Circuit Theory - Solved Assignments - Semester Fall 2005

8/9/2019 Circuit Theory - Solved Assignments - Semester Fall 2005

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.Assignment 1(Fall 2005)

(Solution)CIRCUIT THEORY (PHY301)

MARKS: 35

Due Date: 27/11/2005

Q.1.For the circuit shown in the figure below, all the resistors are given in K Ohms;Find the total resistance RTin the following circuits. Draw the circuit diagram of each stepotherwise you will lose your marks.Draw the circuit diagram of each step otherwise you willlose your marks. Write each step of the calculation to get maximum marks and also mentionthe units of each derived value.

Sol.


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Q.2.In the network given below the power absorbed by the 9resistance is 144 W. Find VS also

find out the value ofIS , IB and IA.

Sol.

P=144W= IA2 x 9

So,IA= 4A ----- (A)

Where by ohms LawVX = 4 x 9 = 36 VIB = 36 /18IB = 2A ------ (B)

From (A) and (B) we haveIS = I A + I B = 6A

VS = 5 x 6 + VX= 30 + 36

VS = 66 V

Q.3.

(a) Determine the current I and its direction in the following circuit.

Draw the circuit diagramof each step otherwise you will lose your marks. Write each step of the calculation to getmaximum marks and also mention the units of each derived value.


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(b) What is the difference between Current division Rule and Voltage division Rule? Elaborateyour points by drawing complete circuit diagrams.

(c) Define Junction, Branch and Loop.

Sol.(a)


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(b)

Voltage divider Rule

The voltage divider rule states that the voltage across a resistor in a series circuit is equal to the value ofthat resistor times the total voltage across the series components divided by the total resistance of theseries components.

A series circuit can be viewed as a voltage divider, because the source voltage is divided among theresistors in the circuit.

In Figure we can easily find the voltage drop across each resistor in the circuit.

1) Find the total currentI = E/RT=12V/12 k = 1 mA

2) Find the IR drop across each resistorV = IR (1mA)(3 k ) = 3VV = IR (1mA)(4 k) = 4VV = IR (1mA)(5 k) = 5V

We can also see that Kirchhoffs Law is satisfied also since the sum of the voltage drops (3V+4V+5V=12V)equals the supply voltage of 12V.You can also see that there is a direct relationship between the value of the resistor and the voltage thatappears across it.

We can use this relationship to come up with the voltage divider rule. In general, for any number ofresistors, the voltage drop across any resistor may be found as:VX= (RX/ R) E

OR

Two resistors are connected as shown in the following diagram:

The output voltage Vout is related to Vin as follows:


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As a simple example, ifR1 = R2 then

Any ratio between 0 and 1 is possible.

Note that this rule only works if the divider is unloaded, i.e. the load resistance is infinite and all of thecurrent flowing through R1 goes into R2. If current flows into a load resistance (through Vout), thatresistance must be considered in parallel with R2to determine the voltage at Vout.

Current Divider Rule

For two parallel resistors the current through the branch equals the resistance of the opposite branchtimes the input current divided by the sum of the two resistors

In parallel networks, the voltage across all parallel elements is the same. However, the currents through

the various elements are typically different.

A parallel circuit can be viewed as a current divider, because the source current is divided among thebranches in the circuit. is used to determine how current entering the circuit is split between the variousparallel resistors connected to it.Current dividers are similar in construction to voltage dividers. A circuit diagram for a current divider isgiven below

(c)A junction: A junction (or node) is a connection point between two or more branches.

A circuit with 3 nodes


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A branch: A branch is a single electrical element or device (resistor, etc.).

A circuit with 5 branches

A loop is a closed path formed by starting at a node, passing through a set of nodes, andreturning to the starting node without passing through any node more than once.

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


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Assignment 2(Fall 2005)

(Solution)

CIRCUIT THEORY (PHY301)

MARKS: 30

Due Date: 05/12/2005

Q.1.Calculates the currents I1,I2, I3 and I4 .Draw the circuit diagram of each step otherwise youwill lose your marks. Write each step of the calculation to get maximum marks and alsomention the units of each derived value.

Sol.

KCL equation at Node V1will be,4 + 2 = V1/5 + V1/10

V1 = 20V

Now KCL equation at Node V2 will be,

5 - 2 = V2/10 + V2/5V2 = 10V

So,I1=V1/5 = 4A, I2=V1/10 = 2A

and

I3=V2/10 = 1A, I4=V2/5 = 2A


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Q.2.

Use nodal analysis to find Voltageat aandb in the given network. Identify and label each node

otherwise you will lose your marks. Write each step of the calculation to get maximum marks andalso mention the units of each derived value.

Sol.

