RRB MCQs

Explanation: For a connected planar network, the number of independent meshes is b-n+1.

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.