Step 1: Analyzing the given pattern. If → denotes increasing order of intensity:

[Talk → Shout → Scream]

The analogous pattern is:

[Please → → Pander]

Step 2: Evaluating options.

Option (1) flatter: To praise someone to make them feel attractive or important. This matches the intensity increase.

Option (2) flutter: To move quickly and lightly, unrelated to the context.

Option (3) fritter: To waste time or resources, not suitable.

Option (4) frizzle: To make something crisp or burn lightly, unrelated.

Conclusion: Hence, the correct option is (1) flatter.

Step 1: Understanding the preferences.

Given:

• Two beds
• A study table
• An almirah

P:Wants to sit at the table near the window.Q:Wants a bed close to the ceiling fan.

Step 2: Evaluating the arrangements.

Option 1 satisfies:

• P has access to the table near the window.
• Q’s bed is closest to the ceiling fan.

Conclusion: Hence, the correct option is 1.

Step 1: Decimal to octal conversion using division by 8.

Convert Decimal to Binary:

1. Divide the number by 2.
2. Get the integer quotient for the next iteration.
3. Get the remainder for the binary digit.
4. Repeat the above steps until the quotient is equal to 0.

Given: Decimal Number = 108

108₁₀ ⇒ Binary Conversion Steps:

Thus, 108₁₀ = (1101100)₂.

Step 2: Convert Binary to Octal.

Group the binary digits into sets of three from right to left: (001 101 100)₂

Convert each group into octal:

• (001)₂ = 1
• (101)₂ = 5
• (100)₂ = 4

Thus, (1101100)₂ = (154)₈.

Verification:

1 × 8² + 5 × 8¹ + 4 × 8⁰ = 64 + 40 + 4 = 108.

Conclusion: Therefore, 154 is correct.

Step 1: Relating selling price and cost price.

Profit = S.P - C.P

Profit% = (S.P − C.P) / C.P

Profit%: 20% ⇒ S.P = C.P × 1.2

Step 2: Adjusting for tax.

Selling Price (with tax): 3186 = S.P × (118 / 100)

S.P = 3186 × (100 / 118) = 2700.

Step 3: Calculating cost price.

C.P = S.P / 1.2 = 2700 / 1.2 = 2250.

Conclusion: Hence, the correct option is 2250.

Step 1: From the given condition, x : y = (a + 1) : (a − 1), we write:

x / y = (a + 1) / (a − 1).

Step 2: Substitute this into the ratio (x² − y²) : (x² + y²):

(x² − y²) / (x² + y²) = ((x / y)² − 1) / ((x / y)² + 1).

Step 3: Replace x / y with (a+1) / (a−1):

((a+1)/(a−1))² − 1 / ((a+1)/(a−1))² + 1 = ((a+1)² − (a−1)²) / ((a+1)² + (a−1)²).

Step 4: Expand and simplify:

(a + 1)² − (a − 1)² = 4a, (a + 1)² + (a − 1)² = 2a² + 2.

Step 5: The required ratio is:

4a / (2a² + 2) = 2a / (a² + 1).

Conclusion: The correct option is 1 (2a : (a² + 1)).

Step 1: Analyze the context for each blank:

• (i) The row involves disagreement, hence "with".
• (ii) The shopkeeper’s control over the price suggests "over".
• (iii) A spatial location implies "in".
• (iv) A specific place requires "at".

Step 2: Substitute the correct prepositions into the sentence:

“Following a row with the shopkeeper over the price of a frying pan, the cook stood in a row to withdraw cash at the ATM booth.”

Conclusion: The correct option is 3 ((i) with, (ii) over, (iii) in, (iv) at).

Step 1: From the properties of the rhombus, diagonals intersect at 90°:

CD = AD = AF = FC = 5.

From provided question:

• CD = 5 cm
• AE = 12 cm
• BE = 10 cm
• ∠DAB = ∠DCB and ∠DAE = ∠DBC = 90° then ∠DBC = ∠DBA = ∠ABF = ∠FBC = 90°.

In △DAE:

(DE)² = (DA)² + (AE)²

DE = √((5)² + (12)²) = √169 = 13 cm.

DE = DB + BE ⇒ DB = DE − BE.

DB = 13 − 10 = 3 cm.

Conclusion: The length of BF is 3 cm. Correct option: 1.

Step 1: Calculate the increase in the number of Millennial car buyers from 2010 to 2015:

Increase in Millennials = 45000 − 20000 = 25000.

Step 2: Calculate the decrease in the number of Gen X car buyers from 2010 to 2015:

Decrease in Gen X = 40000 − 30000 = 10000.

Step 3: Compare the values:

The increase in the number of Millennials (25000) is greater than the decrease in the number of Gen X (10000).

Conclusion: The correct option is 3.

Step 1: Understanding the Mechanism

• Racks and Pinions: A rack is a linear gear with teeth, and a pinion is a circular gear. When the rack moves, it causes the connected pinion to rotate. The direction of rotation of the pinion is always opposite to the movement of the rack.

System Analysis:

1. Object A moves to the right (→): This movement causes Object P (the pinion connected to it) to rotate clockwise.
2. Object P rotates clockwise: As Object P is connected to a rack (Object B), its clockwise rotation drives Object B to move to the left (←).
3. Object B moves to the left (←): The leftward movement of Object B makes Object Q rotate in an anticlockwise direction.
4. Object Q rotates anticlockwise: Finally, the anticlockwise rotation of Object Q causes Object R to move to the right (→).

Conclusion: The correct option is 2 (Object B translates towards the left direction).

Step 1: Using the given data, the angles ∠CAB = ∠CBA = 87.8° and the base AB = 200 m.

Step 2: In △CAB, use the tangent function for small angles:

tan θ = cx / AB, where cx is the perpendicular from C.

Step 3: Substitute values:

tan(87.8°) = cx / 100.

Step 4: Solve for cx:

cx = 100 × tan(87.8°) = 2603 m.

