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Network Theory MCQs

Practice Network Theory MCQs from fundamental concepts through intermediate, advanced and application-oriented questions. Each page contains up to 100 questions with answers and explanations where available.

Question 201

Network TheoryNetwork Topology

A connected network graph has 7 nodes and 11 branches. How many independent cut-set equations can be formed from a tree?

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Correct Answer: B. 6

Explanation: A tree of a connected graph has n-1 twigs, and each twig defines one fundamental cut-set. Thus 7-1=6.

Reference: Standard network theory: fundamental cut-set count

Question 202

Network TheoryNetwork Topology

For an oriented graph, each column of the node-branch incidence matrix corresponding to an ordinary branch contains?

AOne +1 and one -1
BTwo +1 entries only
COne nonzero entry only
DAll zero entries

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Correct Answer: A. One +1 and one -1

Explanation: An oriented branch leaves one node and enters another, so its incidence-matrix column has one +1 and one -1 under a consistent sign convention.

Reference: Standard network theory: incidence matrix structure

Question 203

Network TheoryNetwork Topology

If one row is deleted from the incidence matrix of a connected graph, the resulting reduced incidence matrix has how many rows for n nodes?

An
Bn-1
Cb-n+1
Db

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Correct Answer: B. n-1

Explanation: Deleting the dependent reference-node row leaves a reduced incidence matrix with n-1 rows.

Reference: Standard network theory: reduced incidence matrix

Question 204

Network TheoryNetwork Topology

A tie-set matrix is primarily used to express branch currents in terms of?

AIndependent loop currents
BNode voltages only
CBranch powers
DCut-set voltages only

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Correct Answer: A. Independent loop currents

Explanation: Tie-set or fundamental-loop analysis relates branch currents to a set of independent loop currents.

Reference: Standard network theory: tie-set matrix

Question 205

Network TheoryNetwork Topology

A cut-set matrix is particularly associated with writing independent forms of?

AKCL equations
BKVL equations only
CPower equations only
DTransformer equations

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Correct Answer: A. KCL equations

Explanation: Cut-set equations express current conservation across cuts of a graph and therefore provide independent KCL relations.

Reference: Standard network theory: cut-set matrix

Question 206

Network TheorySource Transformation

A network has a 12 V ideal voltage source in series with 3 ohm. Its equivalent Norton current is?

A2 A
B3 A
C4 A
D36 A

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Correct Answer: C. 4 A

Explanation: Source transformation gives IN=V/R=12/3=4 A, with the same 3 ohm resistance in parallel.

Reference: Standard network theory: source transformation numerical

Question 207

Network TheorySource Transformation

A Norton source of 5 A in parallel with 8 ohm is transformed to a Thevenin source of?

A5 V in series with 8 ohm
B13 V in series with 8 ohm
C40 V in series with 8 ohm
D40 V in series with 1.6 ohm

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Correct Answer: C. 40 V in series with 8 ohm

Explanation: Vth=IN RN=5×8=40 V and Rth=RN=8 ohm.

Reference: Standard network theory: Norton to Thevenin numerical

Question 208

Network TheoryNetwork Theorems

For a linear circuit containing several independent sources, superposition is applied to calculate?

AVoltages and currents, not power directly
BPower directly by adding individual source powers at an element
CEquivalent resistance only
DFrequency only

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Correct Answer: A. Voltages and currents, not power directly

Explanation: Superposition applies to linear response quantities such as voltage and current. Power is nonlinear in voltage/current and cannot generally be superposed directly.

Reference: Standard network theory: superposition limitation

Question 209

Network TheoryNetwork Theorems

While applying superposition, an independent ideal voltage source that is not being considered is replaced by?

AAn open circuit
BA short circuit
CIts internal power rating
DA dependent source

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Correct Answer: B. A short circuit

Explanation: Setting an ideal independent voltage source to zero volts makes it a short circuit.

Reference: Standard network theory: superposition voltage-source suppression

Question 210

Network TheoryCircuit Analysis

A 20 V source is connected to two series resistors of 4 ohm and 6 ohm. The voltage across the 6 ohm resistor is?

A8 V
B10 V
C12 V
D15 V

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Correct Answer: C. 12 V

Explanation: By voltage division, V6=20×6/(4+6)=12 V.

Reference: Standard network theory: voltage divider numerical

Question 211

Network TheoryCircuit Analysis

A total current of 10 A divides between parallel resistors 6 ohm and 3 ohm. The current through the 6 ohm branch is?

A2 A
B3.33 A
C5 A
D6.67 A

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Correct Answer: B. 3.33 A

Explanation: For two parallel resistors, current through 6 ohm is 10×3/(6+3)=3.33 A.

Reference: Standard network theory: current divider numerical

Question 212

Network TheoryCircuit Analysis

Three resistors 2 ohm, 3 ohm and 6 ohm are connected in parallel. Their equivalent resistance is?

A1 ohm
B2 ohm
C3 ohm
D11 ohm

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Correct Answer: A. 1 ohm

Explanation: 1/Req=1/2+1/3+1/6=1, hence Req=1 ohm.

Reference: Standard network theory: parallel resistance numerical

Question 213

Network TheoryCircuit Analysis

A 4 ohm resistor dissipates 36 W. The magnitude of current through it is?

A1.5 A
B3 A
C6 A
D9 A

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Correct Answer: B. 3 A

Explanation: Using P=I^2R, I=sqrt(36/4)=3 A.