Firs of all we will identify and label the nodes in the given network

KCL equation at Node Vawill be,

(Va - 10)/30 + Va/15 + (Va -Vb)/10 = 06Va - Vb = 10 ------------(A)

KCL equation at Node Vbwill be,(Vb - Va)/10 + (Vb 12)/20 + (Vb + 9)/5 = 0

2Va - 7Vb = 24 ------------(B)Solving (A) and (B) leads to

Va = -0.0556 V, Vb =-3.444V

Q.3.

First of all Identify and label each node and then by using nodal analysis find out Voltagesat the eachnode. Write each step of the calculation to get maximum marks and also mention the units ofeach derived value.


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Sol.

Firs of all we will identify and label the nodes in the given network

Sol.

Since 12 Volt voltage source is connected between node V2 and the reference node. It is evident thatV2 -0 = 12 V, V2 = 12 V. Since the voltage sources are connected between node V1 and V2 , and node V2and V3 , we obtain V2 - V1 =10, and V2 V3=20.

Thus,V1= 2 V , V2 = 12 V and V3 = -8V .

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


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(Solution)


MARKS: 40

Due Date: 14/12/2005

Q.1.Use nodal analysis to find Vand the power for the 15 V source (indicate whether the power is absorbed orsupplied) in the given network. Identify and label each node otherwise you will lose your marks. Draw thecomplete circuit diagram. Write each step of the calculation to get maximum marks and also mention theunits of each derived value.

Sol.Fist of all we will identify and label each node,

From the above diagram we know that node V1 is directly connected with15Volt voltage source so the voltage at Node V1will be,

V1= 15Volts -------- (A)And node V2 = V -------- (B)Constraint equation

V2- V3=5V -------(C)

Put the value of V from (B) into (C) we have,V2-V3= 5V2So, V3=-4V2 ------- (D)Now we will write KCL equation at Super Node,

(V2 - V1)/10 + V2/2 + V3/20 + V3/40= 0 ----------- (E)We know V1=15volts put in (E) we have,

(V2 - 1 5)/10 + V2/2 + V3/20 + V3/40= 08V2+V3= 20 -------- (F)

Put the value ofV3 from (D) into (F) we have,


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8V2 - 4V2= 20V2= 5 v o lt s

Put the value of V 2 in (B) we haveV=5 volts

To find the power, we need to find currentForm the figure

Form the above figure I = I1 + I2 = V1/15 + (V 1-V2)/10

= 15/15 + (15-5) / 10= 2A

power for the 15 V source = - 15(2A) =-30W or 30W supplied

Q.2.Use Only Mesh analysis to find Current through each resistance and also find the voltage acrosseach resistance in the given network. Identify and label each mesh and also show each step ofcalculation otherwise you will lose your marks. Draw the complete circuit diagram and also mention theunits of each derived value.

Sol.Fist of all we will label the diagram,

Using Mesh analysis we will write the KVL equationsMesh I

i1= 6A ---------- (A)Mesh II

3 ( i2-i3) + 1 ( i 2-i1) = 04i2 - i1 - 3i3= 0

Put the value ofi1 in above equation we have4i2 - 3i3= 6 ---------- (B)

Mesh III


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2i3 + 4i3 + 3 ( i 3-i2) = 0-3i2 + 9i3=0 ---------- (C)

Solving equation (B) and (C) we havei3= 2/3A, i2= 2A , i1= 6A

But i1, i2 an di3 a r e mes h cu r r en ts no t e l emen t c u r r en ts I1 = i1-i2 = 4 A

I2 = i2-i3 = 4 / 3 A

I3 = i3 = 2 / 3 A

I4 = i3 = 2 / 3 A

V1= I1R = 4 V

V2= I2R = 4 V

V3= I3R = 4 / 3 V

V4= I4R = 8 / 3 V

Q.3.

Use Mesh analysis to find Current through each mesh in the given network also find the current Io.Identify and label each mesh and also show each step of calculation otherwise you will lose your marks.Draw the complete circuit diagram and also mention the units of each derived value.

Sol.Fist of all we will identify the meshes and label the diagram


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Constraint equationI 2 I 1 = 3 ------------ (A)

Mesh 1 and 2 form a Super Mesh or Loop,

KVL equation For the super Mesh,

5I 1+ 1(I 1 - I 3) + 4(I 2 - I 3) + 12 =06I 1+ 4I 2 - 5I 3+ 12 =0 ---------- (B)

For loop 3,

4(I 3 - I 2) + 1(I 3 - I 1) +2I 3+ 6 = 0

- I 1 4 I 2 + 7 I 3+ 6 = 0 ------------(C)

Solving (A)to (C) simultaneously we haveI1= -3.067, I 2 = -.067, I 3= -1.3333

Io = I 1 I 3= -1.7333A

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


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(Solution)


MARKS: 35

Due Date: 26/12/2005

Q.1.Use Mesh analysis to find Current through each mesh in the given network also find the current Io.Identify and label each mesh and also show each step of calculation otherwise you will lose your marks.Draw the complete circuit diagram and also mention the units of each derived value.


