Nt Uvty Solution

8/16/2019 Nt Uvty Solution

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Mangalam College of Engineering | NETWORK THEORY – Solution Key 1

Third Semester B-Tech. Degree Examination Nov 2013

THIRD SEMESTER B-TECH. DEGREE EXAMINATION NOVEMBER 2013

NETWORK THEORY (EC 010 303)

SOLUTION KEY

Part A

1. According to Source Transformation , a current source in series with a resistor isequivalent to a voltage source in parallel with the same resistor.

V s= I s R.

A source transformation is the process of replacing a voltage source v s in series witha resistor R by a current source i s in parallel with a resistor R, or vice versa.It is easy to show that they are indeed equivalent. If the sources are turned off, theequivalent resistance at terminals a-b in both circuits is R.Also, when terminals a-bare short circuited, the short-circuit current flowing from ato b is i sc =vs/R, the circuit on the left-hand side and i sc =is for the circuit on the righthand side. Thus, vs/R = i s in order for the two circuits to be equivalent. Hence,

source transformation requires that

1. A common goal in source transformation is to end up witheither all currentsources or all voltage sources in the circuit. This is especially true if it makes nodalor mesh analysis easier.2. Repeated source transformations can be used to simplify a circuit by allowingresistors and sources to eventually be combined.3. The resistor value does not change during a source transformation, but it is notthe same resistor. This means that currents or voltages associated with the originalresistor are irretrievably lost when we perform a source transformation

2. The property of self-inductance is a particular form of electromagnetic induction.


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Self inductance is defined as the induction of a voltage in a current-carrying wirewhen the current in the wire itself is changing. In the case of self-inductance, themagnetic field created by a changing current in the circuit itself induces a voltage inthe same circuit. Therefore, the voltage is self-induced.

Where:

VL = induced voltage in voltsN = number of turns in the coildø/dt = rate of change of magnetic flux

Mutual inductance is the ability of one inductor to induce a voltage across aneighboring inductor, measured in henrys (H)

Consider two coils with self-inductances L 1 and L 2 that are in close proximity witheach other. Coil 1 has N 1 turns, while coil 2 has N 2 turns. The magnetic fluxφ 1emanating from coil 1has two components: one component φ 11 links only coil 1,and another component φ 12 links both coils. Hence,


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M21 is known as the mutual inductance of coil 2 with respect to coil 1.

The degree to which M approaches its maximum value is described by thecoefficient of coupling, defined as

= Since ≤ , 0 ≤ k ≤ 1

3. Maximum Power Transfer Theorem states that, Maximum power is transferred tothe load when the load resistance equals the Thevenin resistance as seen from the

load (R L = RTh)

Power delivered to load is

Differentiating with respect to R L

Significance:If the load is sized such that its Thevenin resistance is equal to the load

resistance of the network to which it is connected, it will receive maximum power


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from that network. Any change to the load resistance will reduce the power deliveredto the load.

4. The transfer function provides a basis for determining important system responsecharacteristics without solving the complete differential equation. The transferfunction is a rational function in the complex variable s=σ+jω, that is

It is often convenient to factor the polynomials in the numerator and denominator,and to write the transfer function in terms of those factors:

where the numerator and denominator polynomials, N(s) and D(s), have realcoefficients defined by the system’s differential equation and K=b m /a n. The zi’s are

the roots of the equation and are defined to be the system zeros, and the

pi’s are the roots of the equation and are defined to be the system poles.The factors in the numerator and denominator are written so that when s=z i thenumerator N(s) = 0 and the transfer function vanishes, that is

and similarly when s=pi, the denominator polynomial D(s) = 0 and the value of thetransfer function becomes unbounded,

5. Frequency response is the quantitative measure of the output spectrum of asystem or device in response to a stimulus, and is used to characterize the dynamicsof the system. It is a measure of magnitude and phase of The output as a function offrequency, in comparison to the input.A transfer function (also known as the system function or network function and,when plotted as a graph, transfer curve ) is a mathematical representation, in termsof spatial or temporal frequency, of the relation between the input and output of alinear time-invariant system with zero initial conditions and zero-point equilibrium.