Reference: Standard network theory: resistor power numerical

Question 214

Network TheoryAC Circuit Analysis

A sinusoidal voltage has RMS value 230 V. Its peak value is approximately?

A162.6 V
B230 V
C325.3 V
D460 V

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Correct Answer: C. 325.3 V

Explanation: For a sinusoid, Vpeak=sqrt(2)Vrms≈1.414×230=325.3 V.

Reference: Standard network theory: RMS to peak numerical

Question 215

Network TheoryAC Circuit Analysis

A sinusoidal current has peak value 10 A. Its RMS value is approximately?

A5 A
B7.07 A
C10 A
D14.14 A

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Correct Answer: B. 7.07 A

Explanation: For a sine wave, Irms=Ipeak/sqrt(2)=10/1.414≈7.07 A.

Reference: Standard network theory: peak to RMS numerical

Question 216

Network TheoryAC Circuit Analysis

The average value of a pure sinusoidal voltage over one complete cycle is?

AIts RMS value
BIts peak value
CZero
DPeak divided by two

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Correct Answer: C. Zero

Explanation: Positive and negative half-cycle areas cancel, so the average over a full cycle is zero.

Reference: Standard network theory: sinusoidal average value

Question 217

Network TheoryAC Circuit Analysis

For an impedance Z=3-j4 ohm, the magnitude of Z is?

A1 ohm
B5 ohm
C7 ohm
D25 ohm

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Correct Answer: B. 5 ohm

Explanation: |Z|=sqrt(3^2+(-4)^2)=5 ohm.

Reference: Standard network theory: complex impedance numerical

Question 218

Network TheoryAC Circuit Analysis

For Z=3-j4 ohm, the impedance angle is closest to?

A-53.1 degrees
B+53.1 degrees
C-36.9 degrees
D+36.9 degrees

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Correct Answer: A. -53.1 degrees

Explanation: The angle is tan^-1(-4/3)≈-53.1 degrees, indicating capacitive behavior.

Reference: Standard network theory: impedance phase numerical

Question 219

Network TheoryAC Circuit Analysis

An admittance is Y=0.04+j0.03 S. Its magnitude is?

A0.01 S
B0.05 S
C0.07 S
D0.12 S

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Correct Answer: B. 0.05 S

Explanation: |Y|=sqrt(0.04^2+0.03^2)=0.05 S.

Reference: Standard network theory: admittance magnitude numerical

Question 220

Network TheoryAC Power

A load draws 10 A RMS from a 200 V RMS supply at power factor 0.6 lagging. Its real power is?

A600 W
B1200 W
C1600 W
D2000 W

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Correct Answer: B. 1200 W

Explanation: P=VI cos(phi)=200×10×0.6=1200 W.

Reference: Standard network theory: single-phase real power numerical

Question 221

Network TheoryAC Power

For the same 200 V, 10 A, 0.6 lagging load, the reactive power magnitude is?

A800 var
B1200 var
C1600 var
D2000 var

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Correct Answer: C. 1600 var

Explanation: sin(phi)=sqrt(1-0.6^2)=0.8, so Q=VI sin(phi)=200×10×0.8=1600 var.

Reference: Standard network theory: single-phase reactive power numerical

Question 222

Network TheoryNetwork Theorems

While applying superposition, an independent ideal current source that is not being considered is replaced by?

AA short circuit
BAn open circuit
CA voltage source of equal value
DA resistor of 1 ohm

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Correct Answer: B. An open circuit

Explanation: Setting an ideal independent current source to zero amperes makes it an open circuit.

Reference: Standard network theory: superposition current-source suppression

Question 223

Network TheoryAC Power

A load consumes 3 kW and 4 kVAR. Its apparent power magnitude is?

A1 kVA
B5 kVA
C7 kVA
D12 kVA

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Correct Answer: B. 5 kVA

Explanation: |S|=sqrt(P^2+Q^2)=sqrt(3^2+4^2)=5 kVA.

Reference: Standard network theory: complex power triangle numerical

Question 224

Network TheoryThree-Phase Circuits

A balanced three-phase delta-connected load has a phase current of 8 A. The line current magnitude is approximately?

A4.62 A
B8 A
C13.86 A
D24 A

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Correct Answer: C. 13.86 A

Explanation: For a balanced delta connection, IL=sqrt(3)Iph≈1.732×8=13.86 A.

Reference: Standard network theory: delta line current numerical

Question 225

Network TheoryThree-Phase Circuits

A balanced three-phase system supplies 10 kW at 400 V line voltage and unity power factor. The line current is closest to?

A8.33 A
B14.43 A
C25 A
D43.3 A

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Correct Answer: B. 14.43 A

Explanation: P=sqrt(3)VLILcos(phi). At unity power factor, IL=10000/(1.732×400)≈14.43 A.

Reference: Standard network theory: three-phase current numerical

Question 226

Network TheoryThree-Phase Circuits

A balanced three-phase load takes 6 kW at a power factor of 0.6 lagging. Its reactive power is?

A3.6 kVAR
B4.8 kVAR
C6 kVAR
D8 kVAR

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Correct Answer: D. 8 kVAR

Explanation: For cos(phi)=0.6, sin(phi)=0.8 and tan(phi)=4/3. Hence Q=P tan(phi)=6×4/3=8 kVAR.