Using Mesh analysis we will write the KVL equationsThe mesh-current equations are:

Mesh II1 = 5mA --------- (I)

Mesh II5000 (I2 I1) + 10000(I2 - I3) + 20I2 = 0

-I1 + 7I2 -2I3 =0 ----------- (II)Constraint equation

I1 I 3 = 0.4IO ------------ (A)From the above figure we know that

I0=I 2-I 3 ------ (B)Put Io from (B) into (A) we have

I1 I3 = 0.4(I2-I3)


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I1 - 0.4I2- 0.6I3 = 0 -------- (III)Solving I, II and III simultaneously we have

I 1 = 5mA

I2 = 2.6mA

I3 = 6.6mA

I0 = I 2-I 3 = -4mA

Q.2.Find the

(a) Current through each mesh by using Cramers rule in the given network.(b ) Also find the total power developed in the circuit.(c) Check your answer by showing that the total power developed equals the total power dissipated.

Identify and label each mesh and also show each step of calculation otherwise you will lose your marks.Draw the complete circuit diagram and also mention the units of each derived value.


Using Mesh analysis we will write the KVL equationsThe mesh-current equations are:

Mesh I

-230 + 1(I1 -I2) + 2(I1 -I3) + 115 +4I1 = 0 --------- (I)Mesh II

6I2 + 3(I2 - I3) +1(I2 - I1) = 0 ----------- (II)Mesh III

460 + 5I3 -115 + 2(I3 - I1) + 3(I3 -I2) = 0 ---------- (III)Place these equations in standard form:

I1 (1+2+4) + I2 (-1) + I3 (-2) = 115


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I1 (-1) + I2 (6 + 3 +1) + I3 (-3) = 0I1 (-2) + I2 (-3) + I3 (5+2+3) = -345

Place in standard matrices form:

1

2

3

1 2 3

I7 1 2 115

1 10 3 I = 0

2 3 10 -345I

I 4.4A,I 10.6AandI 36.8A

= = =

a) The only components that can develop power in the circuit are the sources:

P230V= - (230V) (4.4A) = -1012W

P115V= - (115V) (-36.8A -4.4A) = 4738W

P460V = (460V)(-36.8A) = -16928W

Pdev = 1012W + 16928W = 17940WFrom part (a) we know that the 115V source is dissipating power; compute the power dissipated by theresistors.

P1= (1) (4.4A +10.6A)2 =225WP4 = (4) (4.4A )2 = 77.44WP6 = (6) (-10.6A )2 = 674.16W

P2 = (2) (4.4AA + 36.8A)2

= 3394.88WP3 = (3) (-10.6A + 36.8A )2 = 2059.32WP5 = (5) (-36.8A )2 = 6771.2W

Pdiss = 4738W +225W +77.44W+674.16W+ 3394.88W + 2059.32W + 6771.2W = 17940W

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


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Assignment 5(Fall 2005)CIRCUIT THEORY (PHY301)

MARKS: 50

Due Date: 13/02/2006

DONT MI SS THESE I m p o r t a n t i n st r u c t i o n s:

Labeled and draw each circuit diagram, other wise you will lose your marks.Write each step ofthe calculation to get maximum marks.

To solve this assignment, you should have good command over first 26 lectures. Upload assignments properly through LMS, (No Assignment w ill be accepted through

email). Write your ID on the top of your solution file. All students are directed to use the font and style of text as is used in this document. Dont use colorful back grounds in your solution files. Use Math Type or Equation Editor etc for mathematical symbols. This is not a group assignment, it is an individual assignment so be careful and avoid copying

others work. If some assignment is found to be copy of some other, both will be awarded zeromarks. It also suggests you to keep your assignment safe from others. No excuse will beaccepted by anyone if found to be copying or letting others copy.

Dont wait for the last date to submit your assignment.Q.1.

Apply Superposition to the circuit given below to find I3.Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram ofeach step and also mention the units of each derived value.


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Q.2.You are given the circuit shown below:

a) Use Source transformation to determine current IO.b) Use Superposition to determine current IO.c) Find the Thevenin equivalent circuit to the left of the terminals a-b and then find IO.

Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of

each step and also mention the units of each derived value.(Note: You must solve this problem with all techniques mention in the problem statement.)

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

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Circuit Theory - Solved Assignments - Semester Fall 2005