Circuit Theory - Solved Assignments - Semester Fall 2003

8/9/2019 Circuit Theory - Solved Assignments - Semester Fall 2003

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Assignment 1 (Fall 2003)(Solution)

CIRCUIT THEORY(PHY301)

MARKS: 30

Due Date: 08/10/2003

*Note: Brain teasers are not a part of the assignment, they dont carry any marks. Those who attemptthese questions will be considered for the best students of the week for this course. So do the best of

your efforts to answer those questions.

Q.1.

If a current of 40 A exists for 1 min, how many coulombs of charge have passed through the

wire. If the current consists of electrons only, calculate the number of electrons passed

through the wire, during this period of time.

(electronic charge = 1.6 10-19 C)

Sol.

Current is the total charge passing through a surface in one second. Hence for this problem,

40C charge will pass through the wire in 1 second. So in 1 min, 2400C charge will pass.

From the electronic charge as give, 1C of charge contains 6.251018, thus 1.51022 electrons

passed through the circuit.

Q.2.

Point A is set at electric potential 2J, and point B has potential 10 J. Now if an electron is

moved from point A to point B, It will release some energy or it will absorb?. Calculate that

energy. If 5C charge is moved from point B to A, calculate the potential difference in Voltsbetween points B and A.

Sol.

Since point A is at low potential compared to point B, electron will absorb energy, which is

equal to the difference of the potentials of two points, that is 8J. Mathematically volt is

defined as the potential difference per unit charge. Thus if 5C charge is moved from point B

to point A, hence there are 8J/5C = 1.6V across points A and B.

Circuit Theory - Solved Assignments - Semester Fall 2003


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Find the total resistance RT in the following circuits, please draw each step.

Sol.

a). Since in this circuit, resistor R5 has no voltages across so it contains no current, hence R2

and R3 are in series. R1, R2, R3, and R4 are in series so total resistance will be 10.8k. as

shown in fig below.

+

-

10.8 K

5.2 K

b). Same circuit can be drawn in the following way.

R1R2

R4R3

2 K 4 K

1.2 K2.1 K


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Now, R1 and R2 are parallel and R3, R4 are parallel. so solving for the parallel resistors.

1.3 K

0.8 K

resulting resistors are in series, hence final solution will be 2.1 k.

2.5 K

c). In this circuit, since resistant network is on both sides of the battery, so we start from both

sides to solve for total resistance. As it is clear, on the left side R1, R2 and R3 are in series

and on the right side R8 and R9 are in series, solving for resistors in series, we get

R3

R5

R6

R76.4 K

5 K

2.3 KV

5 K

4 K 14 K

Resulting 6.4k is parallel to R3, and 14k is parallel to R7, hence

R5

R6

5 K

V1.7 K

5 K

3.1 K

Now resistors on both sides of the battery are in series, so


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6.7 K 8.1 KV

if you look closely, we can rearrange this circuit in a bit different but simple way. as shown

in the following fig.

6.7 K 8.1 KV

Final solution is obvious now, that is 3.7k.

3.7 KV

Q.5. Fill in the following blanks.

1. The valence of an element having 11 electrons is 1.

2. The relative conductivity of silver is 105 percent and that of gold is 70.5 percent.

3. Circuit is said to be short when there is no load resistance in the circuit.

4. According to Ohms law voltages and currents have linear relationship in a circuit.

5. Two resistors are said to be in series if their common node is not connected toa current carrying circuit element.


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*Brain Teasers

1. If we look at the Ohmic relation (relation given by Ohms law), it assumes a constantresistance (no dependence on voltages) of circuit elements. Is it possible for any

electronic component to work against the Ohms law. If so, explain in some detail.2. Insulators dont conduct electric current at room temperature, does it mean that they

are not of any use in electronic circuits?, if no, can you think of any possible use?

Sol.

