Breakdown] None of them are in breakdown. Hence, minimum value of voltage across the combination can be 0 2.8 VV .

Total charge 0eqQ C V

where, 2 31

2 3eq

C CC C

C C

2 5 80

10 F2 5 7eqC

80

2.8 32 C7

Q


. Method 3 :

Here, capacitor 3C has minimum breakdown

voltage and hence the breakdown voltage of capacitor 3C will decide the maximum safe

voltage SV applied across the circuit as shown

in figure.

23

2 3S

CV V

C C

[By VDR]

5

22 5 SV

2.8 VSV

Also, total charge is given by,

eq SQ C V

where, 2 31

2 3eq

C CC C

C C

2 5 80

10 F2 5 7eqC

80

2.8 32 C7

Q



1. For the source X :

From figure,

4 VsV

2 As

s

V

R

Therefore, 2sR

2C 3C

1C2V 3V

V

2C 3C

1C2V 2 V

SV

Scan for

Video Solution

0

1

2

3

1 2 3 4 5 6

Curr

ent

(A)

Voltage (V)

Source Y

Source X

Source Y

Source X 1 �

0

1

2

1 2 3 4 5

Curr

ent

(A)

Voltage (V)

Source X

1.18 1.75




2. For the source Y :

From figure,

3 VsV

3 As

s

V

R

Therefore, 3

13

ss

s

VR

I

The equivalent circuit is shown below,

Applying KVL in above loop,

4 2 3 0I I I

7

4I 1.75 A

Hence, the current through the resistor (1 ) is

1.75 A.

Key Point

(i) Practical independent voltage source

KVL : ( ) ( ) 0s sV i t R V t

( ) ( )s sV t V i t R y mx c

(ii) Practical independent current source

KCL : ( )

( ) 0ss

V tI i t

R

( )

( ) ss

V ti t I

R

y mx c

1

s

mR

and sc I


0

1

2

3

1 2 3

Curr

ent

(A)

Voltage (V)

Source Y

1 �2�

4 V

3V 1�

I

Scan for

Video Solution

LRsV

sR ( )i t

( )V t

A

B

(Small)

V t( )

i t( )

Slope = – Rs

Vs

Practicalvoltagesource

s

s

V

R

Rs = 0 (Ideal)

LR(high)sRsI

( )i t

( )V t

A

B

i t( )

V t( )

Slope = – 1/Rs

Is

Practicalcurrentsource

s sI R

Rs = (Ideal)�

1 A1 R� �

1�1�

1�

1 Volt

I

1.19 1




. Method 1 : Concept of Supermesh :

Applying KVL in loop 3,

3 2 3 11 1( ) 1( ) 0I I I I

1 2 32 1I I I …(i)

From the concept of supermesh,

2 1 1 AI I …(ii)

Applying KVL in loop ABCDEF

1 1 3 2 3 2( ) ( ) 0I I I I I I

1 2 32 2 2 0I I I …(iii)

From equations (i), (ii) and (iii),

1 2 30 A, 1 A, 1 AI I I

Hence, the current through resistor (R) is1 A.

. Method 2 : Concept of Supernode :

KCL of supernode,

00

01 1 1 1

C B CA B A V V VV V V

2 2 0A B CV V V … (i)

KVL of supernode,

1C AV V … (ii)

Applying KCL at node B,

1 01 1

B CB A V VV V

2 1B A CV V V … (iii)

From equation (i), (ii) and (iii),

0 V, 1 V, 1 VA B CV V V

The current through resistor R is given by,

1

1 A1

CVI

R

Hence, the current through resistor (R) is 1 A.

. Method 3 : KVL by using branch current :

Applying KVL in above supermesh,

1 ( 1) 1 1 0I I

1 AI

Hence, the current through resistor (R) is 1 A.

Key Point

Concept of supernode :

If an ideal voltage source (independent or dependent) is connected between two non-reference nodes, then two non-reference nodes form a generalized node or supernode.

Supernode has no voltage of its own.

A supernode requires the application of both KCL and KVL.

Supernode may be regarded as a closed surface enclosing the voltage source and its two nodes.

Node = KCL + Ohm’s law

Super node = KVL + KCL + Ohm’s law

1� 1�

1 R� �1A

2I

3I

1V

1I1�

A C

DEF

I

B

1

1 A 1R � �

1�

1�

1 Volt

AB

C1�

I

This will forma supernode

1 A1 R� �

1�1�

1�

1 Volt

I( 1)I �




Concept of supermesh :

If an ideal current source (independent or dependent) is common between two meshes then we can create supermesh by avoiding the current source and any element connected in series with current source.

A super mesh has no current of its own.

A super mesh requires the application of both KVL and KCL.

Mesh = KVL + Ohm’s law

Super mesh = KVL + KCL + Ohm’s law


. Method 1 : Mesh Analysis :

Applying KVL,

1 0.1 (10 ) 0.1 0x xI I

0.2 2xI

10 mAxI

Hence, the value of current xI is 10 mA.