Reference: Standard network theory: three-phase reactive power

Question 227

Network TheoryThree-Phase Circuits

In the two-wattmeter method for a balanced three-phase load, total real power is equal to?

AW1-W2
BW1+W2
Csqrt(3)(W1+W2)
D3(W1-W2)

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Correct Answer: B. W1+W2

Explanation: For a balanced three-phase load measured by the two-wattmeter method, total real power is the algebraic sum W1+W2.

Reference: Standard network theory: two-wattmeter method

Question 228

Network TheoryThree-Phase Circuits

In the two-wattmeter method, if the balanced load has unity power factor, the two wattmeters ideally read?

AEqual positive values
BEqual and opposite values
COne zero and one positive value
DBoth zero

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Correct Answer: A. Equal positive values

Explanation: At unity power factor phi=0, the readings are proportional to cos(30°-phi) and cos(30°+phi), so they are equal and positive.

Reference: Standard network theory: two-wattmeter unity power factor

Question 229

Network TheoryThree-Phase Circuits

In the two-wattmeter method for a balanced load, one wattmeter reads zero when the power factor is?

A1
B0.866
C0.5
D0

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Correct Answer: C. 0.5

Explanation: One reading becomes zero at phi=60°, for which cos(phi)=0.5.

Reference: Standard network theory: two-wattmeter power factor

Question 230

Network TheoryThree-Phase Circuits

A balanced three-phase load is measured by two wattmeters reading 4 kW and 2 kW. The total real power is?

A2 kW
B4 kW
C6 kW
D8 kW

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Correct Answer: C. 6 kW

Explanation: Total three-phase real power in the two-wattmeter method is W1+W2=4+2=6 kW.

Reference: Standard network theory: two-wattmeter numerical

Question 231

Network TheoryThree-Phase Circuits

For a balanced three-phase delta-connected load, the phase voltage is related to line voltage by?

AVph=VL
BVph=VL/sqrt(3)
CVph=sqrt(3)VL
DVph=3VL

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Correct Answer: A. Vph=VL

Explanation: In a balanced delta connection, each phase is connected directly across two lines, so phase voltage equals line voltage.

Reference: Standard network theory: delta connection

Question 232

Network TheoryThree-Phase Circuits

A balanced three-phase star load has line current 12 A. The phase current is?

A4 A
B6.93 A
C12 A
D20.78 A

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Correct Answer: C. 12 A

Explanation: In a star connection, line current equals phase current, so Iph=IL=12 A.

Reference: Standard network theory: star line and phase current

Question 233

Network TheoryTransient Analysis

For a stable first-order circuit excited by a constant DC source, the natural-response component as t approaches infinity?

AGrows without bound
BApproaches zero
CBecomes equal to the source voltage in every circuit
DOscillates indefinitely

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Correct Answer: B. Approaches zero

Explanation: A stable first-order natural term has the form Ke^(-t/tau), which decays to zero as time increases.

Reference: Standard network theory: natural response decay

Question 234

Network TheoryTransient Analysis

A first-order circuit has time constant 50 ms. After 150 ms, its natural-response magnitude is approximately what percentage of its initial value?

A36.8%
B13.5%
C5.0%
D0.67%

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Correct Answer: C. 5.0%

Explanation: 150 ms equals 3 time constants. e^-3≈0.0498, so about 5% remains.

Reference: Standard network theory: exponential decay numerical

Question 235

Network TheorySecond-Order Circuits

For a series RLC circuit, the condition R^2<4L/C corresponds to which natural-response type?

AOverdamped
BCritically damped
CUnderdamped
DFirst-order

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Correct Answer: C. Underdamped

Explanation: For series RLC, alpha=R/(2L) and omega0=1/sqrt(LC). The inequality R^2<4L/C is equivalent to alpha<omega0, giving an underdamped response.

Reference: Standard network theory: RLC damping condition

Question 236

Network TheorySecond-Order Circuits

For a series RLC circuit, critical damping occurs at resistance R equal to?

Asqrt(L/C)
B2sqrt(L/C)
Csqrt(C/L)
D2sqrt(C/L)

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Correct Answer: B. 2sqrt(L/C)

Explanation: Critical damping requires R/(2L)=1/sqrt(LC), which gives R=2sqrt(L/C).

Reference: Standard network theory: critical resistance

Question 237

Network TheoryAC Power

For a sinusoidal steady-state circuit, complex power S is conventionally written as?

AP-jQ
BP+jQ
CQ+jP
DP/Q

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Correct Answer: B. P+jQ

Explanation: Complex power is S=P+jQ, where P is average real power and Q is reactive power.

Reference: Standard network theory: complex power

Question 238

Network TheoryAC Power

A load has complex power S=3+j4 kVA. Its power factor is?

A0.4 lagging
B0.6 lagging
C0.8 lagging
D1.0

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Correct Answer: B. 0.6 lagging

Explanation: |S|=5 kVA and power factor=P/|S|=3/5=0.6. Positive Q denotes an inductive, lagging load.

Reference: Standard network theory: complex power factor numerical

Question 239

Network TheoryAC Power

A load has complex power S=4-j3 kVA. Its power factor characteristic is?

A0.8 leading
B0.8 lagging
C0.6 leading
D0.6 lagging

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Correct Answer: A. 0.8 leading

Explanation: |S|=5 kVA, so pf=4/5=0.8. Negative reactive power indicates capacitive behavior, hence leading power factor.