1. Literally, resistance is actually the opposition offered by a certain material, in thepath of current. Its one type is what you can call a permanent resistance, which

arises when charges flow through the atomic structure of the material, that

resistance should be independent of the applied voltages, it is actually a property

of a material. Now it is possible that a bit different type of opposition may arise,

which depends upon the voltages. One example is semiconductor PN junction(which well discuss in detail in later lectures), current passing through this

junction feels an opposition which depends upon the applied voltages, as the

voltages increases opposition falls down and the current increases slowly with

respect to the voltages, after certain applied voltages, current starts to increase

linearly with voltages, satisfying Ohms law. But at low voltages, the relation

between current and voltage doesnt follow ohms law, as for that range of

voltages the resistance is not independent but varies with the voltages. Following

graph schematically shows the current voltage relation for a typical PN junction.

OhmicRegion

Non

Ohmic

Re

gion

Voltages

Current

LinearBehavior

Non-LinearBehavior


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2. Insulators are used as a dielectric in capacitors, they actually increases the capacity of

the capacitor. This phenomenon is based upon state electricity, in which electronic region

around the nucleus is deformed while still remain around it, forming a negative and

positive regions. Technically this phenomenon is called polarization.Also read about the

material used in inductors.


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Assignment 2 (Fall 2003)Solution

CIRCUIT THEORY (PHY301)

MARKS: 30

Due Date: 17/10/2003

*Note: Brain teasers are not a part of the assignment, they dont carry any marks. Those who

attempt these questions will be considered for the best students of the week for this course. So do

the best of your efforts to answer those questions.

Q.1.

Find the voltage VAB in the following circuits.

12 V VAB

A

B

2.3 K 1.3 K

2.1 K 2.5 K

6 V

A

B

2 K

1 K

0.5 K

VAB

(A)

(B)

(hint: use the voltage divider rule)

Sol:

a). resistors R1.3 and R2.5 are in series so

adding the two resistors gives 3.8K.

Required voltages VAB are across this

resistance. Using voltage divider rule,

V

2.3K

A

B

3.8K

2.1K

VAB = (3.8/8.2).12 = 5.7 V


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6V

2K

A

B

1.5K

b). Resistors R1 and R1.5 are in series, so

adding them we get 1.5 K. Required voltages

VAB are across this resistance. Using voltage

divider rule,

VAB = (1.5/3.5).6 = 2.6 V

Q.2.Calculate the indicated quantities in the following circuits.

(A)

(B)

24 mA

Vo

6 K

8 K8 K

6 K

+

-

12 mAIo

6 K

4 K

3 K

2 K

12 V

12

12

12

5

Io

5

Is 9 K

3 K

V =6Vo

1 K

5 K2 K+

-

(C )

(D )


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Sol.

a). To find voltages Vo we need to find the current passing through 6K.

R6 and R6 are in series so their resultant will be 12K, now R8, R12 and

R8 are parallel to each other, their resultant will be 3 K. Using

current divider rule, current I passing through R12 will be

I = (3/12).24 mA = 6mA

This current is passing through 6K resistance, so voltages across it

will be

Vo= 6K.6m = 36V

b). Resultant R of R3, R6 and R2 is 4K, which is parallel to 4K, thus

using current divider rule we find the total current I` passing through

R as

I` = (4/8).12mA = 6mA

This current divides further into two components, one passing through

3k and 6K. Thus current Io passing through 3K will be

Io = (6/9).6mA = 4mA

c). Its a bridge network, no resistor is in

series or parallel to any other resistor. Since

R12, R12, and R12 is forming a delta network, we

can change it into Wye network. Please check

this topic Wye-Delta transformation in FAQs on

your LMS. Corresponding resistors in Wye

configuration will be 4, 4 and 4, as shown in

the fig. Now its easy to find the total

resistance R of the network,5 5

44

4

12 V

Io

T

RT = (81/18) + 4 = 8.5

Thus source current I will be 12/8.5 = 1.4 A.

Since the two parallel resistors are equal, so the source current will

divide equally to the two branches.