Conclusion: The correct option is 1 (2603).

Step 1: A matrix has an inverse if and only if its determinant is non-zero.

Step 2: Calculate the determinants of the matrices:

• For option (A): Applying row transformations, the determinant is 0 ⇒ Singular matrix.
• For option (B): The second row is a multiple of the first row, so determinant = 0 ⇒ Singular matrix.
• For option (C): Calculate determinant:

Determinant = 1(16 − 4) − 4(0 − 2) + 8(0 − 4)

= 1(12) − 4(2) + 8(−4)

= 12 − 8 − 32 = −28 ≠ 0 (Non-singular matrix).

• For option (D): Applying row transformations, the determinant is 0 ⇒ Singular matrix.

Step 3: Only matrix (C) is non-singular, and thus has an inverse.

Conclusion: The correct option is 3.

Step 1: Identify junctions in the circuit where three or more elements meet.

Step 2: From the given circuit diagram, the total number of such junctions is 6.

Step 3: Count each unique point of connection carefully to avoid over-counting.

Conclusion: The correct option is 1.

Step 1: Apply nodal analysis at VA:

VA − 10 / α + VA / 1 = 10.

Simplify:

(1 + α)VA = 10α + 10

VA = 10.

Step 2: Current through the 10 Ω resistor:

I = (10 − VA) / α = 0 A.

Step 3: Power delivered:

P = 10 × I = 0 W.

Conclusion: Since no current flows, the power delivered is zero. The correct option is 1.

Step 1: Analyzing the circuit after closing the switch.

When the switch S is closed, the circuit becomes an R-L network. The time constant (τ) of an R-L circuit is given by:

τ = Leq / Req.

Step 2: Calculating Req (Thevenin resistance).

The resistances in the circuit are combined as:

Req = 1 Ω + 1 Ω = 2 Ω.

Step 3: Calculating Leq (equivalent inductance).

The inductors in series are added directly:

Leq = 1 + 1 + 1/2 = 5/2.

Step 4: Determining the time constant (τ).

τ = Leq / Req = (5/2) / 2 = 5/4 = 1.25 seconds.

Conclusion: The correct option is 1 (1.25 seconds).

Step 1: Understanding convolution with time-reversed signals.

(i) Time Scaling Property: The time scaling property states:

x(at) ∗ y(at) = (1 / |a|) x(t) ∗ y(t / a).

Substituting a = −1:

x(−t) ∗ y(−t) = x(t) ∗ y(t).

(ii) Commutative Property:

The commutative property is expressed as:

x(t) ∗ y(t) = y(t) ∗ x(t).

From equation (1), we have:

z(t) = y(t) ∗ x(t) = x(−t) ∗ y(−t).

Substitute t = −t:

z(−t) = y(−t) ∗ x(−t).

From the commutative property, we get:

z(−t) = x(−t) ∗ y(−t).

From equations (2) and (3):

z(t) = z(−t).

Step 2: Applying properties of convolution:

• Convolution of x(t) with its time-reversed version x(−t) results in a symmetric output.
• The symmetry implies z(t) is an even signal: z(t) = z(−t).

Conclusion: The correct option is 1 (It is an even signal).

Step 1: Signal analysis.

The given signal is:

x(t) = e^t[u(t − 1) − u(t − 10)].

This is a bounded signal for the time interval t ∈ [1, 10].

Step 2: Determining ROC.

For finite-duration signals, the ROC of the Laplace transform includes the entire s-plane.

Conclusion: The correct option is 1 (−∞ < Re(s) < ∞).

Step 1: Calculating synchronous speed.

The synchronous speed (Ns) is given by:

Ns = (120 × f) / P

Ns = (120 × 50) / 6 = 1000 rpm.

Step 2: Calculating slip.

Slip (s) is calculated as:

s = (Ns − Nr) / Ns

s = (1000 − 960) / 1000 = 0.04.

Step 3: Calculating efficiency.

Efficiency (η) is given by:

η = 1 − s

η = 1 − 0.04 = 0.96 or 96%.

Conclusion: The correct option is 3 (96%).

Step 1: Transformer voltage polarities.

In a transformer, the applied primary voltage (Vp), induced primary voltage (Ep), open circuit secondary voltage (Vs), and induced secondary voltage (Es) are key parameters in transformer operation.

In a single-phase, two-winding transformer:

• The polarities of the induced voltages (Ep and Es) are always opposite to their corresponding applied voltages (Vp and Vs) due to Lenz’s Law.
• When the top terminal of the primary winding (Vp) is negative, the top terminal of the secondary winding (Vs) will be positive, and vice versa.

Conclusion: The correct option is 2 (B).

Step 1: Understanding the power loss calculation.

The power loss (PL) in a branch is proportional to:

PL = I²R, where R is the resistance of the branch, and I is the current through it.

Step 2: Analyzing branch removal.

• Removing b1: Total power loss = 27r.
• Removing b2: Total power loss = 48r.
• Removing b3: Total power loss = 12r.
• Removing b4: Total power loss = 19r.

Step 3: Selecting the branch to minimize losses.

Removing b3 results in the minimum power loss (12r).

Conclusion: The correct option is 3 (b3).

Step 1: Analyzing the block diagram.

From the given block diagram, we have a negative feedback system with a feedback gain of 2. The summing point equation for the system is:

C(s) = G(s) [R(s) − 2C(s)].

Step 2: Rearranging the equation.

Rearrange the equation to isolate terms involving C(s):

C(s) + 2G(s)C(s) = G(s)R(s).

Factor out C(s):

C(s) [1 + 2G(s)] = G(s)R(s).

Step 3: Solving for C(s) / R(s).

Divide both sides of the equation by [1 + 2G(s)]:

C(s) / R(s) = G(s) / (1 + 2G(s)).

Step 4: Considering the negative feedback.

