JTO MCQs

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.

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

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.

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

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

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

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

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

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.

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

Explanation: Series and parallel connections are dual concepts.

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Explanation: H(0)=10/(1×10)=1.

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

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

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

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

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

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

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

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

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

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.

Explanation: At m=1, total sideband power is Pc/2 and total transmitted power is 1.5Pc, so efficiency=(0.5Pc)/(1.5Pc)=1/3≈33.3%.

Explanation: For single-tone AM, total sideband power=Pc m^2/2=400×0.36/2=72 W.

Explanation: Pt=Pc(1+m^2/2)=400(1+0.18)=472 W.

Explanation: m=(Vmax-Vmin)/(Vmax+Vmin)=(15-5)/(15+5)=10/20=0.5.

Explanation: A balanced modulator cancels the carrier while retaining the product terms that form the two sidebands.

Explanation: The phase-shift method uses quadrature message components and quadrature carrier components so one sideband cancels while the other adds.

Explanation: VSB retains one full sideband and a small portion of the other, reducing bandwidth while easing practical filtering.

Explanation: Video contains significant low-frequency content, making sharp SSB filtering difficult; VSB offers practical filtering with reduced bandwidth.

Explanation: FM modulation index beta=Delta f/fm. Doubling Delta f with fm fixed doubles beta.

Explanation: Carson's rule gives B≈2(Delta f+fm)=2(20+10)=60 kHz.

Explanation: Narrowband FM is characterized by a small modulation index, beta<<1, so higher-order sidebands are negligible.

Explanation: FM and PM have constant envelope in the ideal model, so total average signal power does not vary with modulation index when carrier amplitude is fixed.

Explanation: An FM discriminator produces an output voltage that follows frequency deviations, thereby recovering the message.

Explanation: When two FM signals compete on nearly the same channel, a sufficiently stronger signal can dominate the receiver output.

Explanation: A higher IF increases image separation from the desired RF signal, which can make image rejection easier for a given RF front-end selectivity.

Explanation: A lower IF permits narrower practical IF filters and can improve selectivity, although it worsens image-frequency separation.

Explanation: Noise factor F=10^(NF/10)=10^0.6≈3.98, approximately 4.

Explanation: Noise figure in dB is SNRin(dB)-SNRout(dB)=30-24=6 dB.

Explanation: Equivalent noise temperature is Te=T0(F-1), commonly using T0=290 K.

Explanation: Te=T0(F-1)=290(2-1)=290 K.

Explanation: In Friis' cascade formula, noise contributions of later stages are divided by preceding gains, making a low-noise high-gain first stage especially important.

Explanation: Friis' formula for two stages is F=F1+(F2-1)/G1=2+(5-1)/10=2.4.

Explanation: The available thermal-noise power is N=kTB, where k is Boltzmann's constant.

Explanation: N=kTB≈1.38×10^-23×290×10^6≈4.0×10^-15 W.

Explanation: From N=kTB, noise power is directly proportional to bandwidth, so doubling B doubles N.

Explanation: For M equally likely symbols, H=log2(M). Thus H=log2(8)=3 bits per symbol.

Explanation: H=-sum p log2 p. With two equiprobable outcomes, H=1 bit per symbol, the maximum for a binary source.