Hence the required current will be 0.7A.

d). Resultant of resistors R2K , R1K and R5K is 1.5K .Voltages V9K

across 9K, are divided into voltages V3K across 3K and V1.5K (=6V)

across 1.5K resistance. Thus using voltage divider rule

6 = (1.5/4.5). V9KThis gives,

V9K = 18V

Thus voltages V3K across the resistor 3K will be 12V.

Now the source current IS is divided into a component I3K passing

through 3K, and the other I3K passing through 9K,

I3K = V3K/3K = 4mA

I9K = V9K/9K = 2mA

Thus

IS = I9K + I3K = 6mA

Q.3.

Find the total power absorbed by the following network.

6 K 6 K

18 K2 K

12 K

21 V


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Sol.

Circuit can be rearranged as shown in the fig below

6K 2K

12K

6K 18K

21 V

IS

First we have to find the total current passing

through the network.

Resistors R6K, R2Kand R12K forms a delta network,

its equivalent wye network is shown in the fig.

Corresponding resistors in wye network can be

determined by using the transformation formulas

(Check the FAQs on LMS for this topic).

Total resistance RT can be found easily now.

RT = [(18 + 1.2)(6 + 3.6)/(18+1.2+6+3.6)] + 0.6

RT = 7KSince voltages across the network are 21V, so total

current I passing through the network is given by,

I = 21/7K = 3mA

Total power delivered absorbed by the network is

equal to the total power delivered to the system, which is given by

6 18

1.23.6

0.6

21 V

Io

P = VI = 21.3m = 63.10-3

W = 0.063WQ.4.

The following network is the basic biasing arrangement for the field-effect transistor(FET). 16V (VDD) are the total voltages applied across the network with respect to the

ground. Using these parameters

VGS = -1.75VIG = 0 AID = IS

a) Determine the voltages VG and VS.b) Find the currents I1, I2, ID, and IS.c) Determine VDS.d) Calculate VDG.

Hint:

Imagine some sort of resistance between the nodes G, D

and S.VG is the voltages at node G, with respect to the ground.

If you look closely it is actually the voltages across the

resistor R2, so use the voltage divider rule to find it.VGS is given, hence its easy to find VS, which are the

voltages at node S, with respect to the ground.

Voltages (VG) across R2 is now known, so current I2 caneasily be found. Similarly voltages (VS) across RS is known so current IS can be found.

Currents I1 can be found using KCL. Using same technique you can find IS and ID.

2 M

16 V

VG

IG

I1

IS

I2

ID

VDD

VGS

VS

R2

RS

RD

R1

G

D

S

270 K

1.5 K

2.5 K

+

+

-


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To find VDS and VDG, you first have to find VD, which is the voltages at node D, with

respect to ground. Find voltages across RD first, and then find VD.

Sol.

since IG = 0, so current I1 and I2 are equal, hence resistors R1 and R2

are in series, and the total voltages across them are 16V, thus current

passing through them will be 16/(2.106 + 270.103)=7.10-6A = 7A. Nowvoltages VG are actually across the resistor R2, thus it is given byVG = 270K . 7 = 1.89Vsince VGS = -1.75V,

VGS GV

= V - VS = -1.75

S = VG + VGS = 3.64V

V is now known, hence IS is giv

I = V /R = 3.64/1.5K = 2.43mAS en by

S S S

also ID = 2.43mA

ID is now known, we can find Voltages VRD across RD as

VRD = RD.ID = 2.5K . 2.43m = 6.08V

Now the total voltages across this branch containing the resistors RD,

RS and FET, are 16V, so voltages VD at terminal D of FET with respect to

the ground, is given by 16 - V = 16 6.08 = 9.92V. ThusRDVDS = VD VS = 9.92 3.64 = 6.28V

VDS = VD VG = 9.92 1.89 = 8.03V

Q.5.

In the following network, calculate the value of source voltages VS given that

I4 = 0.5 mA.