Reference: Standard network theory: leading power factor

Question 240

Network TheoryLaplace Transform

The initial-value theorem for a function f(t) with suitable transform F(s) gives f(0+) as?

Alim s->0 of sF(s)
Blim s->infinity of sF(s)
Clim s->0 of F(s)/s
Dlim s->infinity of F(s)/s

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Correct Answer: B. lim s->infinity of sF(s)

Explanation: Under the theorem's applicability conditions, f(0+)=lim as s approaches infinity of sF(s).

Reference: Standard network theory: initial-value theorem

Question 241

Network TheoryLaplace Transform

If F(s)=5/[s(s+5)], the initial value f(0+) obtained using the initial-value theorem is?

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Correct Answer: A. 0

Explanation: f(0+)=lim s->infinity sF(s)=lim 5/(s+5)=0.

Reference: Standard network theory: initial-value theorem numerical

Question 242

Network TheoryLaplace Transform

The final-value theorem gives f(infinity), when applicable, as?

Alim s->infinity sF(s)
Blim s->0 sF(s)
Clim s->0 F(s)/s
Dlim s->infinity F(s)

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Correct Answer: B. lim s->0 sF(s)

Explanation: If the required pole conditions are satisfied, f(infinity)=lim as s approaches zero of sF(s).

Reference: Standard network theory: final-value theorem

Question 243

Network TheoryLaplace Transform

The final-value theorem is not valid for a response whose sF(s) has?

AAll poles strictly in the left-half plane
BA pole in the right-half plane
CNo finite poles
DOnly negative real poles

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Correct Answer: B. A pole in the right-half plane

Explanation: A right-half-plane pole produces a growing mode, so the response does not approach a finite final value and the final-value theorem cannot be used in its standard form.

Reference: Standard network theory: final-value theorem applicability

Question 244

Network TheoryTransient Analysis

The natural response of a linear circuit is produced by?

AStored energy and initial conditions with independent sources set to zero
BIndependent sources only with all stored energy removed
COnly sinusoidal sources
DOnly dependent sources removed

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Correct Answer: A. Stored energy and initial conditions with independent sources set to zero

Explanation: The natural or zero-input response is due to energy initially stored in capacitors and inductors after independent sources are set to zero.

Reference: Standard network theory: natural response

Question 245

Network TheoryTransient Analysis

The zero-state response of a linear circuit is the response when?

AAll independent sources are zero but initial energy is retained
BInitial conditions are zero and external sources act
CBoth sources and initial conditions are zero
DOnly dependent sources are removed

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Correct Answer: B. Initial conditions are zero and external sources act

Explanation: Zero-state response is caused by external excitation with all independent initial energy-storage conditions set to zero.

Reference: Standard network theory: zero-state response

Question 246

Network TheoryTransient Analysis

The complete response of a linear circuit can be expressed as?

ANatural response plus forced response
BResistance plus reactance only
CVoltage response minus current response
DInitial value multiplied by final value

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Correct Answer: A. Natural response plus forced response

Explanation: A linear circuit's complete response is commonly decomposed into natural (transient) and forced components.

Reference: Standard network theory: complete response

Question 247

Network TheorySecond-Order Circuits

A series RLC circuit has L=1 H and C=0.25 F. The resistance required for critical damping is?

A1 ohm
B2 ohm
C4 ohm
D8 ohm

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Correct Answer: C. 4 ohm

Explanation: Rcrit=2sqrt(L/C)=2sqrt(1/0.25)=2×2=4 ohm.

Reference: Standard network theory: critical resistance numerical

Question 248

Network TheorySecond-Order Circuits

In an overdamped second-order circuit, the characteristic roots are typically?

ATwo distinct negative real roots for a stable passive circuit
BA repeated imaginary pair
CA complex-conjugate pair with zero real part
DOne root only at the origin

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Correct Answer: A. Two distinct negative real roots for a stable passive circuit

Explanation: Overdamping gives two distinct real roots; in a stable passive circuit both are negative and produce nonoscillatory decaying exponentials.

Reference: Standard network theory: overdamped roots

Question 249

Network TheorySecond-Order Circuits

In a critically damped second-order circuit, the natural response contains terms of the form?

A(A+Bt)e^(-alpha t)
BA sin(alpha t) only
CAe^(+alpha t) only
DA+Bt with no exponential

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Correct Answer: A. (A+Bt)e^(-alpha t)

Explanation: A repeated real characteristic root at -alpha produces the natural-response form (A+Bt)e^(-alpha t).

Reference: Standard network theory: critically damped response

Question 250

Network TheoryNetwork Duality

In network duality, the dual quantity corresponding to voltage is?

ACurrent
BResistance
CPower
DCharge only

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Correct Answer: A. Current

Explanation: In a dual network, voltage and current interchange roles, just as KVL and KCL interchange.

Reference: Standard network theory: duality

Question 251

Network TheoryNetwork Duality

In network duality, a series connection in the original network corresponds to what in the dual network?

AAnother series connection only
BA parallel connection
CAn open circuit only
DA dependent source only

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Correct Answer: B. A parallel connection

Explanation: Series and parallel connections are dual concepts.

Reference: Standard network theory: series-parallel duality

Question 252

Network TheoryNetwork Duality

The dual of resistance R is represented by which network quantity under the usual normalized duality correspondence?