. Method 2 : Nodal Analysis :


1

10100 100

A AV V mA

1AV V

From figure,

100

Ax

VI

1

100 10 mA


. Method 3 : Source transformation :

Applying source transformation, modified circuit is shown below,

100

20100 100xI

[By CDR]

10xI mA


+–

100� 100�

100� 10 mA1V

xI

+–1V

100� 100�

xI

10 mA100�

10 xI�

+–1V

100� 100�

xI

10 mA100�

A

+–

100� 100�

100� 10 mA1V

xI

100�

xI

10 mA100�100�10 mA

xI

10 mA100�100�10 mA

xI

100�100�20 mA

1.20 10




Key Point If any element is connected in series with ideal current source then there is no effect of that element.

Given : Voltage at Node-1,

11 1 VRV V …(i)

Given circuit is shown in figure,

Applying KCL at Node-1,

1 1 1 241 0

1 0.5R R RV V V V

From equation (i),

21 41 1 0

0.5

V

2 2 VV

Hence, the voltage at Node-2 is 2 V.


Applying KVL in loop ABCD,

1 1 2 3 2 32 6( ) 1( ) 0I I I I I I

1 2 38 7 2 0I I I …(i)

Applying KVL in loop 2,

2 3 2 15 1( ) 6( ) 0I I I I

1 2 36 7 5I I I …(ii)

From the concept of Supermesh,

3 1 2I I … (iii)

From equations (i), (ii) and (iii),

1 2 31A, 2 A, 3 AI I I Hence, the correct option is (A).

R

I

I�

22I1

3�1 �

1RV

0.5�

1 A

1I 2I

14 RV

1 2

3I

+–

2I

6� 1�

2� 1�1I 3I

5V

2 A

A

D

BC

1.21 2

1.22 (A)



1.1 A number in 4-bit two’s complement

representation is 3 2 1 0X X X X . This

number when stored using 8 – bits will be

(A) 3 2 1 00000X X X X

(B) 3 2 1 01111X X X X

(C) 3 3 3 3 3 2 1 0X X X X X X X X

(D) 33 3 3 3 2 1 0X X X X X X X X

1.2 Two 4-bit 2’s complement numbers 1011 and 0110 are added. The result expressed in 4-bit 2’s complement notation is

(A) 0001

(B) 0010

(C) 1101

(D) cannot be expressed in 4-bit 2’s complement

1.3 7E H and 5F H are XORed. The output is multiplied by 10 H. The result is

(A) 0210 H (B) 7E5F H

(C) 5F7E H (D) 2100 H

1.4 An 8-bit 2’s complement representation of an integer is FA (Hex). Its decimal equivalent is

(A) 10 (B) 6

(C) 6 (D) 10

1.5 A number N is stored in a 4-bit 2’s complement representation as

3a 2a 1a 0a

It is copied into a 6-bit register and after a few operations the final bit pattern is

3a 3a 2a 1a 0a 1

The value of the bit pattern in 2’s complement representation is given terms of the original number is N as

(A) 332 2 1a N (B) 332 2 1a N

(C) 2 1N (D) 2 1N

1.6 The result of 10 16(45) (45) expressed in

2’s complement representation is

(A) 011000 (B) 100111

(C) 101000 (D) 101001

1 Number Systems

1999 IITBombay

2003 IITMadras

2004 IITDelhi

2005 IITBombay

2006 IITKharagpur

2008 IIScBangalore




1.7 The binary representation of the decimal

number 1.375 is

(A) 1.111 (B) 1.010

(C) 1.011 (D) 1.001

1.8 The base of the number system for the

addition operation 24 14 41 is_____.

1.9 The representation of the decimal

number (27.625)10 in base-2 number system is

(A) 11011.110 (B) 11101.101

(C) 11011.101 (D) 10111.110

2009 IITRoorkee

2011 IITMadras

2018 IITGuwahati

5.5GATE ACADEMY ® Digital Electronics : Number Systems



1.1 C 1.2 A 1.3 A 1.4 B 1.5 D

1.6 C 1.7 C 1.8 7 1.9 C

RepresentationofSignedBinaryNumbers

Specialcaseof2’scomplementrepresentation

Given : A number (N) of 4-bit having 2’s complement form is 3 2 1 0X X X X

In the 2’s complement representation of a number the MSB bit can be repeated a number of times without affecting magnitude of the number. So, the number (N) represented in 8 bit’s will be

3X 3X 3X 3X 3X 2X 1X 0X


Key Point

If a binary number is represented using N-bits (for 2's complement format) and it has been asked to represent it using (N+M) bits, then just copy MSB M-times. Examples : (i) 1011 bit represents –5 using 4 bits 11011 bit represents –5 using 5 bits 11111011 bit represents –5 using 8 bits (ii) 0011 represents +3 using 4 bits 00011 represents +3 using 5 bits 00000011 represents +3 using 8 bits

Given : Two 4 bit 2’s complement number are 1011 and 0110

. Method 1 .