VS

I1

I5I

2

I3

I4

Va

Vb

1 K

3 K

+

-

6 KX

Y

Z

2 K

6 K3 K

4 K

(hint: use the KVL, KCL and divider rules.)

Sol.

Since I4 is 0.5mA, so Va = 6K . 0.5m = 3V. voltage across 3K is also Va

so I3 is given by 3/3K = 1mA. Now at node y, I2 = I3 + I4, so I2 = 1.5mA,

hence Vb = (1.5m)(2K)= 3V.

Total resistance RXZ between the nodes x and z is given by

RXZ = 2K

and total voltages across x and z,

VXZ = Va + Vb = 6V

Thus using voltages divider rule, consider the circuit consisting of

one loop with series resistors 6K, 4K and RXZ = 2K, with total

voltage VS across them.

VXZ = [2K/(2K + 6K + 4K)].VS = 6V


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this gives

VS = 36V

BRAIN TEASERS

Q.1.

Here is a paradox. According to the ohms law voltages V across a resistor R is theproduct of its resistance and current I passing through it. Now we know that current can

only pass between the two nodes, if there is some potential difference across them, that is,there must be some voltages across the nodes to have current between them. In shortcircuits, we know that current is maximum (practical infinite), this is because the path

have zero resistance, now in that situation ohms law says that voltages across that path

must be zero, as the resistance is zero (V = RI). Now if there is no voltages across thatpath, current must also be zero, but current is maximum in that situation.

How would you resolve this paradox.

Sol.

For the short circuit, current passing through it, is actually

controlled by the external voltage source (e.g. battery). Hence,

although potential difference across nodes, between which the circuit

is short, is zero, current flows through them due to the external

circuit. Short circuit will act as a path with zero resistance. Ohms

law will not be applicable between these nodes.

Q.2.

Find the current passing through 1K and 1.5K in the following network.

12V

2K

5K

1.5K1K

2.1K 3K

Sol.

Look closely at the circuit, there is a short circuit parallel to the

resistors 1K and 1.5K, so no current will pass through them.

Q.3.

Did you ever think about the basis of Kirchhoffs Laws?, Can you think of any situationwhere you cant use these laws. Try to give a mathematical argument in support of your

point.

Sol.

Kirchhoffs voltage law is actually based upon energy conservation

principle, that is total energy absorbed by the circuit must be equal

to the energy delivered to it. Consider a single loop circuit with some

independent voltage sources. Now energy conservation requires that

total power delivered must be equal to the total power absorbed, so

Pabsorbed = Pdelivered


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since same current I is flowing through all elements in the loop

IVdrop = IVsource

this gives

Vdrop = Vsource

This is the Kirchhoffs voltage law.

Similarly, Kirchhoffs current law is based upon the conservation of

charge principle, that is, number of charges entering a node must be

equal to the number of charges leaving it. Since these laws are based

upon universal principles, so they must be satisfied in any circuit.

----Good luck----


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Assignment 3 (Fall 2003)CIRCUIT THEORY (PHY301)

MARKS: 30

Due Date: 30/10/2003

Q.1.

Use nodal analysis to find VO in the following circuit.

4mA 2mA3K

4K

4K

6K

VO

Sol.

4mA 2mA3K

4K

4K

6K

VO

V1 V2

Consider the above fig. reference node is selected, and node voltagesare labeled as V1 and V2. Directions of unknown currents are chosen

according to the rule, all unknown currents will be considered as

flowing outwards from a particular node. Writing the KCL equations for

the two nodes,

At node 1

-4m + V1/3K + (V1-V2)/4K = 0

4V1 + 3V1 3V2 = 48

7V1 3V2 = 48 (a)

At node 2,

(V2-V1)/4K + 2m + V2/4K = 0

V1 2V2 = 8 (b)

Solving equations (a) and (b) simultaneously, yields

V1 = 6.54V

V2 = -0.73V

Required voltages VO is given by

VO = V2-V1 = -0.73 6.54 = -7.27V

Q.2.

Find the voltages V1, V2 and VO in the following network.