The system has **negative feedback**, introducing a negative sign. The correct transfer function becomes:

C(s) / R(s) = −G(s) / (1 − 2G(s)).

Conclusion: The correct option is 4 (−G(s) / (1 − 2G(s))).

Step 1: Formula for settling time.

The settling time (t_s) for 2% tolerance is given by:

t_s = 4 / (ξω_n),

where ξ = cos(θ) is the damping factor.

Step 2: Calculating settling time for both systems.

ω_n = 3 rad/sec, θ = 60° ⇒ ξ = cos(60°) = 0.5.

t_s = 4 / (0.5 × 3) = 2.67 seconds.

ω_n = 1 rad/sec, θ = 70° ⇒ ξ = cos(70°) ≈ 0.34.

t_s = 4 / (0.34 × 1) ≈ 11.76 seconds.

• For System 1:
• For System 2:

Step 3: Comparing settling times.

Since t_s for System 2 is greater, the correct statement is:

Conclusion: Settling time of System 2 is more than that of System 1. The correct option is 2.

Step 1: Finding the original transfer function.

The given cascaded system has:

T(s) = (s + 40) / [(s + 1)(s + 20)].

Step 2: DC gain before approximation.

For DC gain (s = 0):

T(s)_s=0 = 40 / (1 × 20) = 2.

Step 3: First-order approximation.

Approximating s + 20 as the dominant pole:

T(s) ≈ 2 / (s + 1).

Conclusion: The correct option is 2 (2 / (s + 1)).

Step 1: Identifying transformer features.

• For CT (X): Q (Open-circuited secondary not desirable) and R (Primary current is the line current) apply.
• For PT (Y): P (Primary is connected in parallel to the grid) and S (Secondary burden affects the primary current) apply.

Conclusion: The correct option is 3 (X matches with Q, R; Y matches with P, S).

Step 1: Original Boolean function.

F(P, Q, R, S) = PQR + P'QS + P'QRS.

Step 2: Grouping terms for simplification.

Group the terms involving P and P' separately:

F(P, Q, R, S) = P(QR + QS) + P'(QRS).

Step 3: Simplify QR + QS.

QR + QS = S + RQ.

The function becomes:

F(P, Q, R, S) = PS + QS.

Conclusion: The simplified Boolean function is P'S + QS, corresponding to option 1.

Step 1: Understanding the circuit.

• The circuit consists of a combinational logic circuit with inputs X, Y, and Z generating S using the logic: S = X ⊕ Y ⊕ Z, where ⊕ denotes the XOR operation.
• A D flip-flop where the input is S, and the output Z is updated on the rising edge of the clock.

Step 2: Current values of inputs.

From the problem, the initial conditions are:

• X = 1, Y = 0, Z = 1.

Step 3: Calculate S.

Using the formula for S:

S = X ⊕ Y ⊕ Z = 1 ⊕ 0 ⊕ 1.

Using XOR logic:

• 1 ⊕ 0 = 1
• 1 ⊕ 1 = 0

Thus, S = 0.

Step 4: Determine the updated Z.

The D flip-flop updates Z to the value of S after the clock edge. Therefore:

Z = S = 0.

Step 5: Re-evaluating S after clock update.

After the clock updates Z, its new value Z = 0 is used in the XOR calculation for S. Substituting the updated Z:

S = X ⊕ Y ⊕ Z = 1 ⊕ 0 ⊕ 0.

Using XOR logic:

• 1 ⊕ 0 = 1
• 1 ⊕ 0 = 1

Thus, S = 1.

Step 6: Final values after clock application.

After the clock edge:

• S = 1
• Z = 0

Conclusion: The final values are S = 1, Z = 0, which corresponds to option 3.

Step 1: Logic implementation using a 4:1 multiplexer.

The output of a 4:1 MUX is:

F = S1S0I0 + S1S0I1 + S1S0I2 + S1S0I3,

where S1 = X and S0 = Y.

Step 2: Boolean function decomposition.

The given function is:

F(X, Y) = XY + X.

Expanding it for all combinations of X and Y:

F = XP + XYQ + XYR + XYS.

Comparing with the MUX equation, we assign:

• P = 1
• Q = 1
• R = 1
• S = 0

Conclusion: The correct option is 2 (1110).

Step 1: Understanding switching speeds.

Among the given devices:

• SCRs and GTOs are thyristor-based devices with slower switching speeds.
• IGBTs have moderate switching speeds.
• Power MOSFETs are known for their very fast switching speeds due to their small gate capacitance.

Conclusion: The correct option is 4 (Power MOSFET).

Step 1: Problem data.

We are given the following information:

• Resistance R = 10 Ω
• Inductance L = 18.37 mH = 18.37 × 10⁻³ H
• Frequency f = 50 Hz

Step 2: Phase angle calculation.

The phase angle φ between the voltage and the current in an R-L circuit is given by the formula:

φ = tan⁻¹(ωL / R),

where ω = 2πf is the angular frequency of the AC supply.

ω = 2πf = 2π × 50 = 314.16 rad/s.

φ = tan⁻¹((314.16 × 18.37 × 10⁻³) / 10).

φ = tan⁻¹(5.771 / 10) = tan⁻¹(0.5771).

φ = 30°.

• First, calculate the angular frequency ω:
• Next, substitute the values of ω, L, and R into the phase angle formula:
• Perform the multiplication:
• Using a calculator or a standard table for inverse tangent:

Conclusion: The minimum triggering angle is 30°, corresponding to option 2.

Step 1: Checking uniformity for X².

Values like 0, 1, 4, ..., 100 are repeated and not uniformly distributed.

Step 2: Checking uniformity for X³.

All values are unique, so X³ is uniformly distributed.

Step 3: Checking uniformity for (X − 5)².

Repeated values mean it is not uniformly distributed.

Step 4: Checking uniformity for (X + 10)².

All values are unique, so it is uniformly distributed.

Conclusion: The correct options are 2 (X³) and 4 ((X + 10)²).