AConductance G
BInductance L
CCapacitance C
DReactance only

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Correct Answer: A. Conductance G

Explanation: Resistance and conductance are dual quantities because voltage-current and series-parallel roles are interchanged.

Reference: Standard network theory: resistance-conductance duality

Question 253

Network TheoryNetwork Duality

Under network duality, an inductor is dual to a?

AResistor
BCapacitor
CVoltage source
DTransformer

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Correct Answer: B. Capacitor

Explanation: Inductance and capacitance are dual energy-storage elements.

Reference: Standard network theory: inductor-capacitor duality

Question 254

Network TheoryNetwork Duality

The dual of Kirchhoff’s voltage law is?

AOhm's law
BKirchhoff's current law
CFaraday's law
DMaximum power theorem

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Correct Answer: B. Kirchhoff's current law

Explanation: KVL constrains loop voltages, while its dual KCL constrains node currents.

Reference: Standard network theory: KVL-KCL duality

Question 255

Network TheoryNetwork Theorems

The substitution theorem permits a branch of a network to be replaced by another element or source provided the replacement maintains the same?

ABranch voltage and current
BTotal number of nodes only
COperating frequency only
DPower factor of every other branch only

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Correct Answer: A. Branch voltage and current

Explanation: At a specified operating condition, substitution is valid when the replacement preserves the original branch terminal voltage and current.

Reference: Standard network theory: substitution theorem

Question 256

Network TheoryNetwork Synthesis

Which polynomial is Hurwitz?

As^2-3s+2
Bs^2+3s+2
Cs^2-1
Ds^2-2s+5

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Correct Answer: B. s^2+3s+2

Explanation: s^2+3s+2=(s+1)(s+2), whose roots -1 and -2 lie in the left-half plane.

Reference: Standard network theory: Hurwitz polynomial numerical

Question 257

Network TheoryNetwork Synthesis

For a positive-real rational function, any poles on the j-omega axis must be?

AMultiple with complex residues
BSimple with real nonnegative residues
CAlways double poles
DLocated only in the right-half plane

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Correct Answer: B. Simple with real nonnegative residues

Explanation: A positive-real function may have simple imaginary-axis poles, and their residues must be real and nonnegative.

Reference: Standard network theory: positive-real pole condition

Question 258

Network TheoryNetwork Synthesis

For a positive-real function F(s), complex poles and zeros must occur in?

AUnrelated locations
BComplex-conjugate pairs
COnly the right-half plane
DOnly at s=0

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Correct Answer: B. Complex-conjugate pairs

Explanation: Because a positive-real network function is real for real s, nonreal poles and zeros occur in complex-conjugate pairs.

Reference: Standard network theory: conjugate roots

Question 259

Network TheoryNetwork Synthesis

An RC driving-point impedance has poles and zeros on the negative real axis that characteristically?

AInterlace
BAlways coincide
CAppear only as complex pairs
DLie only on the positive real axis

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Correct Answer: A. Interlace

Explanation: For an RC driving-point impedance, realizability imposes alternating or interlacing negative-real poles and zeros with the appropriate ordering.

Reference: Standard network theory: RC impedance synthesis

Question 260

Network TheoryNetwork Synthesis

For an RL driving-point impedance, poles and zeros of a realizable simple ladder form are characteristically located on the?

APositive real axis
BNegative real axis
Cj-omega axis only
DUnit circle

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Correct Answer: B. Negative real axis

Explanation: RL driving-point functions have real nonpositive critical frequencies; their finite poles and zeros interlace on the negative real axis.

Reference: Standard network theory: RL impedance synthesis

Question 261

Network TheoryNetwork Synthesis

A Foster realization is especially associated with expansion of a driving-point function into?

APartial fractions
BA Fourier series only
CA determinant of the incidence matrix
DA power series in time

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Correct Answer: A. Partial fractions

Explanation: Foster synthesis decomposes an immittance into partial-fraction terms that map to simple circuit sections.

Reference: Standard network theory: Foster synthesis principle

Question 262

Network TheoryNetwork Synthesis

In Cauer-I synthesis of a driving-point impedance, continued-fraction extraction generally begins with behavior near?

As=infinity
Bs=1 only
Ct=0 in the time domain
DThe load current zero crossing

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Correct Answer: A. s=infinity

Explanation: Cauer-I realization is obtained by continued-fraction expansion about infinity, successively extracting ladder elements.

Reference: Standard network theory: Cauer-I synthesis

Question 263

Network TheoryNetwork Synthesis

In Cauer-II synthesis, the continued-fraction expansion is associated with behavior near?

As=0
Bs=infinity only
Cs=j only
Dt=infinity directly

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Correct Answer: A. s=0

Explanation: Cauer-II form is obtained through continued-fraction expansion about the origin.

Reference: Standard network theory: Cauer-II synthesis

Question 264

Network TheoryNetwork Synthesis

For a pure reactance one-port made only of ideal L and C elements, the real part of its impedance on the j-omega axis away from poles is?

APositive and frequency dependent
BNegative and frequency dependent
CZero
DAlways equal to 1 ohm

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Correct Answer: C. Zero

Explanation: An ideal lossless LC network absorbs zero average power, so its impedance is purely imaginary away from singularities.

