Network Theory MCQ – 52 previous year questions with solutions

The paper uses the signed impedance angle: Z = R − jX_C, so tan φ = −X_C/R. (As a pure magnitude, the current leads by tan⁻¹(X_C/R).)

In a parallel RLC circuit the source current is the phasor sum of I_R, I_L and I_C. The phase angle is the angle between the common voltage and that resultant current.

B — R + j0. A pure resistance has only a real part and zero imaginary (reactive) part; equivalently R∠0°.

An ideal inductor stores energy in its magnetic field and returns it to the source. Voltage and current are 90° apart, so the average real power over a complete cycle is zero.

The equivalent resistance is not simply R; it is obtained after adding the two admittances. Complex-conjugate multiplication gives |4+j3|²=25. Adding admittances cancels ±j3/25 exactly. Key relation: Yeq = 1/(4+j3)+1/(4−j3)=8/25=0.32 S; Zeq=1/Yeq=3.125 Ω; I=V/Zeq=32 A; PF=cos0°=1.

Exam focus: For parallel complex branches, convert Z to Y first; it prevents sign and denominator mistakes. Common trap: Directly adding parallel impedances or assuming the equivalent resistance equals the real part of each branch.

D — 1/Ztotal = 1/Z1 + 1/Z2. In a parallel circuit admittances add, so the reciprocal relation gives Ztotal = Z1Z2/(Z1 + Z2).

C — 4 A. The resistor and capacitor currents are 90° apart, so I² = IR² + IC² → IC = √(25 − 9) = 4 A.

C — The applied voltage. All branches of a parallel circuit are connected between the same two nodes, so they share the same voltage; branch currents differ.

D — Zero. In a pure capacitor current leads voltage by 90°, so P_avg = VI cos 90° = 0; energy is only stored and returned.

A — Zero. In a pure inductor the current lags the voltage by 90°, so power factor = cos 90° = 0 (zero lagging).

B — An inductive circuit. Net reactance X = X_L − X_C is positive, so current lags voltage.

For a coil Z = R + jωL. As ω → ∞, ωL dominates R, so the impedance becomes nearly purely inductive and the voltage leads current by 90°.

B — Z = R + j(X_L − X_C) Ω. In series, R, +jX_L and −jX_C add as complex quantities.

The phase difference identifies the element before any calculation: voltage is at +45° and current at −45°, so voltage leads current by 90°, which is the ideal-inductor relation. Then Xₗ = 628/10 = 62.8 Ω and L = Xₗ/ω = 62.8/314 = 0.2 H = 200 mH.

If a circuit response x depends linearly on independent sources u₁, u₂, …, then x(u₁+u₂+…) = x(u₁)+x(u₂)+…. Controlled/dependent sources stay in the circuit because their value is tied to another circuit variable; turning them off changes the network itself and destroys the correct controlling relationship.

Exam focus: When applying superposition, write “independent OFF, dependent ON” before starting the circuit reduction. Common trap: A very common error is to short/open a dependent source or to add individual source powers.

Superposition is a linear-circuit method used to find voltages or currents caused by multiple independent sources. Analyze one independent source at a time, replace other independent voltage sources by shorts and independent current sources by opens, and keep dependent sources active. Key relation: x = x₁ + x₂ + … for a linear response; P = V²/R = I²R is not a superposable quantity.

Exam focus: Deactivate only independent sources; never suppress a dependent source. Common trap: Adding powers from separate source cases is a classic trap.

The network is linear, so superposition can be used by analyzing the response to each sinusoidal source at its own frequency and then adding the time-domain currents. Because the two sources operate at different frequencies and the reactive impedances are frequency-dependent, a single phasor-domain source transformation cannot combine the complete problem without separate frequency analysis. Thus statement 1 is false and statement 2 is true.

Thevenin's theorem replaces a linear two-terminal network by one voltage source Vth in series with an equivalent impedance. For a network with independent sources, Rth/Zth is found by suppressing independent sources; dependent sources remain and may require a test-source method. Key relation: V_th=V_OC; Z_th seen at the port with independent sources suppressed.

Exam focus: In AC, use impedances—not just resistances. Common trap: Dependent sources are not turned off when finding Thevenin resistance.

B — Thevenin's theorem: V_th (open-circuit voltage) in series with R_th. Norton's theorem uses a current source in parallel instead.