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Part B

6. The idea of superposition rests on the linearity property.The superposition principle states that the voltage across (or current through) anelement in a linear circuit is the algebraic sum of the voltages across (or currents

through) that element due to each independent source acting alone.The principle of superposition helps us to analyze a linear circuit with more thanone independent source by calculating the contribution of each independent sourceseparately. To apply the superposition principle, we must keep two things in mind:

1. Consider one independent source at a time while all other independentsources are turned off. i.e. Replace every voltage source by a short circuit,and every current source by an open circuit.

2. Dependent sources are left intact because they are controlled by circuitvariables.

Steps to Apply Superposition Principle:1. Turn off all independent sources except one source. Find the output(voltage or current) due to that active source using nodal or mesh analysis.

2. Repeat step 1 for each of the other independent sources.3. Find the total contribution by adding algebraically all the contributions

due to the independent sources.7. In electrical circuit theory, the zero state response (ZSR), also known as the forced

response is the behavior or response of a circuit with initial state of zero. The ZSRresults only from the external inputs or driving functions of the circuit and not fromthe initial state. The ZSR is also called the forced or driven response of the circuit.The ZIR results only from the initial state of the circuit and not from any externaldrive. The ZIR is also called the natural response , and the resonant frequencies ofthe ZIR are called the natural frequencies .The total response of the circuit is the superposition of the ZSR and the ZIR, or ZeroInput Response

8. (Question ,not expected from the prescribed syllabus)Reactive loads such as inductors and capacitors dissipate zero power, yet the

fact that they drop voltage and draw current gives the deceptive impression thatthey actually do dissipate power. This “phantom power” is called reactive power , and

it is measured in a unit called Volt-Amps-Reactive (VAR), rather than watts. Themathematical symbol for reactive power is Q. The actual amount of power beingused, or dissipated, in a circuit is called true power (Real power) , and it is measuredin watts (P). The combination of reactive power and true power is called apparent power , and it is the product of a circuit's voltage and current, without reference tophase angle. Apparent power is measured in the unit of Volt-Amps (VA) and issymbolized by S.


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True power, P = I 2R ; P = E2/R (Measured in watts)

Reactive power, Q = I 2X ; Q = E2/X (Measured in Volts-Amps-Reactive)

Apparent power, S = I 2Z ; S = E2/Z (Measured in Volt – Amps)

9.

Initial value theorem states that, if a function f(t) and its derivative f’(t) are Laplacetransformable, then

Proof:By differentiation property of Laplace transform

If we let s → ∞, the integrand in above equation vanishes due to the dampingexponential factor

Final value theorem states that, if a function f(t) and its derivative f’(t) are Laplacetransformable, then

Proof:By differentiation property of Laplace transform

s→ 0; then

10. The equations for Y-parameters isI1 = y11 V1 + y12 V2 (1)I2 = y21 V1 + y22 V2 (2)The equations for T-parameters isV1 = A V2 – B I2 (3)I1 = C V2 – D I2 (4)


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From 3

= − + (5)Substituting I 2 in 4

= − −1 + = − (6)

Comparing 6 and 1

y11 = D/B y 12 = -[AD-BC]/B y 21 =-1/B y 22 = A/B

Part C

11. Applying KCL to node 1

2 + −1 = 5 3v 1 – 2v 2 = 10 (1)Node 2 and node 3 combines to form a super node

−1 +2 + −103 +2 = 0 -6v 1 + 9v 2 + 6v 3 = 20 (2)

v2 – v3 = 20 (3)

v1 = 13.03

v2 = 14.5

v3 = -5.5

Current through 3Ω resistor is

I = (10-(-5.5))/3 = 5.16 A

12. To Thevenize the network, 14Ω is to be removes making V X = 0So 0.1VX also will be zero.So VTh = 10 VRTh = 5+8 = 13 ΩSo drawing the equivalent and connecting back 14Ω