Reference: Standard network theory: reactance function

Question 265

Network TheoryNetwork Synthesis

If Z(s)=sL is the driving-point impedance of an ideal inductor, Z(0) equals?

A0
BL
C1/L
DInfinity

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Correct Answer: A. 0

Explanation: At s=0, Z(s)=sL becomes zero, consistent with an ideal inductor behaving as a short circuit at DC steady state.

Reference: Standard network theory: inductor driving-point impedance

Question 266

Network TheoryNetwork Synthesis

If Z(s)=1/(sC) is the driving-point impedance of an ideal capacitor, the magnitude of Z(s) as s approaches zero tends to?

AZero
BC
C1/C
DInfinity

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Correct Answer: D. Infinity

Explanation: As s approaches zero, 1/(sC) becomes unbounded, consistent with a capacitor acting as an open circuit at DC steady state.

Reference: Standard network theory: capacitor driving-point impedance

Question 267

Network TheoryNetwork Functions

For a passive one-port, the driving-point admittance Y(s) is the reciprocal of?

ADriving-point impedance Z(s)
BTransfer impedance only
CLoop-current matrix
DIncidence determinant only

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Correct Answer: A. Driving-point impedance Z(s)

Explanation: At the same port, Y(s)=I(s)/V(s)=1/Z(s), provided the reciprocal is defined.

Reference: Standard network theory: driving-point immittance

Question 268

Network TheoryNetwork Theorems

The compensation theorem is useful for determining changes in network currents and voltages caused by a change in?

AA branch impedance
BThe number of graph nodes only
CSource frequency with every impedance fixed
DReference-node labeling

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Correct Answer: A. A branch impedance

Explanation: The compensation theorem evaluates incremental response when a branch impedance changes in a linear bilateral network.

Reference: Standard network theory: compensation theorem

Question 269

Network TheoryNetwork Theorems

If a branch carrying current I has its impedance increased by Delta Z, the compensating source magnitude used in the compensation theorem is proportional to?

AI/Delta Z
BI Delta Z
CI squared divided by Delta Z
DDelta Z/I squared

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Correct Answer: B. I Delta Z

Explanation: The compensating voltage magnitude is I Delta Z, with polarity chosen according to the theorem's replacement convention.

Reference: Standard network theory: compensation theorem relation

Question 270

Network TheoryNetwork Synthesis

For a polynomial to be Hurwitz, all of its roots must lie in which region, allowing simple j-omega-axis roots only under the standard special conditions?

AClosed left-half plane
BOpen right-half plane
COnly positive real axis
DOnly at the origin

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Correct Answer: A. Closed left-half plane

Explanation: A Hurwitz polynomial has no roots in the open right-half plane; permitted imaginary-axis roots are simple under the standard network-synthesis definition.

Reference: Standard network theory: Hurwitz polynomial

Question 271

Network TheoryNetwork Functions

A zero of a driving-point impedance Z(s) becomes what in the corresponding admittance Y(s)=1/Z(s), assuming no cancellation?

AA zero
BA pole
CA constant
DA branch current source

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Correct Answer: B. A pole

Explanation: Taking the reciprocal interchanges poles and zeros, so a zero of Z becomes a pole of Y.

Reference: Standard network theory: impedance-admittance poles and zeros

Question 272

Network TheoryNetwork Functions

If Z(s)=(s+2)/(s+5), the corresponding admittance Y(s) is?

A(s+2)/(s+5)
B(s+5)/(s+2)
C1/[(s+2)(s+5)]
Ds+7

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Correct Answer: B. (s+5)/(s+2)

Explanation: Admittance is the reciprocal of impedance, hence Y(s)=(s+5)/(s+2).

Reference: Standard network theory: immittance conversion numerical

Question 273

Network TheoryNetwork Synthesis

A rational function whose numerator degree exceeds its denominator degree by more than one cannot directly represent the driving-point impedance of a finite passive RLC network because it violates the usual?

APositive-real realizability requirements
BKCL sign convention only
CNode numbering rule
DTransformer turns-ratio rule

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Correct Answer: A. Positive-real realizability requirements

Explanation: Finite passive RLC driving-point immittances are positive-real rational functions with constrained high-frequency behavior; an excessive degree difference is incompatible with those realizability conditions.

Reference: Standard network theory: realizability condition

Question 274

Network TheoryNetwork Synthesis

For a positive-real function F(s), the real part Re[F(j omega)] at frequencies where F is finite must be?

ANonnegative
BStrictly negative
CAlways zero
DPurely imaginary

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Correct Answer: A. Nonnegative

Explanation: A key boundary property of a positive-real function is Re[F(j omega)]>=0 wherever the function is finite.

Reference: Standard network theory: positive-real frequency-axis condition

Question 275

Network TheoryMatrix Network Analysis

For a network with node-voltage vector V and nodal admittance matrix Y, the standard matrix equation is?

AYV=I
BYI=V
CVY=I squared
DY+V=I

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Correct Answer: A. YV=I

Explanation: Nodal analysis in matrix form is YV=I, where I is the vector of externally injected node currents.

Reference: Standard network theory: nodal admittance matrix

Question 276

Network TheoryMatrix Network Analysis

In a passive reciprocal resistive network, the nodal admittance matrix is typically?

ASkew-symmetric
BSymmetric
CAlways diagonal
DSingular after choosing a reference node in every case

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Correct Answer: B. Symmetric

Explanation: For a reciprocal resistive network, mutual admittance terms satisfy symmetry, so the reduced nodal admittance matrix is symmetric.