1011 2’s complement representation of –5

0110 2’s complement representation of +6

Adding both numbers

5 6 1

+1 is represented in 4 bits 2’s complement notation as

1 0001


. Method 2 .

So, the result of addition in 4 bits 2’s complement representation is 0001.


Given :

7E H 0111 1110

5F H 0101 1111

XORed 0010 0001 21H

Converting 21 H to decimal number,

21 H 1 02 16 1 16 10(33)

Converting 10 H to decimal number,

1 010 H 1 16 0 16 10(16)

According to the question 21 H 10 H

10 10 10(33) (16) (528)

Converting decimal number 528 to hexadecimal number,

Scan for Video

Explanation

Scan for Video

Explanation

1 0 1 10 1 1 0

1 0 0 0 1+

end around carry is discarded

Answers NumberSystems

Explanations NumberSystems

1.1 (C)

1.2 (A)

1.3 (A)




= 210 H 21H 10H 210H


Given : 8-bit number in 2’s complement representation is FA H.

16 2(FA) (11111010)

As sign bit (MSB) is 1, so the number is negative. Hence, the binary of actual numbers will be 2’s complement of the binary shown above. The actual number is given as – [decimal equivalent of 2’s complement of (11111010)] i.e. – [decimal equivalent of (00000110)] = – 6 Hence, the correct option is (B).

Key Point

1. Short trick to find 2’s complement : Move from right to left, keep bits till first

‘1’ as it is, then complement each bit. Example : To find 2’s complement of

101011100100

2’s complements = 010100011100

2. In signed data representation, MSB

represent sign bit.

3. To represent any positive number in 1’s or in 2’s complement representation, first write binary of the given number and then place sign bit (MSB) equal to 0.

Example :

4. To represent any negative number in 1’s

or 2’s complement representation, write the positive number first and then take its 1’s or 2’s complement.

Example

To represent – 5 in 1’s and 2’s

complement representation

+ 5 0101

– 5 1010 (in 1’s complement)

– 5 1011 (in 2’s complement)

5. To find value of any number represented in 1’s or 2’s complement representation

(i) If MSB = 0 Number is positive

Just take the decimal of given binary

(ii) If MSB = 1 Number is negative

Given binary is 1’s or 2’s complement of actual.

Copy MSB once and decimal equivalent will be 1’s or 2’s complement of given binary with negative sign.

Example :

If number in 1’s complement is 101001

Copy MSB once 1101001

Decimal equivalent equal to

– (decimal equivalent of 0010110) = – 22

If number in 2’s complement is 101001

Copy MSB once 1101001

Decimal equivalent equal to

– (decimal equivalent of 0010111) = – 23

16 528

33

2 1

016

Scan for

Video Solution

101011100100

First ‘1’

Move

010100011100

Same

Complemented

5(in 4 bits) 0 101� �

signbit binary of 5

1.4 (B)

5.7GATE ACADEMY ® Digital Electronics : Number Systems



Given : (i) 4-bit 2’s complement representation of

number N as,

N 3a 2a 1a 0a

(ii) 2’s complement number after some operations using 6 bits is

3a 3a 2a 1a 0a 1

In the 2’s complement representation of a number the MSB bit can be repeated any number of times without affecting magnitude of the number. So, the number N can be represented in 6 bits as,

3 3 3 2 1 0N a a a a a a

If the bits of the number is shifted to left by one then LSB 0 and the number is multiplied by 2.

If we add binary 1,

Then, 3 3 2 1 0 3 3 2 1 02 1 0 1 1N a a a a a a a a a a


Key Point

(i) A number gets multiplied by 2n if n times shift left operation is performed.

(ii) A number is divided by 2n if n times shift right operation is performed.

Given : 10 16(45) (45)N … (i)

Conversion from Hexadecimal to decimal :

016(45) 4 16 5 16 69

From equation (i), 10 10(45) (69)N

10( 24)N

2’s complement representation of – 24 24 011000

24 2’s complement of 011000

= 101000 Hence, the correct option is (C).

Given decimal number 1.375

The binary representation 2(1.011)


Given : 24 14 41 Let, the base is b

So, (24) (14) (41)b b b 1 0 1 0 1 02 4 1 4 4 1b b b b b b

2 4 4 4 1b b b 3 8 4 1b b 7b Hence, the base is 7.

Given : (27.625)10

10 2(27) (11011)

10 2(0.625) (0.101)

10 2(27.625) (11011.101)


a3 a3 a3 a2 a1 a0 0

a3 a3 a2 a1 a0 02N �

0.375 2 0.750� �

0.75 2 1.50� �

0.50 2 1.00� �

0

1

1

Scan for

Video Solution

272

132

62

32

1 1

0

1

10.625 2�

0.25 2�

0.50 2�

�

�

�

1.25

0.50

1.00 1

0

1

1.5 (D)

1.6 (C)

1.7 (C)

1.8 7

1.9 (C)

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