Test: ESE Electrical - 1 - Electrical Engineering (EE) MCQ

• In order for 600 C charge to be delivered to the 100 V source, the current must be anticlockwise.
• Applying KVL we get:
  v₁ + 60 - 100 = 10 x 20
  v₁ = 240 V

Going from 10 V to 0 V:

• 10 + 5 + E + 1 = 0
• E = -16V
• Minus sign indicates that the polarity of battery should be reversed.

Waveform: The curve obtains by plotting the instantaneous value of any electrical quantity such as voltage, current, or power.

Cycle: One complete set of positive and negative values or maximum or minimum value of the alternating quantity is called a cycle.

Note: One-half cycle of the wave is called Alternation.
Amplitude: the maximum value of the positive or negative alternative quantity is called Amplitude.

• For sinusoidal current, the unit of the amplitude is amperes.
• For sinusoidal voltage, the unit of the amplitude is volts.
• It is also known as peak value or crest value.
• Peak-Peak Value = 2 × Amplitude
• In the given figure amplitude of the waveform is V_m.
  

Time period (T): It is the time required to complete one cycle. It measured in seconds.
Frequency: The number of cycles completed per second is called frequency.
f = 1 / T
It measured in Hz or radian/sec.

Concept:
Considered a sinusoidal Alternating wave of voltage and current

From the waveform:
v = V_msin⁡(ωt)
And,
i = I_msin⁡(ωt + ϕ)
Where V_m and I_m are the maximum value of instantaneous voltage and current respectively.
v, i is the instantaneous value of voltage and current at any instant t.
ω is the angular frequency in radian/second.
And, ω = 2πf  
f is the frequency in Hz
From the above three equations instantaneous value of voltage and current can be written as:
v = V_msin⁡(2πft)
And,
i = I_msin⁡(2πft + ϕ)
Calculation:
Sinusoidal expression of the wave is given as 5 sin (4πt – 60°)
Comparing the above expression with the given equation:
as:
v = V_msin⁡(2πft)
Or,
i = I_msin⁡(2πft + ϕ)
Hence, ω = 4π
We know that,
ω = 2πf
∴ 4π = 2πf
∴ f = 2 Hz

Concept:
Considered a sinusoidal Alternating wave of voltage and current

From the waveform:
v = V_msin⁡(ωt)
And,
i = I_msin⁡(ωt + ϕ)
Where V_m and I_m are the maximum value of instantaneous voltage and current respectively.
v, i is the instantaneous value of voltage and current at any instant t.
ω is the angular frequency in radian/second.
And, ω = 2πf  
f is the frequency in Hz
From the above three equations instantaneous value of voltage and current can be written as:
v = V_msin⁡(2πft)
And,
i = I_msin⁡(2πft + ϕ)
Calculation:
Sinusoidal wave is given as v = 12 cos (50t + 30°)
Comparing to the given equation
v = V_msin(ωt ± θ)
ωt = 2πft = 50t
f = 2π50
We know that,

=0.1257sec

R-Ccauses a positive phase shift in voltage

Let I₃ be the clockwise loop current in center loop
 V₂ = 4(I₂ + I₃) ⇒ I₃ = 0.25V2 - I₂
⇒ I₁ = 0.35V₂ - I₂  .........(i)
V₁ = 4I₁ - 0.2V₁ + V₂
1.2V₁ = 4(0.35V₂ - I₂) + V₂ = 2.4V₂ - 4I₂
⇒ V₁ = 2V₂ - 3.33I₂ .........(ii)

As the line voltage V_RY = V∠0° is taken as a reference phasor. Then the source voltage V_YB is V∠-120°.

If the system is a three-wire system, the currents flowing towards the load in the three lines must add to zero at any given instant.

D = εE. E = (1/4π)/(2Xεo) = 4.5 X 10⁹ units.

Electric flux density is given by the ratio between number of flux lines crossing a surface normal to the lines and the surface area. The direction of D at a point is the direction of the flux lines at that point.

Energy is area under the |x(t)|² = A²
Area = (A²)*(T)

Note that y[n] =  x[n] + x[n + N/2] has a period of N/2 and N has been assumed to be even,

Nyquist Sampling Theorem: 
A continuous-time signal can be represented in its samples and can be recovered back when sampling frequency fs is greater than or equal to twice the highest frequency component of the message signal, i.e.
f_s ≥ 2f_m
Therefore when we want to convert continuous signals to discrete signals, the sampling must be done at the Nyquist rate.
Calculation:
Given that,
f_m = 3.1 kHz 
⇒ f_s ≥ 2f_m
⇒ f_s ≥ 2 × 3.1 = 6.4 kHz

We know that if x(n)=x1(n)+jx2(n) then x1(n)= (x(n)+x*(n))/2
On applying DFT on both sides of the above equation, we get
X1(k)= 1/2 {DFT[x(n)]+DFT[x*(n)]}
We know that if X(k) is the DFT of x(n), the DFT[x*(n)]=X*(N-k)
⇒X1(k)= 1/2 [X*(k)+X*(N-k)].

An FIR filter has an linear phase if its unit sample response satisfies the condition
h(n)= ±h(M-1-n) n=0,1,2…M-1.

We know that the roots of the polynomial H(z) are identical to the roots of the polynomial H(z -1). Consequently, the roots of H(z) must occur in reciprocal pairs.

• There is no-control action in open loop systems. Hence, statement 1 is false.
• Generally, all control systems operated by present timing mechanism are open loop because they are more stable than closed loop control system. Hence, statement-2 is true.
• Due to various advantages of negative feedback systems (except reduction in gain), feedback signal is usually negative in a closed loop control system. Hence, statement-3 is true.
• In open-loop control systems stability can be ensured because they are more stable than closed-loop control system.
• Thus, only statements 2 and 3 are true.

On combining the blocks G₂ and G₃ which are in parallel, the given block diagram will be reduced as shown below.

Number of sign changes in the first column of Routh’s array indicates the number of closed loop poles or roots of characteristic equation lying in the right half of s-plane.

The root locus is defined as location of closed loop poles when system gain k is varied from zero to infinity.

With no-pole at origin, initial slope will be OdB/decade.
For n-pole at origin, initial slope will be -20 ndB/decade.