A — Only linear, bilateral AC and DC circuits. Thevenin's theorem replaces a linear two-terminal network by an equivalent voltage source with a series resistance/impedance.

In AC, ZL must equal the complex conjugate of source/Thevenin impedance. At the resistive maximum-power point, half of the source power is dissipated internally, so efficiency is 50%. Key relation: DC: R_L=R_th. AC: Z_L=Z_th*.

Exam focus: Maximum power and maximum efficiency are different design objectives. Common trap: Do not claim the maximum-power condition minimizes circuit losses.

Source transformation preserves the terminal voltage-current behavior of a linear source-resistance pair. A voltage source Vs in series with R is equivalent to a current source Is=Vs/R in parallel with the same R, and conversely Vs=IsR. Key relation: Vs = Is R; Is = Vs/R.

Exam focus: Keep the resistance value unchanged when converting the source form. Common trap: Do not place the transformed resistor in the wrong series/parallel position.

The controlling current is signed: i_2 = −2 A. Hence v_x = 15i_2 = −30 V. The two source voltages add algebraically in the marked polarity, so v = v_x + 10 = −20 V.

All four branches share the same two nodes, so one node-voltage equation solves the circuit. Let the top-node voltage be V. Then i₀ = V/2, the dependent source current is i₀/4 = V/8 and the 8 Ω current is V/8. KCL gives V = 8 V, hence P₈Ω = V²/8 = 8 W.

C — One less than the number of nodes. After choosing one reference node, the remaining (n − 1) node voltages are the unknowns.

C — Kirchhoff's Current Law. Nodal analysis writes KCL at each non-reference node with node voltages as unknowns.

With iAB = 5 A, first determine the equivalent resistance seen between A and B from the resistor network. The network reduction gives approximately 6.842 Ω, so VAB = iAB Req ≈ 5 × 6.842 = 34.21 V.

For a connected network, the number of independent meshes is L = b - n + 1. If 4 mesh-current equations are required, 4 = b - n + 1, giving b = n + 3.

A — Super mesh. When a current source is shared by two adjacent meshes, they are combined into a supermesh and the source gives the constraint equation.

For a series RLC circuit, XL=XC at resonance, so the impedance is purely resistive and minimum, current is maximum, and power factor is unity. In an ideal parallel resonant circuit the input impedance is high, not minimum. Key relation: ω₀=1/√(LC); f₀=1/(2π√LC); at resonance XL=XC.

Exam focus: Series resonance: Z minimum, I maximum. Parallel resonance: input Z maximum (ideal case). Common trap: Reactive voltages can be large even though net reactive voltage is zero.

For a high-Q series RLC circuit, bandwidth is Δf = R/(2πL). With R=0.1 Ω and L=15 mH, Δf≈1.061 Hz. Half-power frequencies lie approximately symmetrically around 50 Hz, giving about 49.47 Hz and 50.53 Hz; the nearest option is 49.57 Hz and 50.53 Hz.

Electrical resonance occurs when the net reactive effect cancels. For a series RLC circuit, XL=XC at resonance, so the impedance is purely resistive and minimum, current is maximum, and power factor is unity. Key relation: ω₀=1/√(LC); f₀=1/(2π√LC); at resonance XL=XC.

Exam focus: Series resonance: Z minimum, I maximum. Parallel resonance: input Z maximum (ideal case). Common trap: Reactive voltages can be large even though net reactive voltage is zero.

A — As a purely resistive circuit. At series resonance X_L = X_C and they cancel, so Z = R (minimum) and current is maximum and in phase with voltage.

For a resonant network, bandwidth is inversely proportional to quality factor: BW = f₀/Q. Therefore a wider bandwidth requires a lower Q.

B — Q = ωL/R, with ω evaluated at resonance (ω = ω₀). The quality factor equals the voltage across either reactive component divided by the supply voltage at series resonance.

A is the intended option if the stated phase current of 10 A and the 440 V line voltage are used: P = √3 × 440 × 17.32 × 0.8 ≈ 10.56 kW. However, the supplied voltage, impedance and current are not mutually consistent.

A balanced three-phase system has three equal-magnitude phase quantities displaced by 120°. For a balanced load, the instantaneous total three-phase power is constant and the phasor sum of phase currents is zero, so a neutral conductor carries no current in the ideal balanced case. Key relation: P=√3 V_L I_L cosφ.