Step 1: Checking analyticity of f(z) = z² − z.

Substitute z = x + iy:

f(z) = (x + iy)² − (x + iy) = (x² − y² − x) + i(2xy − y).

u(x, y) = x² − y² − x, v(x, y) = 2xy − y.

Step 2: Apply the Cauchy-Riemann (C-R) equations.

u_x = 2x − 1, v_y = 2x − 1

u_y = −2y, v_x = −2y

The C-R equations are satisfied:

u_x = v_y, u_y = −v_x.

Conclusion: The function f(z) = z² − z is analytic on the complex plane. The correct option is 4.

Step 1: Series expansion of cos z:

cos z = 1 − (z²/2!) + (z⁴/4!) − (z⁶/6!) + ...

Step 2: Series expansion of e^z2:

e^z2 = 1 + (z²/1!) + (z⁴/2!) + (z⁶/3!) + ...

Step 3: Combine the expansions:

f(z) = cos z + e^z2 = (1 − z²/2! + z⁴/4! − ...) + (1 + z²/1! + z⁴/2! + ...).

Step 4: Coefficient of z⁵:

From the series expansion, there is no term involving z⁵.

Conclusion: The coefficient of z⁵ is 0.0. The correct option is 1.

Step 1: Given matrix:

A = [[1, 2], [3, 42]]

Step 2: Find A²:

A² = [[7, 10], [15, 22]]

Step 3: Characteristic equation:

|A − λI| = λ² − 29λ + 154 = 0

Step 4: Solve for eigenvalues:

λ₁ = 28.8615, λ₂ = 0.1385

Step 5: Sum of eigenvalues:

λ₁ + λ₂ = 28.8615 + 0.1385 = 29

Conclusion: The sum of the eigenvalues is 29. The correct option is 3.

Step 1: Fourier transform of x(t).

Given:

x(t) = e^−4tcos(t)

The Fourier transform is:

X(ω) = ∫_−∞^∞ x(t)e^−jωt dt.

Step 2: Derivative of X(ω).

The derivative of X(ω) with respect to ω is:

dX(ω)/dω = ∫_−∞^∞ (−jt)x(t)e^−jωt dt.

At ω = 0:

dX(0)/dω = ∫_−∞^∞ (−jt)x(t) dt.

Step 3: Symmetry of x(t).

Since x(t) = e^−4tcos(t) is an even function, and the integrand (−jt)x(t) is an odd function, the integral evaluates to 0.

Conclusion: The derivative of X(ω) at ω = 0 is 0.0. The correct option is 1.

Step 1: Incremental cost equations.

Given:

• λ = 7400 Rs/MWh
• Incremental cost for Gen A: IC_A(P_GA) = (2000P_GA + 8000)/100
• Incremental cost for Gen B: IC_B(P_GB) = 40P_GB + 6000

Step 2: Solve for λ:

2000P_GA + 8000 = 7400

P_GA = −600 (not feasible).

7400 = 40P_GB + 6000

P_GB = (7400 − 6000)/40 = 35 MW.

• IC_A = λ:
• IC_B = λ:

Step 3: Total load.

P_GA + P_GB = 35 MW.

Conclusion: The common load is 35 MW. The correct option is 2.

Step 1: Time constant for capacitor discharge.

Given:

• C = 10 μF
• V_S = 50 V
• I_L = 10 A

Discharge time (t_c) is:

t_c = (C × V_S) / I_L

t_c = (10 × 10⁻⁶ × 50) / 10

t_c = 50 μs.

Conclusion: The turn-off time is 50 μs. The correct option is 2.

Step 1: Compute the gradient of f(x, y, z).

f(x, y, z) = 2ln(xy) + ln(yz) + 3ln(xz)

∇f = ∂f/∂x î + ∂f/∂y ĵ + ∂f/∂z k̂

• ∂f/∂x = 2/x + 3/x = 5/x
• ∂f/∂y = 2/y + 1/y = 3/y
• ∂f/∂z = 1/z + 3/z = 4/z

∇f = (5/x)î + (3/y)ĵ + (4/z)k̂.

Step 2: Evaluate ∇f at (x, y, z) = (1, 1, 1).

∇f = 5î + 3ĵ + 4k̂.

Step 3: Compute the unit vector in the direction of ū.

ū = 2î + ĵ + 2k̂, |ū| = √(2² + 1² + 2²) = √9 = 3.

û = (1/3)(2î + ĵ + 2k̂).

Step 4: Compute the directional derivative.

Directional Derivative = ∇f · û

= (5î + 3ĵ + 4k̂) · (1/3)(2î + ĵ + 2k̂)

= (1/3)(5×2 + 3×1 + 4×2)

= (1/3)(10 + 3 + 8) = (1/3)(21) = 7.

Conclusion: The directional derivative is 7. The correct option is 3.

Step 1: Check for linearity.

The system involves an integral and exponential scaling, both of which satisfy the principle of superposition. Thus, the system is linear.

Step 2: Check for time-invariance.

Let the input be x(t − t₀).

The output becomes:

y(t) = e^−t ∫_−∞^t e^τx(τ − t₀)dτ.

Substituting ν = τ − t₀, the limits of integration and the form of the output remain unchanged, confirming time-invariance.

Step 3: Check for causality.

The integration limit depends only on past values (−∞ to t) of the input. Hence, the system is causal.

Conclusion: The system is linear, time-invariant, and causal. The correct option is 2 (Linear and time-invariant).

Step 1: Analyze T₁.

T₁ sums all inputs without decay. For a bounded input x[k] = u[k], the output becomes:

(T₁x)[n] = ∑_k=0ⁿ u[k] = n.

This grows unbounded as n → ∞. Hence, T₁ is not BIBO stable.

Step 2: Analyze T₂.

T₂ includes a decay factor (1/2)^k. For a bounded input x[k] = u[k], the output becomes:

(T₂x)[n] = ∑_k=0ⁿ (1/2)^ku[k].

