Sinusoidal Steady State Analysis MCQ Quiz - Objective Question with Answer for Sinusoidal Steady State Analysis - Download Free PDF

Concept

The inverse Laplace Transform of 

\({k\over s+a}=ke^{-at}\)

where, k = Constant

Calculation

Given, \(\rm I(s)=\frac{1.5}{s+4}\)

On comparison, k = 1.5 and a = 4

\({1.5\over s+4}=1.5e^{-4t}\)

Concept

The average and RMS value of the full and half-wave rectifier are given by:

Full wave rectifier:

\(V_{o(avg)}={2V_m\over \pi}\)

\(V_{o(rms)}={V_m\over \sqrt{2}}\)

Half-wave rectifier:

\(V_{o(avg)}={V_m\over \pi}\)

\(V_{o(rms)}={V_m\over 2}\)

Calculation

Given, V_m = 20V

The average output voltage is given by:

\(V_{o(avg)}={2\times 20\over \pi}\)

\(V_{o(avg)}=\frac{40}{\pi}V\)

Concept

The current through a capacitor is given by:

\(I_C=C{dV_C\over dt}\)

where, \({dV_C\over dt}=\) Rate of change of capacitor voltage

The voltage across a capacitor is given by:

\(V_c(t)={1\over C } \int I_c(t) dt\)

Concept

When an inductor is represented in the frequency domain, it is represented as sL and the initial current across the inductor is also considered.

After source transformation:

V(s) = sL × I(s) − Li(0)

Power Triangle

In the given figure, the given terms represent the type of power:

P = Active Power

Q = Reactive Power 

S = Apparent Power

The power factor is given by:

\(cosϕ={P\over S}\)

Calculation

Given, cos ϕ = 0.8

S = 10 kVA

The active power is given by:

\(0.8={P\over 10}\)

P = 8 kW

Calculation,

Given,

cos ϕ₁ = 0.8

∴ sin ϕ₁ = 0.6

cos ϕ₂ = 0.6

∴ sin ϕ₂ = 0.8

Asuume,

|I₁| = |I₂| = |I|

In complex form the phasor quantity is written as,

i = I(cos ϕ ± jsin ϕ)

Both current can be written as,

i₁ = |I|(0.8 + j0.6)

i₂ = |I|(0.6 + j0.8)

Resultant current (i) will be phasor sum of i₁ and i₂,

i = |I|(0.8 + j0.6) + |I|(0.6 + j0.8) = |I|(1.4 + j1.4)

Phase angle can caculated as,

\(\tan ϕ = \frac{imaginarry\ component}{real\ component }=\frac{1.4}{1.4 } = 1\)

ϕ = tan^-1 (1) = 45° 

Hence,

Power factor = cos 45° = 0.707

Alternate Method

When two circuits having the same magnitudes of impedances are joined in parallel. The power factor of the combination is 

\(\cos ϕ = \frac{1}{{\sqrt {1 + \left( {{{\cos }^2}{ϕ _1} + {{\cos }^2}{ϕ _2}} \right)} }}\)

The power factor of the first circuit = 0.8 (cos ϕ₁)

Power factor of second circuit = 0.6 (cos ϕ₂)

The power factor of parallel combination,

\(\cos ϕ = \frac{1}{{\sqrt {1 + \left( {{{\cos }^2}{ϕ _1} + {{\cos }^2}{ϕ _2}} \right)} }}\)

Kirchhoff’s Current Law (KCL): It states that the algebraic sum of currents entering a node or closed boundary is zero.

Mathematically, KCL implies that

\(\mathop \sum \limits_{n = 1}^N {i_n} = 0 \)

Where N is the number of branches connected to the node

And in is the nth current entering or leaving the node.

By this law current entering a node may be regarded as positive and current leaving a node may be regarded as negative or vice-versa.

Hence, the sum of currents entering a node is equal to the sum of the currents leaving the node.

Calculation:

Given that, three currents i1, i2, and i3 meet at a node,

From the above concept,

i1,+ i2, + i3 = 0

or,  i3 = - (i1 + i2)

Since the current is given in phasor form, hence the addition of current i₁ and i₂ can be done by using the parallelogram method,

We have,

i₁ = 10 sin (400t + 60°) A .... (1)

i₂ = 10sin (400t - 60°) A .... (2)

The phasor diagram can be drawn as,

By using the parallelogram method,

 |ir| = \(\sqrt{i_1^2+i^2_2+2i_1i_2cosθ}\) .... (3)

Where, it is the resultant current

We have, θ = (60° + 60°) = 120°

From equation (1), (2) & (3),

|i_r| = \(\sqrt{10^2+10^2+200cos120 ^\circ}\)

|ir| = \(\sqrt{10^2+10^2+200\times -0.5}=10\)

Since i₁ & i₂ has the same phase angle as well as the magnitude,

∴ i_r = 10sin 400t A

Since the sum of all current is zero, hence,

i₃ = - i_r = - 10sin 400t A

Concept:

The impedance(Z) of a circuit can be written as:

Z = R ± jX

R = Resistance

X = Reactance

The admittance(Y) can be written as:

Y = G ± jB   ---(1)

G = Conductance

B = Susceptance

\(Y = \frac{1}{Z}\)

Calculation:

Z = 3 + j4

\(Y = \frac{1}{Z} = \frac{1}{{3 + j4}} \)

\(= \frac{{\left( {3 - j4} \right)}}{{\left( {3 + j4} \right)\left( {3 - j4} \right)}} \)

\(Y= \frac{3}{{25}} - \frac{{j4}}{{25}}\)

Comparing this with Equation (1), we get:

\(G= \frac{3}{{25}}\)

Mistake Point:

The conductance will not be 1/R. For a given impedance, the conductance will be:

\(G=\frac{R}{|Z|^2}\)

R = Resistance 

|Z|² = Square of the magnitude of impedance

Concept:

The average value of a periodic waveform is given as

\({F_{avg}} = \frac{1}{T}\mathop \smallint \limits_0^T f\left( t \right)dt\)

It is also calculated by calculating the area of the given waveform.