Exam focus: Balanced load ⇒ I_R+I_Y+I_B=0. Common trap: Do not apply zero-neutral-current result to unbalanced or harmonic-rich loads.

In a delta connection each phase is connected directly between two line conductors, so the phase voltage is exactly the line-to-line voltage: VL = Vph.

B — The load keeps working but becomes unbalanced. A delta has alternative paths between line terminals, so with one phase open it runs as open-delta (V-V) with reduced, unbalanced capacity.

A — Interchanging any pair of lines. Swapping two phase connections reverses their cyclic order and the direction of the rotating magnetic field.

D — The line currents are not equal. In delta, phase voltage = line voltage (balanced supply), but unequal branch impedances make phase and hence line currents unequal.

A first-order RC charging circuit responds exponentially because the capacitor voltage cannot change instantaneously. The time constant τ=RC sets the speed; at t=τ the capacitor reaches about 63.2% of its final voltage. Key relation: v_C(t)=V(1−e^(−t/RC)); i(t)=(V/R)e^(−t/RC); τ=RC.

Exam focus: One time constant = 63.2%; five time constants ≈ settled. Common trap: Charging current decreases with time; it does not increase toward steady state.

In a first-order RL circuit driven by a DC step, inductor current cannot change instantaneously. The time constant depends on both inductance and resistance, so changing either alters the transient speed. Key relation: i(t)=V/R·(1−e^(−t/τ)); τ=L/R.

Exam focus: Initial inductor current is continuous; final DC inductor acts as a short circuit. Common trap: Do not write τ=L alone; resistance is part of the time constant.

For an RL switching response i(t)=I∞+(I0+−I∞)e⁻ᵗ/τ, the given expression identifies I∞=2.5 A, the initial current as 4 A, and τ=1/4 s. Applying the pre-switch and post-switch DC equivalents and τ=L/Req gives R1=3 Ω and R2=5 Ω.

For an RL circuit the time constant is τ=L/R. It increases when inductance increases and when resistance decreases, because larger L resists current change while larger R dissipates energy faster.

Second-order dynamics are governed by two independent energy-storage states and are described by natural frequency and damping ratio. An underdamped response oscillates with exponentially decaying amplitude, critical damping gives the fastest non-oscillatory response, and overdamping gives a slower non-oscillatory response. Key relation: Characteristic form: s²+2ζω_n s+ω_n²=0.

Exam focus: ζ1 overdamped. Common trap: Overdamped responses do not sustain oscillations.

Damping is determined from the homogeneous second-order equation, not from the forcing term. Comparing s² + s + 1 with s² + 2ζωₙs + ωₙ² gives ωₙ = 1 rad/s and ζ = 0.5. Since 0 < ζ < 1, the series RLC response is underdamped and its natural response decays while oscillating.

For a linear two-port, terminal voltages are expressed in terms of terminal currents. Open-circuiting a port forces its current to zero and isolates the required coefficient. Key relation: V1 = Z11 I1 + Z12 I2; V2 = Z21 I1 + Z22 I2. Z11=(V1/I1)|I2=0, Z12=(V1/I2)|I1=0.

Exam focus: Memorize the condition “Z → open circuit (I=0)” and “Y → short circuit (V=0)”. Common trap: Do not label Z12=Z21 as symmetry; that equality is the reciprocity condition.

Z-parameters express two-port terminal voltages as linear combinations of port currents. Reciprocity requires Z12=Z21; symmetry additionally requires Z11=Z22. Key relation: V1=Z11I1+Z12I2; V2=Z21I1+Z22I2.

Exam focus: Reciprocal: Z12=Z21. Symmetric: Z11=Z22. Common trap: Do not confuse reciprocity with symmetry.

GPSC revised the final answer to option C. For Z-parameters, Z12 = V1/I2 with I1 = 0. With port 1 open, no current flows through the 10 Ω left series arm, so the voltage at port 1 equals the junction voltage produced by I2 through the 5 Ω shunt branch. Hence Z12 = 5 Ω.

KVL: the algebraic sum of voltages around a closed loop is zero. Here +10 − 6 − 4 = 0 V.

KCL follows from conservation of electric charge. In the lumped-circuit model, charge does not accumulate indefinitely at an ideal node, so the algebraic sum of currents entering and leaving the node is zero.