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2mA

4mA 3K

2K

+

-

2K VO

V1 V

2

6K

12K

ol.

t node 1;

-2m

and 2K in series, so using

rule,

Find the voltage VO, in the following circuit.

S

Considering above fig. we write the KCL equations for the two nodes as

A

4m + V1/3K + (V1-V2)/6K = 2m

V2 = -12 (a)3V1

At node 2;

4K /12K =(V2-V1)/6K + V2/ + V2

VV1 3 2 = 12 (b)solving (a) and (b) simultaneously we get,

V1 = -6V

V2 = -6V

ross the two resistors 2KSince V2 is ac

voltage divider

Vo

.3.

= (2/4).(-6) = -3V

Q

1K

1K1K

+

-

VO

1K

2mA

12V

4mA

2mA

3K

2K

+

-

2K VO

V1 V2

6K

12K


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Sol.

1K1K

+

-

VO

1K

1K

2mA

12V

V1

V2

V3

At node 1;

(V1-V2)/1 + (V1-V3)/1 = 2

V1 - V2 + V1 - V3 = 2

2V1 - V2 - V3 = 2 (a)

At super node:

(V3-V1) + V3 + V2 + (V2-V1)= 0

-2V1 + 2V3 + 2V2 = 0

-V1 + V3 + V2 = 0 (b)

Now voltage source 12V between V2 and V3 has put a constrain on

the values of V2 and V3, mathematically,

V2 V3 = 12V (c)

Solving (a), (b) and (c) simultaneously gives,

V1 = 2V

V2 = 7V

V3 = -5VQ.4.

Find the voltage across 8 resistor in the following circuit.

8

10

1K

2A

12V


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Sol.

Consider the above fig. Voltage across resistor 8 resistor,

considering the supposed direction of the current, is (v1 - 12), hence

current passing through this resistor is (v1 - 12)/8. Now writing the

KCL equations for the two nodes,

at node 1,

(V1 - 12)/8 + (V1 V2)/1K = -2 (a)

at node 2,

(V V )/1K + V /10 = 2 (b)2 1 2solving the two equations (a) and (b), yields,

V1 = -3.81V

8

10

1K

2A

12V

V1 V2

V2 = 19.76V

Voltage across 8 resistor is (12 V1) = 12 + 3.81 = 15.81V

Q.5.Find IO in the following circuits.

4mA

24V

V1 V2 V3

6

10

4 12

20 403A(A)

(B)

4A

16V

Io

Io

Sol.A). Consider the following fig.


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4mA

24V

V1

V2

V3

6

10

4 12

Io

At node 1,

V1/6 + (V1-V2)/10 = -4m (a)

AT supernode,

(V2-V1)/10 + V2/4 + V3/12 = 0 (b)

constrain due to voltage source 24V, gives

V2 V3 = 24 (c)

Solving these three equations, we get

V1 = 1.91V

V2 = 5.06V

V3 = -18.94VRequired current I passing through resistor 12 iso Io = -V3/12 = 18.94/12 = 1.58A

B).

20 40

3A

4A

16V

Io

V1 V2

Considering the above fig, at supernode,

V1/20 + 3 3 + V2/40 4 = 0

V1/20 + V2/40 4 = 0 (a)

Constraint due to the voltage source 16V,

V2 V1 = 16 (b)

Solving (a) and (b) gives

V1 = 48V

V2 = 64V

required current Io is given by

Io = -V1/20 = -48/20 = -2.4A

----- Good Luck -----


20/35


CIRCUIT THEORY (PHY301)

MARKS: 30

Due Date: 12/11/2003

Q.1. Use loop analysis to find VO in the following circuits.