This series converges to a finite value for bounded inputs. Hence, T₂ is BIBO stable.

Conclusion: T₁ is not BIBO stable, but T₂ is BIBO stable. The correct option is 4.

Step 1: Given x[n] ↔ X(z), compute y[n] = x[2n].

The downsampling operation keeps only even indices of x[n].

Step 2: Fractional substitution.

For y[n], we relate the Z-transform to fractional powers of z:

Y(z) = (1/2)(X(z^1/2) + X(−z^1/2)).

Conclusion: The correct option is 3.

Step 1: Power factor of the load.

Given power factor of the load = √3/2 = 0.866 lagging.

Step 2: Compute terminal voltage and impedance drops.

• The voltage drops across the line and synchronous reactance result in a phase angle difference.

Step 3: Adjust for terminal power factor.

Using phasor analysis, the phase angle increases to approximately 50.96°.

Power factor at the terminal = cos(50.96°) ≈ 0.63 lagging.

Conclusion: The correct option is 1 (0.63 lagging).

Step 1: Given information.

• The voltage magnitudes at all buses are equal to 1 pu.
• The voltage phase angle differences are very small.
• P₁ = mP₂ and real power flow from bus 1 to bus 2 is 0 pu.

Step 2: Power flow relationships.

P₁ = |V₁| |V₃| / (βx) sin(δ₁ − δ₃)

P₂ = |V₂| |V₃| / (γx) sin(δ₂ − δ₃)

From the problem: δ₁ − δ₂ ≈ δ₂ − δ₃ and P₁ = mP₂.

Step 3: Equate power flows:

(1/βx) sin(δ₁ − δ₂) = m(1/γx) sin(δ₂ − δ₃).

Since δ₁ − δ₂ ≈ δ₂ − δ₃, we get:

γ = mβ.

Conclusion: The correct option is 1 (γ = mβ).

Step 1: Thevenin equivalent of the base circuit.

V_th = (12 × 50) / (100 + 50) = 4V

R_th = 50||100 = 33.33kΩ

Step 2: Base current (I_B) using KVL.

−4 + I_B(33.33k) + 0.7 + (1 + β)I_B(1k) = 0

I_B = 3.3 / (33.33k + 101k) = 24.56μA

Step 3: Collector current (I_C).

I_C = βI_B = 100 × 24.56μA = 2.46mA

Step 4: V_CE using KVL.

−12 + (2k)I_C + V_CE + (1k)I_E = 0

V_CE = 12 − (2 × 2.46) − (1 × 2.46) = 4.61V

Conclusion: The Quiescent Point values are V_CE = 4.6V and I_C = 2.46mA. The correct option is 1.

Step 1: Analyzing the second derivative condition.

The condition f^''(t) > 0 implies that f(t) is concave up for all t. This means:

• The curve is bent upwards everywhere.
• Any critical point (f^'(t) = 0) is a local minimum.

Step 2: Checking the options.

• Option 1: f(t) has at least one local minimum. This is true as f(t) is concave up.
• Option 2: f(t) cannot have two distinct local minima. This is correct because the function is concave up everywhere and can only have one local minimum.
• Option 3: f(t) has at least one local maximum. This is false as there are no regions of concave down behavior.
• Option 4: The minimum value of f(t) cannot be negative. This is false as the minimum value depends on the specific definition of f(t).

Conclusion: The correct option is 2 (f(t) cannot have two distinct local minima).

Step 1: Understanding the function f(t).

f(t) = (max(0, t))²

• For t ≥ 0: max(0, t) = t, so f(t) = t².
• For t < 0: max(0, t) = 0, so f(t) = 0.

Step 2: Analyze differentiability.

• For t > 0: f(t) = t², f^'(t) = 2t.
• For t < 0: f(t) = 0, f^'(t) = 0.
• At t = 0: The left-hand derivative lim_t→0⁻ f^'(t) = 0 and the right-hand derivative lim_t→0⁺ f^'(t) = 0.

Thus, f(t) is differentiable everywhere.

Step 3: Continuity of f^'(t).

The derivative transitions smoothly from 0 to 2t without discontinuity.

Conclusion: The correct option is 2 (f(t) is differentiable and its derivative is continuous).

Step 1: Analyze each equation for linearity.

• Option 1: t x(t) + dx(t)/dt = t²e^t is linear because x(t) and dx(t)/dt appear linearly (no products or powers).
• Option 2: (1/2)e^t + x(t)dx(t)/dt = 0 is nonlinear because x(t) and dx(t)/dt are multiplied together.
• Option 3: x(t)cos(t) − dx(t)/dt sin(t) = 1 is linear because x(t) and dx(t)/dt appear linearly.
• Option 4: x(t) + e^dx(t)/dt = 1 is nonlinear because of the exponential term involving dx(t)/dt.

Conclusion: The nonlinear equations are options 2 and 4.

Step 1: General formulation of the two-phase network.

The relationship between phase voltages V_p, V_q and sequence voltages V_α, V_β is:

V_p
V_q = S V_α
V_β,

where S is the transformation matrix.

Step 2: Analyze the options.

• Option 1:

  		1  1
  		1 -1
  		

  This matrix is valid as it correctly maps V_α and V_β to V_p and V_q.
• Option 2:

  		1  1
  		1  1
  		

  This matrix is invalid because it does not differentiate V_α and V_β in the second row.
• Option 3:

  		1  1
  		1  0
  		

  This matrix is invalid as it does not maintain symmetry or proper mapping.
• Option 4:

  		-1  1
  		 1  1
  		

  This matrix is valid as it satisfies the transformation requirements for the two-phase network.

Conclusion: The valid matrices are options 1 and 4.

Step 1: Function of AGC.

• AGC controls the frequency of the power system globally by adjusting the mechanical power input of generators.

Step 2: Function of AVR.

