Test: KCL & KVL in AC Circuits - 1 - Electrical Engineering (EE) MCQ

Concept

For a DC signal

• The average value over one time period is non zero
• The non-zero value is known as the DC value
• 
• 'A 'is the DC component of the signal

Calculation

Given the signal v(t) = 5 + sin(2000πt)

Now 

we have, 

The average value of sinusoidal signal over one time period is always 0

Hence the DC component of the signal v(t) is 5

Now observe the AC component in signal v(t)

sin(2000πt) is the AC component

The angular frequency ω = 2000π 

Also ω = 2π / T ; here T is the Time period of the signal

2π / T= 2000π

T = 10^-3 s 

T =1 ms

The Time period of the signal is 1 ms

Therefore the correct answer is option 4

The correct answer is 'option C'

Concept:

The power in an AC circuit is given by the expression

ϕP=VIcosϕ........(1)

here, ϕ is the angle between voltage and current.

Solution:

It is given to us that ϕ = 90⁰

From eqn(1)

P = 0 Watts

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 = Vmsin⁡(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

Since there is a voltage source present between the two nodes, the concept of supernode will be applied.

By KCL, using supernode at Node (1) and (2), we can write:

24 = j2 V₁ + (1 - j) V₂      ---(1)

By KVL, V₁ – V₂ = 10 ∠0°

V₁ = V₂ + 10       ---(2)

Substituting (2) in (1), we get:

24 = j2 (V₂ + 10) + (1 - j) V₂

24 = j2 V₂ + j20 + (1 - j) V₂

24 = j20 + (1 + j) V₂

V₂ = 2 – j22 V

By applying KCL at node 1,

i₁ = i₂ + i₃

i₃ = i₁ – i₂

= 3 cos ωt – 4 sin ωt

= 

= 5 cos (ωt + θ)

Given that, i₃ = I₃ cos (ωt + θ)

⇒ I₃ = 5A.

Concept:

Considered a sinusoidal Alternating wave of voltage and current

From the waveform:

v = V_msin⁡(ωt)

And,

i = I_msin⁡(ωt + ϕ)

Where Vm and Im 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 = Vmsin(ωt ± θ)

ωt = 2πft = 50t

f = 2π50

We know that,

=0.1257sec

Concept:

Current through an open circuit or through a circuit with impedance tending to infinity is zero.

Calculation:

We are given a circuit shown below:

The impedance seen from the voltage source is Z_eq (say) which can be calculated as:

For I = 0, Z_eq → ∞ --- (1)

Concept:

When current source i(t) connected across an inductor(L), then the voltage across the inductor is given as

When voltage source v(t) connected across a capacitor(C), then the current across the capacitor is given as

Calculation:

Given i(t) = (8e^-250t - 4e^-1000t) A for t ≥ 0, L = 0.5 mH

Then the voltage across the inductor is given as

At t = 0 its value is given as

V_L = 0.5 × 10^-3 × (- 2000 + 4000)

VL = 1 V

In delta connection, V_L = V_ph = 400 Ω

 I_Line = √3 I_phase = √3 × 1 = 1.732 A