Test: ESE Electrical - 9 - Electrical Engineering (EE) MCQ

When we consider the 16V source, we short the 10V source and open the 15A and 3A source. From the resulting series circuit we can use voltage divider to find Vx.
Vx = 16*20/(20+80) = 3.2A.

Equivalent capacitance between terminals a-b is:

Max. value of current:

Assuming voltage as the reference phasor:

The phase difference between v and i is: 

Since v leads i therefore, the circuit element is an inductor.

 = 0.8264 or θ = 34.26°,

In wye or star connection, similar ends of the three phases are joined together within the alternator. The common terminal so formed is referred to as the neutral point or neutral terminal.

The voltage V_BR is V_BR = 400 ∠ -240⁰V. The impedance Z₃ is Z₃ = 10 ∠ -90⁰Ω
⇒ I_B = (400 ∠ 240^o)/(10 ∠ -90^o) = (-34.64-j20)A.

The electromotive force is given by Vemf = -dλ/dt. Thus Vemf = -dλ/dt = -(3cos t – 5sin t) = -3cos t + 5sin t.

The induced emf is given by Vemf = -dλ/dt = -Ndψ/dt. Thus emf will be -100 x 0.3 = -30 units.

A discrete time signal can be obtained from a continuous time signal by replacing t by nT, where T is the reciprocal of the sampling rate or time interval between the adjacent values. This procedure is known as sampling.

We know that the Discrete Fourier transform of a signal x(n) is given as

Thus we get Nth rot of unity W_N= e^-j2π/N

The formula for calculating N point DFT is given as

From the formula given at every step of computing we are performing N complex multiplications and N-1 complex additions. So, in a total to perform N-point DFT we perform N² complex multiplications and N(N-1) complex additions.

The FS coefficients of x(t) are

Therefore, the FS coefficients of Ev{ x(t)} are

Period of sampling train, T_s = 20 ms

If frequency of x(t) is fx, then after sampling the signal, the sampled signal has the frequency,
f_s - f_x = 50 - f_x and f_s + f_x = 50 + f_s
Now, the sampled signal is applied to and ideal low pass filter with cut off frequency.
f_c = 25 Hz
Now, the o/p of filter carried a single frequency component of 20 Hz
∴, only (f_s−f_x) component passes through the filter, ie.
f_s - f_x < 25
and f_s - f_x = 20
50 - f_x­ = 20
f_x = 50 - 20 = 30 Hz

The above given diagram is the basic butterfly computation in the decimation-in-time FFT algorithm.

We know that,

For A: Two conjugate poles on imaginary axis i.e. it’s impulse response = sin at (marginally stable).
For B: Two poles at origin i.e. impulse response = t u(t) (unstable system).
For C: Two complex conjugate poles with negative real parts. So, its impulse response = er^-at sin bt (stable system).
For D: Two imaginary complex double roots. So, impulse response = t sin bt. (unstable system).

Hence, only 1 is correctly matched.

If damping of system increases, peak overshoot Mp decreases and vice-versa.

The Routh’s array is

Since there are two sign changes in first column of Routh’s array, therefore the system is unstable.
Number of roots with positive real part = 2
Number of roots with negative real part = 2
Number of roots with zero real part = 0

From the root sensitivity, standpoint, a system should not be operated at the breakaway points because root sensitivity at breakaway point is infinite.

at breakaway points

Here, α = closed loop poles(s)
and β = system gain (K)

• In control systems, the constant M-circles represent curves in the G (gain) plane that correspond to constant magnitudes (M) of the closed-loop transfer function of a linear system. For stability analysis, it is essential to keep the magnitude of the closed-loop transfer function (M) less than one to ensure stability.
• The M = 1 line in the G-plane represents the boundary between stable and unstable regions. When the value of M is less than one, the constant M-circles lie to the right of the M = 1 line in the G-plane. The reason for this is that any point in the G-plane that leads to a magnitude (M) less than one indicates a stable closed-loop system.
• Therefore, the correct answer is D: right of the M = 1 line.