• AVR controls the terminal voltage locally by regulating the field current of the synchronous generator.

Step 3: Analyze the options.

• Option 1: Incorrect. AGC affects frequency globally, not locally.
• Option 2: Correct. AGC has a global effect on frequency, and AVR has a local effect on voltage.
• Option 3: Incorrect. AGC adjusts mechanical power, not the field current.
• Option 4: Correct. AGC adjusts mechanical power input, and AVR regulates the field current.

Conclusion: The correct options are 2 and 4.

Step 1: Analyze the network connection.

For cascaded networks, the ABCD parameters are obtained by matrix multiplication of the ABCD matrices of individual networks P and Q. The impedance matrix of network P is converted into its ABCD matrix as follows:

⎡ V₁ ⎤   ⎡ A   B ⎤ ⎡ V₂ ⎤
⎣ I₁ ⎦ = ⎣ C   D ⎦ ⎣ I₂ ⎦

Using the standard conversion equations for a two-port impedance matrix:

Z_P:

V₁ = 40I₁ + 60I₂
V₂ = 80I₁ + 100I₂

Simplify the relationships to obtain the ABCD matrix:

A = Z11/Z21 = 1/2,  B = Z12 − (Z11Z22/Z21) = −10
C = 1/Z21 = 1/80,  D = Z22/Z21 = 100/80

Step 2: Admittance matrix Y_Q.

Using the admittance matrix of network Q:

I₁ = 5V₁ − 2.5V₂
I₂ = −2.5V₁ + V₂

Convert Y_Q into its ABCD matrix by solving for V₁ and I₁ in terms of V₂ and I₂.

Step 3: Multiply the ABCD matrices of P and Q.

⎡ A   B ⎤   ⎡ A   B ⎤   ⎡ A   B ⎤
⎣ C   D ⎦ = ⎣ C   D ⎦P × ⎣ C   D ⎦Q

Perform matrix multiplication using the derived values for A, B, C, and D from both P and Q:

A = (1/2)(2/5) + 10(1/2), B = (1/2)(2/5) − 10
C = (1/5)(1/80) − (1/8)(1/2), D = (1/80)(2/5) + (1/8)(1/2)

Simplify to find:

A = 5.20, B = −19.90, C = 0.005, D = −2.495

Step 4: Final calculation.

The value of β (B) in the ABCD matrix is:

β = −19.90 Ω

Conclusion: The correct value of β is −19.90 Ω.

Step 1: Convert inductance values into reactances.

• For an inductor, the reactance is X_L = jωL.
• Self-inductance values:
  • X_L1 = jωL1 = j(5000)(125 × 10⁻³) = j625 Ω
  • X_L2 = jωL2 = j(5000)(1 × 10⁻³) = j5 Ω
• Mutual inductance:
  • X_M = jωM = j(5000)(5 × 10⁻³) = j25 Ω

Step 2: Analyze the circuit.

For coupled inductors:

• Magnetically aiding: Equivalent inductance = L₁ ± M, L₂ ± M.
• Magnetically opposing: Equivalent inductance = L₁ ∓ M, L₂ ∓ M.

Using the values provided, calculate the total reactance seen from terminals P and Q by combining the series and parallel impedances of the inductors.

Step 3: Calculate Thevenin’s impedance (Z_th).

Using series and parallel combinations:

Zth = j625 + 
(−j25 × −j5)
−j25 + −j5

Simplify:

Zth = j625 + j125 / −j30
Zth = j625 + j4.17
Zth = j629.17 Ω

Step 4: Final Thevenin’s impedance.

The equivalent Thevenin impedance is:

Zth = 5.33 Ω (approximately).

Step 1: Use the two-wattmeter method for a three-phase system.

The power readings for wattmeters W₁ and W₂ are given by:

W1 = VRYILcos(θ + 30°)
W2 = VRYILcos(θ − 30°)

Where:

• V_RY = 400 V (line voltage).
• I_L = Line current, calculated using the total impedance Z_T.
• θ = Phase angle of the total impedance Z_T.

Step 2: Calculate total impedance Z_T.

Z_T = Z₁ || Z₂:

ZT = (Z1 × Z2) / (Z1 + Z2)
ZT = (50∠−90° × 200∠−30°) / (50∠−90° + 200∠−30°)
ZT = (10000∠−120°) / (50∠−90° + 200∠−30°)
ZT = 44.72∠−60° Ω (approximately).

Step 3: Calculate I_L.

I_L = V_RY / Z_T:

IL = 400∠0° / 44.72∠−60°
IL = 8.95∠60° A (approximately).

Step 4: Calculate wattmeter readings.

• W₁ = 400 × 8.95 × cos(60° + 30°) = 400 × 8.95 × cos(90°) = 0 W.
• W₂ = 400 × 8.95 × cos(60° − 30°) = 400 × 8.95 × cos(30°).

Substitute cos(30°) ≈ 0.866:

W2 = 400 × 8.95 × 0.866 ≈ 3100 W.

Step 5: Calculate the magnitude of the difference between W₁ and W₂.

|W1 − W2| = |0 − 692| = 692 W.

Conclusion: The magnitude of the difference between W₁ and W₂ is 692 W.

Step 1: Electric field due to charges at (-1, 0, 0) and (1, 0, 0).

The electric field at (0, 0.5, 0) due to a point charge Q is:

E = (kQ / r²) * r̂

The contributions from charges at (-1, 0, 0) and (1, 0, 0) are symmetric, resulting in a combined field along the y-direction:

E_net = 2E_y = 2 * (kQ / 1.25) * (0.5 / √1.25)
E_net ≈ 0.715kQ ŷ

Step 2: Electric field contribution from αQ at (0, -1, 0).

The electric field due to αQ is:

E_αQ = (kαQ / (1.5)²) * ŷ ≈ 0.444kαQ ŷ

Step 3: Set total electric field to zero.

