Test: ESE Electrical - 7 - Electrical Engineering (EE) MCQ

On shorting the voltage sources:
RL=3||2+4||5.

On shorting the voltage sources:
R_L = 3||2+4||3 
= 1.20 + 1.71
= 2.91 ohm.
The two nodal equations are:
(V_A-10)/3 + VA/2 = 0
(V_B-20)/4 + VB/3 = 0
On solving the two equations, we get V_A=4V, V_B=8.571V.
V_AB = V_A - V_B 
= 4V – 8.571V 
= -4.57V.
E_th = 4.57V
The maximum power transferred = E_th2/4RL. Substituting the given values in the formula, we get P_max = 1.79W

v_a = 2(3 + 1) + 3 (1) = 11 V

α = W_o, critically damped response

s = -2, -2
i(t) = (A + Bt)e^-2t, A = -2

At t = 0. ⇒ B = -2

X_L = 2*π*f*L = 10 ohm
Z² = (R²+X_L²)
Therefore the total impedance
Z = 12.2 ohm
V = IZ
therefore,
I= V/Z = 100/12.2 = 8.2A

Due to initial condition, at t = 0 capacitor will act as a constant voltage source (at t = 0, capacitor acts as short-circuit). Hence, option (d) is correct.

pf = cos θ = 0.9 ⇒ θ = 25.84°
Q = S sin θ ⇒ 

As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the R impedance is I_R = V_BR∠0⁰/Z∠Ø = (V/Z)∠-Ø.

The term power is defined as the product of square of current and the impedance. So the power in the R phase = 20² x 17.32 = 6928W.

The Lorentz force is given by F = qE + q(v x B), it is the sum of electric and magnetic force. On substituting q = 2.5, E = 5, v = 1.5 and B = 7.25, F = 2.5(5) + 2.5(1.5 x 7.25) = 39.68 units.

The force on a conductor is given by F = BIL sin θ, where B = 3.75, I = 8, L = 0.12 and θ = 300. We get F = 3.75 x 8 x 0.12 sin 30 = 1.8 units.

y(t) = tx(t)
y₁(t) = t.x₁(t) = r₁(t)
y₂(t) = tx₂(t) = r₂(t)
y₁(t) + y₂(t) = t(x₁(t) + x₂(t))
= r₁(t) + r₂(t)    ∴ linear
y(t) = t.x(t)
y( t - t_o) = (t - t_o) x ( t - t_o)
and for delayed input signal,
y(t) = t x (t - t_o)
y(t) ≠ y( t-t_o)
∴ Time varying signal

First we express X(z) in terms of positive powers of z, in the form X(z)=z³/((z+1)〖(z-1)〗² )
X(z) has a simple pole at z=-1 and a double pole at z=1. In such a case the approximate partial fraction expansion is
(X(z))/z = z²/((z+1)〖(z-1)〗² ) =A/(z+1) + B/(z-1) + C/〖(z-1)〗²
On simplifying, we get the values of A, B and C as 1/4, 3/4 and 1/2 respectively.
Therefore, we get X(z)= z/(4(z+1)) + 3z/(4(z-1)) + z/(2〖(z-1)〗² ) .

The partial fraction expansion for the given X(z) is
X(z)= 2z/(z-1)-z/(z-0.5)
In case when ROC is |z|<0.5,the signal is anti causal. Thus both the terms in the above equation are anti causal terms. So, if we apply inverse z-transform to the above equation we get x(n)= [-2+0.5ⁿ]u(-n-1).

Given x(t) = u(sint)
Drawing waveform

We know for waveform like y(t) shown

DC component is zero and have sine odd harmonics.
Performing upward shift in y(t) by adding dc value of 0.5, then
x(t) = y(t) + 0.5
So, x(t) → DC + sine odd harmonics

We know that from Fourier Series coefficient of square periodic signal, power is proportional to (1/n²).

Concept:
Nyquist rate: The minimum sampling rate is often called the Nyquist rate. The Nyquist sampling rate is two times the highest frequency of the input or message signal.
f_s = 2f_m
Where,
f_s is the minimum sampling frequency or Nyquist rate
f_m is the highest frequency of the input or message signal.
Calculation:
sinc(700t) + sinc(500t)

f₁ = 700π/2π = 350 Hz
f₂ = 500π/2π = 250 Hz
f_m = max (f_1, f₂) = max (350, 250) = 350 Hz
Sampling frequency,
f_s = 2f_m = 2 × 350 = 700 Hz
Sampling time period = 1/700 sec

We know that, X(k) = F₁(k)+W_N^k F₂(k)
We know that F₁(k) and F₂(k) are periodic, with period N/2, we have F₁(k+N/2)= F₁(k) and F₂(k+N/2)= F₂(k). In addition, the factor W_N^k+N/2= -W_N^k.
Thus we get, X(k+N/2)= F₁(k)- W_N^k F₂(k).

• When the FIR system has linear phase, the unit sample response of the system satisfies either the symmetry or asymmetry condition, h(n)= ±h(M-1-n)
• For such a system the number of multiplications is reduced from M to M/2 for M even and to (M-1)/2 for M odd. Thus for the structure given in the question, M is odd.

• We have to do 3 multiplications for every second order equation.
• So, we have to do 6 multiplications if we combine two second order equations and we have to perform 3 multiplications by directly calculating the fourth order equation.
• Thus the number of multiplications are reduced by a factor of 50%.

• In a control system, feedback refers to the process of measuring the output of the system and comparing it to the desired or reference input. This information is then used to adjust the system's behavior in order to reduce any error between the output and the desired input. By continuously monitoring and correcting the system's output, feedback helps to minimize the error and improve the system's performance.
• Additionally, feedback also helps to reduce the sensitivity of the system to parameter variations. In a control system, there are often uncertainties or variations in the system's parameters (such as variations in component values or environmental conditions). These parameter variations can affect the system's behavior and introduce errors. However, by using feedback, the control system can continuously adapt and adjust its response to compensate for these variations, thus reducing the sensitivity of the system to parameter changes.
• Therefore, the correct answer is C: "reduces the error and the sensitivity of the system to parameter variation."

Loop is the part of the network in which the branch starts from the node and comes back to the same node and non touching loop must not have any node in common.

Given, 


(Using final value theorem) 
or, 

The characteristic equation is
1 + G (s) H (s) = 0
or, 

or, s³ + 10s² + (21 + K )s + 13 K= 0

For stability, 13K > 0 or K > 0
Also, 

or, K < 70
Hence, 0 < K< 70 (For stability).

Given, s² + 2s + 10 + K(s² + 6s + 10) = 0

Here, number of open loop poles, P = 2. Number of open loop zero, Z = 2
∴ P - Z = 0.
Angle of asymptotes are given by:

Sines P - Z = 0, therefore there are no angle of asymptotes for the root locus of the given system.

The bandwidth is defined as the band of frequencies tying between-3 dB points. Hence, statement-2 is false.
Higher the value of resonant frequency of the system, faster is the time response. Hence, statement-3 is false.