Reference: Standard network theory: reciprocal nodal matrix

Question 277

Network TheoryMatrix Network Analysis

In a nodal admittance matrix, a diagonal element Yii is generally formed from?

AThe negative sum of all conductances incident at node i
BThe sum of conductances connected to node i
COnly the largest branch resistance
DThe product of all branch impedances

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Correct Answer: B. The sum of conductances connected to node i

Explanation: The diagonal entry is the sum of admittances incident on the corresponding node under the standard nodal formulation.

Reference: Standard network theory: nodal matrix diagonal

Question 278

Network TheoryMatrix Network Analysis

For two distinct nodes i and j directly connected by admittance y, the off-diagonal nodal-matrix entries Yij and Yji are normally?

A+y
B+2y
C-y
D1/y

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Correct Answer: C. -y

Explanation: Under the standard nodal convention, a branch admittance between two non-reference nodes contributes -y to the corresponding off-diagonal entries.

Reference: Standard network theory: nodal matrix off-diagonal

Question 279

Network TheoryMatrix Network Analysis

A two-node reduced nodal matrix is [[5,-2],[-2,4]] siemens. If the injected-current vector is [6,2] A, which equations are correct?

A5V1-2V2=6 and -2V1+4V2=2
B5V1+2V2=6 and 2V1+4V2=2
C5V1-2V2=2 and -2V1+4V2=6
DV1+V2=8 only

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Correct Answer: A. 5V1-2V2=6 and -2V1+4V2=2

Explanation: Direct multiplication of YV=I gives 5V1-2V2=6 and -2V1+4V2=2.

Reference: Standard network theory: nodal matrix numerical

Question 280

Network TheoryMatrix Network Analysis

For the equations 5V1-2V2=6 and -2V1+4V2=2, the node voltage V1 is?

A1.00 V
B1.75 V
C2.25 V
D3.50 V

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Correct Answer: B. 1.75 V

Explanation: Solving simultaneously: the second equation gives V1=2V2-1. Substitution yields V2=1.375 V, hence V1=1.75 V.

Reference: Standard network theory: nodal matrix solution numerical

Question 281

Network TheoryTwo-Port Networks

A reciprocal two-port has z12=3 ohm. What must z21 be?

A-3 ohm
B0 ohm
C3 ohm
D1/3 siemens

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Correct Answer: C. 3 ohm

Explanation: Reciprocity in z-parameters requires z12=z21, hence z21=3 ohm.

Reference: Standard network theory: z-parameter reciprocity numerical

Question 282

Network TheoryTwo-Port Networks

A symmetrical two-port has z11=8 ohm. What must z22 be?

A0 ohm
B4 ohm
C8 ohm
D16 ohm

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Correct Answer: C. 8 ohm

Explanation: For a symmetrical two-port represented by z-parameters, z11=z22.

Reference: Standard network theory: z-parameter symmetry numerical

Question 283

Network TheoryTwo-Port Networks

For a two-port with z11=6 ohm, z12=z21=2 ohm and z22=5 ohm, the determinant of the z-parameter matrix is?

A14 ohm squared
B26 ohm squared
C30 ohm squared
D34 ohm squared

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Correct Answer: B. 26 ohm squared

Explanation: The determinant is z11z22-z12z21=6×5-2×2=26 ohm squared.

Reference: Standard network theory: z-matrix determinant numerical

Question 284

Network TheoryTwo-Port Networks

If a two-port z-parameter matrix has determinant Delta_z, the y-parameter matrix, when the inverse exists, is obtained by?

AMultiplying every z-entry by Delta_z
BMatrix inversion of Z
CAdding the four z-parameters
DTransposing Z only

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Correct Answer: B. Matrix inversion of Z

Explanation: The admittance matrix is Y=Z^{-1}, provided det(Z) is nonzero.

Reference: Standard network theory: z-to-y conversion

Question 285

Network TheoryTwo-Port Networks

For Z=[[4,2],[2,3]] ohm, the element y11 of Y=Z^{-1} is?

A0.125 S
B0.25 S
C0.375 S
D0.50 S

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Correct Answer: C. 0.375 S

Explanation: det(Z)=4×3-2×2=8. Z^{-1}=(1/8)[[3,-2],[-2,4]], so y11=3/8=0.375 S.

Reference: Standard network theory: z-to-y conversion numerical

Question 286

Network TheoryTwo-Port Networks

For the same Z=[[4,2],[2,3]] ohm, the transfer admittance y12 is?

A-0.25 S
B+0.25 S
C-0.50 S
D+0.50 S

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Correct Answer: A. -0.25 S

Explanation: From Z^{-1}=(1/8)[[3,-2],[-2,4]], y12=-2/8=-0.25 S.

Reference: Standard network theory: transfer admittance numerical

Question 287

Network TheoryNetwork Functions

A pole of multiplicity two at s=-a produces a time-domain natural-response term containing?

AOnly e^(-at)
Bt e^(-at) in addition to e^(-at)
Csin(at) only
DA constant only

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Correct Answer: B. t e^(-at) in addition to e^(-at)

Explanation: A repeated pole of order two generates terms proportional to e^(-at) and t e^(-at).