12V VO

+

-

3

22

4

12V VO

+

-

2mA

4 k2 k

2 k

(A)

(B)

Sol. a)

12V VO

+

-

3

22

4

I1 I2

Currents I1 and I2 are assigned in clockwise direction for

the two loops, as shown in the fig above. Now writing the

KVL equation for the two loops,

Loop 1;

2kI1 + 3k(I1 I2) = 12

5I1 3I2 = 12m (a)Loop 2;

3k(I2 I1) + 6kI2 = 0

3I1 9I2 = 0 (b)

Solving the two equations (a) and (b) simultaneously, we

get

I1 = 0.6A

I2 = 1mA

Now voltage across 4k is 4k(I2) = 4V


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b)

12V VO

+

-

4 k2 k

2 k

I1 I2

Since current source is shared by the two loops, we make

supermash by removing the current source, as shown in fig.

Now writing KVL for the supermash,

2I1 + 6I2 = 12m (a)

Presence of a current source has put a constraint on the

values of I1 and I2, mathematically

I2 I1 = 2m (b)

solving the two equations (a) and (b), we get

I1 = 0A

I2 = 2mA

Now voltage across 2k is 2k(I2) = 4V

Q.2. Use loop analysis to find IO in the following circuits.

2mA

IO

IO

12V

4mA

2k

4 k

6k

12V

4mA

2mA

1k

1k

1k1k

1k

1k

(A)

(B)

Sol.

a)


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I1 I2I3

IO

12V

2k

4 k

6k

Taking I1 = -2mA and I3 = 4mA. I2 = IO is the required

current to find, as shown in the fig above. KVL equation

for the loop 2 is given by

2k(I2 I1) + 4kI2 + 6k(I2 I3) = 12

-I1 + 6I2 3I3 = 6m

putting the values of I1 and I3 in above equation

I2 = 2.66mA

b)

I1 I

2

I3 I4

IO

12V

1k

1k

1k1k

1k

1k

Setting I2 = 2mA. Current source 4mA is a common source for

loop 3 and 4, making a supermesh by joining these two

loops, as shown in fig. Writing KVL equations for loop 1

and supermesh

for supermesh;

I3 + (I3-I1) + (I4-I2) + I4 = 0

2I3 I1 + 2I4 = 2m (a)

for loop 1;

I1 + (I1-I2) + (I1-I3) = 12

3I1 I3 = 14m (b)

Also due to the constraint imposed by the current source

4mAI4 I3 = 4m (c)

Solving the three equations simultaneously

I1 = 4.5mA

I3 = -0.36mA

I = 3.6mA4Required current IO = I4 I2 = 3.6m 2m = 1.6mA


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Q.3. Use both Nodal and Loop analysis to find voltage VO in thefollowing circuit.

12V

VO

4mA

6k

2k

2k

4k

Sol.

Nodal Analysis

12V

V1 V2

V3

VO

4mA

6k

2k

2k

4k

At super node;

(V1V2)/2 + (V3-V2)/4 + V3/6 = 48

6V1 9V2 + 5V3 = 48 (a)

At node 2;

(V2-V1)/2 + (V2-V3)/4 + V2/2 = 0

-2V1 + 5V2 V3 = 0 (b)

Constraint due to the presence of voltage source

V3 V1 = 12 (c)

Solving three equations simultaneouslyV1 = 1.71V

V2 = 3.43V

V3 = 13.71V

Required voltage VO = V3 = 13.71V


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Loop Analysis

I1

I2

I3

12V

VO

6k

2k

2k

4k

Consider the above fig.

Setting I3 = 4mA

for loop 1;

4k(I1-I2) + 2k(I1-I3) = 12

6I1 4I

2 2I

3= 12m

6I1 4I2 = 20m

3I1 2I2 = 10m (a)

for loop 2;

4k(I2-I1) + 2k(I2-I3) + 6kI2 = 0

-4I1 + 12I2 2I3 = 0

4I1 12I2 = -8m

I1 3I2 = -2m (b)

Solving the two equation simultaneously

I1 = 4.8571mA

I2 = 2.28571mA

Thus voltage across 6k resistor is 6k(I2)=6k2.28571m =13.71V


25/35

Assignment 5 (Fall 2003)CIRCUIT THEORY (PHY301)

MARKS: 30

Due Date: 05/01/2004

Q.1. Using superposition, determine the current through 6resistance in the following Circuit.