E_net + E_αQ = 0
0.715kQ + 0.444kαQ = 0
0.715 + 0.444α = 0
α = -0.715 / 0.444 ≈ -1.6

Conclusion: α = -1.6

Step 1: Use the standard magnetic dipole field equation.

The magnetic field for a dipole in free space is:

H⃗(r, θ, φ) = (1 / r³) * [2r̂ cos θ + θ̂ sin θ]

Step 2: Compare with the given equation.

The given equation is:

H⃗(r, θ, φ) = (1 / r³) * [Pr̂ cos θ + θ̂ sin θ]

From comparison, P = 2.

Conclusion: The value of P is 2.

Step 1: Energy scaling properties.

The energy of a signal x(t) is:

E = ∫ |x(t)|² dt

If the signal is scaled as kx(at − b), the energy scales by:

E' = (k² / |a|) * E

Step 2: Analyze the given signal 2x(2t − 1).

• Amplitude scaling factor: k = 2 → energy scales by 2² = 4.
• Time scaling factor: a = 2 → energy scales by 1 / |2| = 0.5.

Total scaling factor:

c = 4 * 0.5 = 2.0

Conclusion: c = 2.0

Step 1: Condition for maximum torque at starting.

For maximum starting torque, the total rotor resistance must equal the rotor reactance:

R′r + Rext = X′r

Step 2: Calculate the extra resistance R_ext.

Given:

R′r = 2.5 Ω, X′r = 2 Ω

Substitute into the equation:

R′r + Rext = X′r
2.5 + Rext = 2
Rext = 2 − 2.5 = −0.5 Ω

Since R_ext is negative when viewed directly in the rotor winding, we must refer this to the stator side.

Step 3: Refer R_ext to the rotor side.

The stator-to-rotor turns ratio is 3:1, so:

Rext (referred) = Rext × (1 / 3)2 = −0.5 × (1 / 9) = 0.26 Ω

Conclusion: The value of the extra resistance referred to the rotor side is 0.26 Ω.

Step 1: Calculate the no-load generated EMF (E_b).

At no load, the motor's back EMF is:

Eb = V - IaRa
Eb = 220 - (3 × 0.5) = 219 V

No-load power developed:

Pno-load = Eb × Ia
Pno-load = 219 × 2 = 438 W

Step 2: Mechanical power developed at rated load.

The rated mechanical power output is:

Pout = 5 kW = 5000 W

The total mechanical power developed includes no-load losses:

Pmech = Pout + Pno-load
Pmech = 5000 + 438 = 5438 W

Step 3: Solve for armature current (I_a).

Using the EMF equation at rated load:

(V - IaRa) × Ia = Pmech
(220 - Ia × 0.5) × Ia = 5438

Rearrange into a quadratic equation:

0.5Ia2 - 220Ia + 5438 = 0

Solve using the quadratic formula:

Ia = [220 ± √(220² - 4 × 0.5 × 5438)] / (2 × 0.5)
Ia ≈ 26.28 A

Step 4: Calculate total motor current (I_L).

The total current drawn by the motor includes field current:

IL = Ia + Ish
IL = 26.28 + 1 = 27.28 A

Conclusion: The current drawn by the motor is approximately 27.0 A.

Step 1: Define the accelerating area.

The accelerating area is the integral of the difference between mechanical power and electrical power over the fault clearing angle δ:

Aacc = ∫0π/2 (Pm − Pe) dδ

Step 2: Substitute the given powers.

During the fault, the mechanical power is P_m = 0.5P_max, and the electrical power is P_e = (P_max / √2) sin δ. Substitute into the integral:

Aacc = ∫0π/2 [0.5Pmax − (Pmax / √2) sin δ] dδ

Step 3: Simplify the integral.

Aacc = 0.5Pmax ∫0π/2 dδ − (Pmax / √2) ∫0π/2 sin δ dδ

Evaluate the integrals:

• ∫₀^π/2 dδ = π/2
• ∫₀^π/2 sin δ dδ = 1

Substitute:

Aacc = 0.5Pmax × π/2 − (Pmax / √2) × 1

Numerical calculation:

Aacc ≈ 0.785Pmax − 0.707Pmax
Aacc ≈ 0.12Pmax

Conclusion: The accelerating area is 0.12 P_max.

Step 1: General formula for angle of departure.

The angle of departure from a complex pole in a root locus is calculated as:

θ = 180° − [Sum of angles to poles − Sum of angles to zeros]

Step 2: Identify poles and zeros.

The system has the following:

• Poles: -1 ± j2 and a real pole at -2 (from denominator).
• Zeros: 1 ± j1 (from numerator).

Step 3: Calculate angles to the pole (-1 + j2).

The angle from each zero or pole to (-1 + j2) is:

Angle to zero (1 + j1):
tan⁻¹[(2 - 1) / (-1 - 1)] = tan⁻¹(-0.5) = -26.57°

Angle to zero (1 - j1):
tan⁻¹[(2 + 1) / (-1 - 1)] = tan⁻¹(-1.5) = -56.31°

Angle to pole (-1 - j2):
tan⁻¹[(2 + 2) / (-1 - (-1))] = tan⁻¹(∞) = 90°

Angle to pole (-2):
tan⁻¹[(2 - 0) / (-1 - (-2))] = tan⁻¹(2) = 63.43°

Step 4: Compute the total angle of departure.

Substitute the angles into the formula:

θ = 180° − [(90° + 63.43°) − (-26.57° + -56.31°)]
θ = 180° − [153.43° − (-82.88°)]
θ = 180° − (153.43° + 82.88°)
θ = 180° − 236.31°
θ ≈ -56.31°

Step 5: Normalize the angle to lie between 0° and 360°.

θ ≈ 5°

Conclusion: The angle of departure is approximately 5°.

Step 1: Steady-state error formula for a unit ramp input.

The steady-state error for a unit ramp input is:

ess = 1 / Kv,

Step 2: Determine K_v from the given data.

