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


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10 Questions MCQ Test - Test: KCL & KVL in AC Circuits - 1

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Test: KCL & KVL in AC Circuits - 1 - Question 1

An AC voltage is given by v(t) = 5 + sin(2000πt), what is the time period and the dc component in v(t)?

Detailed Solution for Test: KCL & KVL in AC Circuits - 1 - Question 1

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

Test: KCL & KVL in AC Circuits - 1 - Question 2

If the voltage and current in an A.C. circuit is 90° out of phase, then the power in the circuit will be -

Detailed Solution for Test: KCL & KVL in AC Circuits - 1 - Question 2

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

Test: KCL & KVL in AC Circuits - 1 - Question 3

In the case of a sinusoidal current, the unit of the amplitude is:

Detailed Solution for Test: KCL & KVL in AC Circuits - 1 - Question 3

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 Vm.

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.

Test: KCL & KVL in AC Circuits - 1 - Question 4

A sinusoid wave is expressed as 5 sin(4πt – 60°). Find the frequency.

Detailed Solution for Test: KCL & KVL in AC Circuits - 1 - Question 4

Concept:

Considered a sinusoidal Alternating wave of voltage and current

From the waveform:

v = Vmsin⁡(ωt)

And,

i = Imsin⁡(ωt + ϕ)

Where Vm and Iare 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 = Vmsin⁡(2πft)

And,

i = Imsin⁡(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 = Vmsin⁡(2πft)

Or,

i = Imsin⁡(2πft + ϕ)

Hence, ω = 4π

We know that,

ω = 2πf

∴ 4π = 2πf

∴ f = 2 Hz

Test: KCL & KVL in AC Circuits - 1 - Question 5

In the circuit shown in the figure, the value of node voltage ��2 is

Detailed Solution for Test: KCL & KVL in AC Circuits - 1 - Question 5

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

*Answer can only contain numeric values
Test: KCL & KVL in AC Circuits - 1 - Question 6

Three currents i1, i2 and i3 meet at a node as shown in the figure. If i1 = 3 cos (ωt) ampere, i2 = 4 sin (ωt) ampere and i3 = I3 cos (ωt + θ) ampere, the value of I3 in ampere is


Detailed Solution for Test: KCL & KVL in AC Circuits - 1 - Question 6

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.

Test: KCL & KVL in AC Circuits - 1 - Question 7

What will be the period of the sinusoid, v(t) = 12 cos (50t + 30°)?

Detailed Solution for Test: KCL & KVL in AC Circuits - 1 - Question 7

Concept:

Considered a sinusoidal Alternating wave of voltage and current

From the waveform:

v = Vmsin⁡(ωt)

And,

i = Imsin⁡(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 = Vmsin⁡(2πft)

And,

i = Imsin⁡(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

*Answer can only contain numeric values
Test: KCL & KVL in AC Circuits - 1 - Question 8

In the circuit shown, the current I flowing through the 50 Ω resistor will be zero if the value of capacitor C (in μF) is ______.


Detailed Solution for Test: KCL & KVL in AC Circuits - 1 - Question 8

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 Zeq (say) which can be calculated as:

For I = 0, Zeq → ∞ --- (1)

Test: KCL & KVL in AC Circuits - 1 - Question 9

Consider a current source i(t) connected across a 0.5 mH inductor, where i(t) = 0 A for t < 0 and i(t) = (8e-250t - 4e-1000t) A for t ≥ 0. The voltage across the inductor at t = 0 s is

Detailed Solution for Test: KCL & KVL in AC Circuits - 1 - Question 9

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

VL = 0.5 × 10-3 × (- 2000 + 4000)

VL = 1 V

*Answer can only contain numeric values
Test: KCL & KVL in AC Circuits - 1 - Question 10

Three 400 Ω resistors are connected in delta and powered by a 400 V (rms), 50 Hz balanced, symmetrical R - Y - B sequence, three-phase three-wire mains. The rms value of the line current (in amperes, rounded off to one decimal place) is


Detailed Solution for Test: KCL & KVL in AC Circuits - 1 - Question 10

In delta connection, VL = Vph = 400 Ω

 ILine = √3 Iphase = √3 × 1 = 1.732 A

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