Sol.

Considering the effect of the 44 voltage source in the

figure below,

RT=R1+R2||R3 = 30 + 16 ||6

= 96/22

= 34.36

I = V1/RT=44/34.36=1.28A

Using the current division rule,

I3 = R2I/R2+R3

= (16)(1.28)/22

= 0.93 A

Considering the effect of the 38V source in the

figure below,


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RT=R3+R1||R2 = 6 + 30 ||16 = 6 + 10.43

= 16.43

I3 = V2/RT= 38/16.43

= 2.31A

Total current through the 6 resistance,

I3= I

3-I

3= 2.31A 0.93 A

= 1.38 A (direction of I3)

Q.2. Find the current through 2 resistor of the network given belowby using superposition theorem.

Sol.

Considering the effect of the 12V source in the figurebelow,

I1 = V1/R1+R2

= 12V / 2 +4

=12/6 =2AConsidering the effect of the 6V source in the figure

below,

I

1

= V2

/R1

+R2

= 6V / 2 +4

=6/6 =1A

Considering the effect of the 3-A source in the figure

below,


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I

1 = R2(I)/R1+R2= (4)(3A) / 2 +4

= 12A/6 = 2A

Total current through the 2W resistance

I1 = I1 + I

1 - I

1

= 1A + 2A - 2A

= 1A

Q.3. Find the Thevenin equivalent circuit for the network in theshaded area.

Sol.

First step: Removing RL

Second step: Calculating Vth


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In this case since an open circuit exists between the two

terminals, the current is zero between these terminals and

through the 2ohm resistance. The voltage drop across R2 is

therefore,

V2=I2R2=(0)R2 = 0V

and Vth= V1=I1R1=IR1

=(12A)(4ohm)= 48VThird step: Calculating R

th

Rth=R1+R2 = 4 + 2 =6 ohm

Fourth step:After calculating V

thand R

th, re-inserting the load

resistance RLin the circuit in series with R

thand

considering the Vth

as a battery in series with

these two resistances

Substituting the Thevenin equivalent circuit in the

network external to the resistor R3

------ Good Luck -----


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MARKS: 30

Due Date: 17/01/2004

Q.1.

Assume that a silicon diode in the figure below requires a minimum of1mA to be above

the knee of its I-V characteristic.

(1) What should be the value ofR to establish 5mA in the circuit?

(2) With the voltage value ofR calculated in (1) what is the minimum value to

which the voltage Ecould be reduced and still maintain the diode current

above the knee?

Sol.

(1) If I is equal to 5mA we know that the voltage across

the diode will be 0.7V.

Therefore

E = IR +0.7

R = E - 0.7/I = (5-0.7)/5 x 10-3= 860

(2) In order to maintain the diode current above knee .I

must be at least 1mA.

I=E-0.7/R 10-3 A

Therefore, since R= 860

E-0.7/860 10-3 A

E (860 x10-3)+0.7

E 1.56 V

Q.2.

Determine the values of VD, IT and VR for the circuits given below.

(a)

(b)


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Sol. (a)

Because the arrow in the schematic symbol is pointing

toward the positive terminal of the source, we know that

the diode is reverse biased. Therefore,

a. The full voltage is dropped across D1.VD1=VS=5V

b. D1 will not allow conduction. Therefore, IT=0Ac. Since there is no current through R1, there is no

voltage drop across the component (VR1=0V)

Sol. (b)Because the arrow in the schematic symbol is pointing

toward the negative terminal of the source, we know that

the diode is forward biased. Therefore,

a. VD1 =0V, leaving the total applied voltage to be

dropped across R1.

b. VR1=VS=5Vc.IT is determined by the source voltage and R1.

IT= VR1/R1

= 5V/1K

= 5mA

Q.3.

Find the Norton equivalent circuit for the network in the shaded area of the figure:

First step: Replacing RLwith a short circuit to find I

N.