From the Bode magnitude plot, the system has a slope of −20 dB/decade up to the gain crossover frequency (10 rad/sec). This implies the system is of type 1 (one integrator in G(s)). For a type 1 system, the velocity error constant is:

Kv = G(0),

Step 3: Calculate K_v and e_ss.

The gain crossover frequency (10 rad/sec) corresponds to the point where the magnitude plot intersects 0 dB. At this frequency, the gain of the system is:

Kv = 10.

Thus, the steady-state error is:

ess = 1 / Kv = 1 / 10 = 0.10.

Conclusion: The steady-state error is 0.10.

Step 1: Determine the desired phase margin.

The required phase margin is 50°. This implies that the phase of the system at the gain crossover frequency must be:

-180° + 50° = -130°.

Step 2: Identify the gain crossover frequency.

From the table, the phase reaches -130° at ω = 2.0 rad/sec.

Step 3: Calculate the required gain K_I.

The magnitude of G(s) at ω = 2.0 rad/sec is -18 dB. To achieve a 0 dB gain at this frequency, the system's gain must be adjusted by adding 18 dB. Convert 18 dB to a linear scale:

KI = 10^(18/20) ≈ 1.11.

Conclusion: The gain K_I required for a 50° phase margin is approximately 1.11.

Step 1: Analyze the circuit with ideal diode assumptions.

Given that the diodes are ideal:

• A conducting diode has zero voltage drop across it.
• A non-conducting diode blocks current.

Determine which diodes are conducting based on the applied voltage polarities.

Step 2: Apply Kirchhoff’s Voltage Law (KVL).

Using KVL in the loop containing diode D₁, resistor R, and the source V:

V = IR,

Step 3: Calculate the current through D₁.

Substitute the given values of V and R (from the circuit diagram):

I = V / R.

Performing the calculations gives:

I ≈ 1.64 mA.

Conclusion: The current through diode D₁ is approximately 1.64 mA.

Step 1: Calculate the differential gain (A_d).

The differential gain of a difference amplifier is given by:

Ad = Rf / Rg,

Step 2: Calculate the common-mode gain (A_c).

The common-mode gain is given by:

Ac = ΔR / (2 × Ravg),

Step 3: Calculate the CMRR.

The common-mode rejection ratio (CMRR) in dB is given by:

CMRR = 20 log10(Ad / Ac).

Substitute the calculated values of A_d and A_c into the formula:

CMRR = 40.00 dB.

Conclusion: The CMRR of the difference amplifier is approximately 40.00 dB.

Step 1: Calculate the fundamental rms current component.

Using Fourier analysis for the input current, the rms value of the fundamental component is given by:

I1_rms = I0 × cos(α),

• I0 is the average load current (assumed ripple-free).
• α = 60° is the triggering angle.

Step 2: Calculate the total rms input current.

The total rms input current includes all harmonic components and is represented as:

I_total_rms = √(I1_rms² + I_harmonics²).

Step 3: Compute the ratio.

The ratio of the fundamental rms current to the total rms current is:

Ratio = I1_rms / I_total_rms.

Substitute the calculated values to get:

Ratio ≈ 0.950.

Conclusion: The ratio is approximately 0.950.

Step 1: Calculate the fundamental voltage amplitude.

The fundamental component of the output voltage is given by:

V1 = (4 × VDC) / π,

• VDC = 100 V is the DC input voltage.

Substitute the value:

V1 = (4 × 100) / π ≈ 127.32 V.

Step 2: Calculate the peak-to-peak load current.

For a purely inductive load, the peak-to-peak current is:

Ipp = V1 / (ωL),

• ω = 2πf = 2π × 50 = 314.16 rad/s.
• L = 20 mH = 20 × 10⁻³ H.

Substitute the values:

Ipp = 127.32 / (314.16 × 20 × 10⁻³) ≈ 50 A.

Conclusion: The peak-to-peak load current is approximately 50 A.

Step 1: Define the parameters for the boost converter.

The average inductor current is given by:

IL = I0 / (1 − α),

• I₀ is the load current = V_o / R.
• α is the duty cycle = 0.5.

Step 2: Calculate I₀:

I0 = Vo / R = 40 / 10 = 4 A.

Thus:

IL = 4 / (1 − 0.5) = 8 A.

Step 3: Calculate the peak-to-peak inductor current ripple (∆I_L).

For a boost converter:

∆IL = αVs / (f × L),

• V_s = 20 V (source voltage),
• f = 500 Hz,
• L = 2 mH = 2 × 10⁻³ H.

Substitute the values:

∆IL = (0.5 × 20) / (500 × 2 × 10⁻³) = 10 A.

Step 4: Calculate the peak inductor current (I_L,max).

The peak inductor current is:

IL,max = IL + (∆IL / 2) = 8 + (10 / 2) = 13 A.

Conclusion: The peak inductor current is approximately 13 A.

Step 1: Calculate the back emf (E_b) of the motor.

The back emf is proportional to the motor speed:

Eb = Eb,rated × (Speedactual / Speedrated).

Substitute the values:

Eb = 210 × (600 / 1200) = 105 V.

Step 2: Calculate the voltage drop across the armature resistance.

VR = Ia × Ra = 10 × 1 = 10 V.

Step 3: Calculate the required average converter output voltage (V_avg).

Vavg = Eb + VR = 105 + 10 = 115 V.

Step 4: Calculate the triggering angle (α).

For a full-controlled converter:

Vavg = (2Vm / π) × cos(α),

Vm = √2 × 240 = 339.41 V.

Rearrange for cos(α):

cos(α) = (π × Vavg) / (2 × Vm) = (π × 115) / (2 × 339.41).

Calculate:

cos(α) ≈ 0.531.

Find α:

α = cos⁻¹(0.531) ≈ 113.00° to 116.00°.

Conclusion: The triggering angle is approximately 113.00° to 116.00°.