Reference: Standard network theory: repeated poles

Question 288

Network TheoryNetwork Functions

A complex-conjugate pole pair at s=-2 plus or minus j5 produces a natural response with an envelope proportional to?

Ae^(+2t)
Be^(-2t)
Ce^(-5t)
DA constant

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Correct Answer: B. e^(-2t)

Explanation: The real part of the poles determines the exponential envelope, so the oscillation decays as e^(-2t).

Reference: Standard network theory: complex poles

Question 289

Network TheoryNetwork Functions

For H(s)=K/[(s+2)(s+5)], what value of K gives a DC gain of 1?

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Correct Answer: C. 10

Explanation: H(0)=K/(2×5)=K/10. Setting H(0)=1 gives K=10.

Reference: Standard network theory: normalized DC gain numerical

Question 290

Network TheoryNetwork Functions

For H(s)=(s+4)/(s+8), the high-frequency gain approaches?

A0
B0.5
C1
D4

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Correct Answer: C. 1

Explanation: The numerator and denominator have equal degree and equal leading coefficients, so the limit as |s| tends to infinity is 1.

Reference: Standard network theory: high-frequency gain numerical

Question 291

Network TheoryNetwork Functions

A network function has a pole at the origin. Its corresponding natural mode includes which behavior?

AA nondecaying constant term
BOnly an exponentially decaying term
COnly a growing sinusoid
DNo time-domain contribution

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Correct Answer: A. A nondecaying constant term

Explanation: A simple pole at s=0 corresponds to a constant time-domain mode rather than an exponentially decaying mode.

Reference: Standard network theory: pole at origin

Question 292

Network TheoryNetwork Functions

For F(s)=1/[s(s+2)], the residue associated with the pole at s=0 is?

A0
B0.5
C1
D2

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Correct Answer: B. 0.5

Explanation: For a simple pole at zero, the residue is lim(s->0) sF(s)=1/(s+2)|s=0=1/2.

Reference: Standard network theory: residue numerical

Question 293

Network TheoryNetwork Functions

For F(s)=1/[(s+2)(s+3)], the residue at s=-2 is?

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Correct Answer: B. 1

Explanation: The residue at s=-2 is lim(s->-2)(s+2)F(s)=1/(s+3)|s=-2=1.

Reference: Standard network theory: residue at simple pole

Question 294

Network TheoryNetwork Functions

For F(s)=1/[(s+2)(s+3)], the residue at s=-3 is?

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Correct Answer: A. -1

Explanation: The residue at s=-3 is lim(s->-3)(s+3)F(s)=1/(s+2)|s=-3=-1.

Reference: Standard network theory: second residue numerical

Question 295

Network TheoryLaplace Transform

The inverse Laplace transform of 1/[(s+2)(s+3)] is?

Ae^(-2t)+e^(-3t)
Be^(-2t)-e^(-3t)
C2e^(-2t)-3e^(-3t)
De^(-5t)

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Correct Answer: B. e^(-2t)-e^(-3t)

Explanation: Partial fractions give 1/(s+2)-1/(s+3), hence f(t)=e^(-2t)-e^(-3t) for t>=0.

Reference: Standard network theory: partial fractions numerical

Question 296

Network TheoryNetwork Functions

For a stable causal rational network function, moving a pole closer to the imaginary axis generally makes its associated transient?

ADecay faster
BDecay more slowly
CDisappear instantly
DBecome independent of pole location

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Correct Answer: B. Decay more slowly

Explanation: A pole with a smaller-magnitude negative real part produces a slower exponential decay and a longer transient.

Reference: Standard network theory: pole location and transient

Question 297

Network TheoryNetwork Functions

For a stable pair of complex poles -sigma plus or minus j omega_d, increasing sigma while keeping omega_d fixed causes the oscillatory transient to?

ADecay faster
BDecay slower
COscillate at zero frequency
DGrow without bound

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Correct Answer: A. Decay faster

Explanation: The envelope is e^(-sigma t); increasing positive sigma increases the decay rate while omega_d fixes the oscillation frequency.

Reference: Standard network theory: damping from pole location

Question 298

Network TheoryNetwork Functions

A transfer function has poles at -1 and -10. Which pole normally dominates the long-time response if its mode is present?

APole at -10
BPole at -1
CBoth always decay at identical rates
DNeither pole affects settling

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Correct Answer: B. Pole at -1

Explanation: The pole closest to the imaginary axis, -1, has the slowest exponential decay and therefore dominates the long-time response.

Reference: Standard network theory: dominant pole

Question 299

Network TheoryNetwork Functions

For H(s)=10/[(s+1)(s+10)], the DC gain is?

A0.1
B1
C10
D100

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Correct Answer: B. 1

Explanation: H(0)=10/(1×10)=1.

Reference: Standard network theory: DC gain numerical

Question 300

Network TheoryNetwork Functions

A one-port network has driving-point impedance Z(s)=(s^2+4s+3)/(s^2+2s+2). At omega=1 rad/s, what is the real part of Z(j omega)?

A1 ohm
B2 ohm
C3 ohm
D4 ohm

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Correct Answer: B. 2 ohm

Explanation: At s=j, the numerator is (j)^2+4j+3=2+4j and the denominator is (j)^2+2j+2=1+2j. Thus Z(j)=(2+4j)/(1+2j)=2 ohm, so its real part is 2 ohm.

Reference: Standard network theory: driving-point impedance evaluation

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