\({F_{avg}} = \frac{A}{T}\)

Where A is the area

RMS value of a periodic waveform is given as

\({F_{rms}} = \sqrt {\frac{1}{T}\mathop \smallint \limits_0^T {{\left[ {f\left( t \right)} \right]}^2}dt} \)

Where T is the time-period of the given waveform.

Calculation:

Time period of the given waveform (T) = π

The area of the given waveform is,

\(A = \left( {\frac{1}{2} \times \alpha \times {F_m}} \right) + {F_m}\left( {\pi - 2\alpha } \right) + \left( {\frac{1}{2} \times \alpha \times {F_m}} \right)\)

= F_m (π – α)

Concept:

Equation of current is I = Im sin ωt

I_m = Maximum current

ω = Angular frequency = 2πf

f = Frequency

Calculation:

Given, I = 100 A

f = 50 Hz

100 = 200 sin 2 × π × 50 × t

100 × π × t = 30° = π/6

t = 1/600 s   

t = 1.666 ms

Concept:

RMS (Root mean square) value:

• RMS value is based on the heating effect of wave-forms.
• The value at which the heat dissipated in AC circuit is the same as the heat dissipated in DC circuit is called RMS value provided, both the AC and DC circuits have equal value of resistance and are operated at the same time.

• RMS value 'or' the effective value of an alternating quantity is calculated as:

  \({V_{rms}} = \sqrt{\frac{1}{T}\mathop \smallint \limits_0^T {v^2}\left( t \right)dt} \)

  T = Time period

Note:

• Average or mean value of alternating current is that value of steady current which sends the same amount of charge through the circuit in a certain interval of time as is sent by alternating current through the same circuit in the same interval of time

• The ratio of the maximum value (peak value) to RMS value is known as the peak factor or crest factor.
• \(Peak\;factor = \frac{{maximum\;value}}{{rms\;value}}\)
• The ratio of RMS value to the average value is known as the form factor.
• \(Form\;factor = \frac{{rms\;value}}{{average\;value}}\)

Calculation:

For a rectangular wave, the RMS value is equal to the average value. RMS value is also equal to peak value.

For given waveform RMS value is 100 V.

IMPORTANT EVALUATIONS:

Concept:

The Laplace transform resistance, inductor, and capacitance are given by:

1. Resistance: R
2. Inductor: sL
3. Capacitor: \({1\over sC}\)

​Calculation:

The circuit diagram in the Laplace domain is given below:

Applying voltage division rule across capacitor:

\(V_{out}(s) = V_{in}\times{{1 \over s}\over {1 \over s}+2s}\)

Concept:

KCL in DC circuits:

According to Kirchhoff’s current law (KCL), the algebraic sum of the electric currents meeting at a common point is zero. I.e. the sum of currents entering a node is equal to the sum of currents leaving the node. It is based on the conservation of charge.

KCL in AC circuits:

Kirchhoff’s current law as applied to the ac circuit is defined as the phasor sum of currents entering the node is equal to the phasor sum of currents leaving the node.

Calculation:

By applying KCL at the node,

I₁ = I₂ + I₃

I₁ = 1∠10° + 1∠70°

= (cos 10 + j sin 10) + (cos 70 + j sin 70)

= (cos 10 + cos 70) + j(sin 10 + sin 70)

= 1.3268 + j1.1133 A

⇒ I₁ = 1.732∠ 40° A

The correct answer is option 4):​\(\sqrt {\frac{5}{{18}}} A\)

​Concept:

The magnitude of the total current through the circuit

I = \(\sqrt{I_R^2+(I_L -I_C)^2} \)

where

I_R is the resistive component of current

I_L  is the Inductive component of current

I_C  is the Capacitive component of current

IR = \(V\over R\)

V is the voltage

R is the resistance

I_C = \(V\over X_c\)

X_c is the capacitive reactance

I_L = \(V\over X_L\)

X_L  is the inductive reactance

Calculation:

The impedance of a network is given as

I = \(\sqrt{I_R^2+(I_L -I_C)^2} \)

I_R = \(20 \over 120 \)

= \(1\over 6\)

I_c = \(20 \over 40\)

= \(1\over 2\)

I = \(\sqrt(\frac{1}{36}+\frac{1}{4})\)

= \(\sqrt {{4+36} \over 36\times 4}\)

=\( \sqrt{10 \over 36} \)

= \(\sqrt {\frac{5}{{18}}} A\)

Given that,

V = 200 Sin (2000t + 50°) V

i = 4 cos (2000t + 13.2°) A

We know that,

sin (90 + ϕ) = cos ϕ 

⇒ V = 200 sin (2000t + 50°) V  --------- (i)

⇒ i = 4 sin (2000t + 13.2°+ 90°) A

⇒ i = 4 sin (2000t + 103.2°) A    ------- (ii)

From, (i) and (ii) we can observe that,

Current 'i' is leading voltage V by 53.2° 

Hence, the element is practical capacitor