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Second step: As it is shown in the above figure indicating

that the short circuit connection between terminals a and b

is in parallel with R2 and eliminates its effect. IN is

therefore the same as through R1, and the full battery

voltage appears across R1 since

V2=I2R2

=(0)(6)

= 0VTherefore

IN = V/R1

= 9V/3 = 3A

Third step:

To calculate RN we will short circuit all voltage

sources.

As shown in the above figure we have

RN=R1|| R2 = 3 ||6 = 2

Fourth Step:

After calculating IN and RN, re-inserting the load

resistance RL in the circuit in parallel RN and

considering the IN current source parallel with these

two resistances.

------ Good Luck -----


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MARKS: 30

Due Date: 26/01/2004

Q.1.

A silicon diode has a saturation current 1pA. Using the values of n given here and

assuming the temperature is 25oC(room temperature) find the current in the diode.

(a) It is reverse biased by 0.1 V (n = 2)

(b) It is forward biased by 0.5 V (n=1)

Sol.

We first calculate the thermal voltage at

T= 273 + 25oC= 298

VT =kT/q

= (1.38 x 10-23) (298) / 1.6 x 10-19

= 0.0257 V

(a) Since the diode is reverse biased, we substitute

V = -0.1 V in the following equation we have

I=IS(eV/nT-1)

= (1pA)( e(-0.1)/2(0.0257V)-1)

= (1pA)(0.143-1)= -0.857 pA

(b) Since the diode is forward biased, we substitute

V = 0.5 V

I=IS(eV/nT-1) where 1pA10-12A

= (1pA)( e(0.5)/(0.0257V)-1)

= (1pA)( 2.814 x 108-1)

= (10-12A)( 2.814 x 108)= 0.2814 mA

Note that the value of the exponential term (2.814 x 108)

in this case is so much large than 1 that for the practical

purpose I=ISeV/nT

Q.2.

Determine which diodes are forward biased and which are reverse biased in each of theconfiguration give arguments to verify your point of view.


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Sol.

a. In (a) the anode is grounded and therefore at 0V.The cathode side is positive by virtue of 5V

source connected to it through resistor R. The

cathode is therefore positive w.r.t. the anode

i.e. the anode is more negative than the cathode

so the diode is reverse biased.

b. In (b) the anode side is more positive than thecathode side (10V > +5V) so the diode is forward

biased. Current flow from 10V source, through the

diode and into the 5V source.

c. In (c) the anode side is more negative than thecathode side, so the diode is reverse biased. No

current flows in the circuit, so no drop across R.

Q.3.

Determine the peak Load current for the circuit shown in the figure below.

Sol.

The input voltage is given an rms value. This value

is converted to a peak value as follows:

V1(pk) = V1(rms)/0.707

= 150/0.707V

1(pk)= 212.2 V

pk

Now, the load voltage and current are found,

after fining peak voltage, as

V2(pk)

= N2/N1 V1(pk)

= (1/10)(212.2 Vpk

)


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V2(pk)

= 21.22Vpk

Finally, the load voltage and current and current

values are found as:

VL(pk)

= V2(pk)

VF

= 21.22 0.7

VL(pk)

= 20.52Vpk

and the current will beIL(pk)

= VL(PK)

/RL

= 20.52Vpk/5.1

IL(pk)

= 4.02 mApk

------ Good Luck -----


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Assignment 8 (Fall 2003)CIRCUIT THEORY(PHY301)

MARKS: 30Due Date: 20/02/2004

Q.1.

Determine the values of IC and IE for the circuit shown in figure below

Q.2.(a) For the transistor circuit in the figure, what is VCE when VIN =0V ?(b)What minimum value of IB is required to saturate this transistor if DC is

200?

(c) Calculate the maximum value of RB when VIN =5V.

Q.3.

Find IC and VEC in the following circuit.

G d L k

Documents

Circuit Theory - Solved Assignments - Semester Fall 2003