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Current through 14Ω = 10/(13+14) = 0.294 A 13. For t0Applying KVL

12 () +() = 12 Homogeneous solution,ih(t) = K e -12t Particular solutionip(t) =ASubstituting in the equation12A + 0 = 12 ; A = 1Total solutioni(t) = i h(t) + i p(t)

= K e-12t + 1

Applying initial conditionsi(0) = K +1K = 0i(t) = 1 A

14. Applying KVL to mesh 1j4 I 1 + j3 I 2 –j15 (I 1-I2) = 10-j11 I 1 + j18 I 2 = 10 (1)Applying KVL to mesh 2-j15(I 2-I1) +j2 I 2 + j3 I 1 + 10 I 2 = 0j18 I 1 + (10-j13) I 2 = 0

= − 11 10 18 0 − 11 18 18 10− 13 = − 180181 − 110 = 0.441− 0.7

V10 = 10 I 2 = 4.41-j7.3 V 15. Power factor

cos θ = 0.707


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So the phase angle between voltage and current isθ = cos -1(0.707) = 45 0 Hence the current equation isi(t) = I m sin(50t+45 0)True power = V eff Ieff cos θ = 200 W

Ieff = 200/Veff cos θ = 200/[(100/ √ 2) X 0.707)] = 4 AIm = 4 X √ 2 = 5.66 AThe current equation isi(t) = 5.66 sin(50t+45 0)The impedance of the circuit is

= =(100/√ 2) ≺ 20(5.66/√ 2) ≺ −25 = 17.67 ≺ 45 = 12.5+ Since Z = R + jX L R = 12.5Ω and X L = 12.5 Ω XL = 12.5 = wLL = 12.5/50 = 250 mH

16. Given current through (3+j4) is zero. So V 3 = V4 So KCL at node 3 −4 + 3 + −3+ 4 = 0 Solving by substituting V 1 = 20 and V 3 = V4 V3 = 0.34 j2.6 VKCL at node 4

−5 + − 5+ −3+ 4 = 0 Solving by taking V 3 = V4 and substituting its valueV2 = -2.26-j2.94 V

17. (a)Initial value theorem

lim→ () = lim→∞ () f(t) = e -t (sin 3t + cos 5t)

lim→ () = 1

() = 3( +1) +9+ +1( +1) +9 lim→∞ () = 3/1+1+9+

1+11+1+9 = 1


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Final value theorem

lim→∞ () = lim→() lim→∞ () = 0

lim→() = 3( +1) +9+ ( +1)( +1) +9 = 0

(b)Z = [(4/s)||s] + 3 = +3 = +3 = 18. Finding each parameter

() = ()() = 25 + 2 =

() = ()() =

() = ()() = 15 + /2 = 19.

The z-parameter equations are given byV1 = z11 I1 + z12 I2 (1)V2 = z21 I1 + z22 I2 (2)The y-parameter equations are given byI1 = y11 V1 + y12 V2 (3)I2 = y21 V1 + y22 V2 (4)


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Given z 11 = 6, z 12 = 3, z 21 = 3, z22 = 4Δz = z11 z22 - z12 z21 =15y11 = 4/15 = 0.267 y12 = y21 = -3/15 = 0.2 y22 = 6/15 = 0.4 The ABCD parameters are given by


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A = 6/3 = 2 B = 15/3 = 5 C = 1/3 = 0.33 D = 4/3 = 1.33

20. Substituting s=jw

() =( +5) ( +2) Converting to general form

() =2.5(1+ 5 )

( )(1+ 2 )

Magnitude response

|C(w)| dB = 20 log(2.5) + 20 log 1+ 5 - 20 log(jw) – 20 log 1+ 2 Phase responseΦ(w) = tan -1(w/5) – 90 – tan -1(w/2)


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Nt Uvty Solution

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