Alternating Current MCQ Quiz - Objective Question with Answer for Alternating Current - Download Free PDF

Concept:

The Phasor diagram of a purely inductive circuit is:

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• In a purely inductive circuit, the current lags voltage by 90°
• The power factor in a purely inductive circuit is given by cosϕ, where ϕ is the angle between current and voltage.
• Therefore, cosϕ = cos 90° = 0

Explanation:

The power in a purely inductive circuit is given by:

P = VI cosϕ

where P = Active power

V = Voltage

I = Current

cosϕ = Power factor

∵ cosϕ = 0

∴ P = VI × (0)

P = 0

Therefore the power is zero in a purely inductive circuit.

Additional Information The reactive power in a purely inductive circuit is given by:

Q = VI sinϕ

∵ cosϕ = 0

∴ sinϕ = 1

Q = V× I × (1)

Q = V× I = Maximum

• A pure inductive circuit is an energy-absorbing circuit, hence it only absorbs lagging reactive power.
• It does not dissipate any active power.

Concept:

Reactive power (Q) = V I sinϕ

Where,

V = Voltage in volt

I = Current in ampere

ϕ = phase difference between voltage & current

Calculation:

Voltage (V) = 230 V

Current (I) = 5 A

Phase difference (ϕ) = 30°

Q = 230 × 5 × sin30°

Concept:

In a star-connected circuit,

\({V_L} = \sqrt 3 {V_P}\)  & \({I_L} = {I_P}\)

In a delta-connected circuit,

\({V_L} = {V_P}\)  & \({I_L} = \sqrt 3 {I_P}\)

Where,

VL = Line voltage

VP = Phase voltage

IL = Line current

IP = Line current

Calculation:

For 200 V phase voltage in a star connected circuit, then its line voltage is

\({V_L} = \;\sqrt 3 \times 200 = 346.4\;V\)

Concept

The complex power is given by:

S = VI^*

Also, P = S cosϕ 

and Q = S sinϕ 

where, S = Apparent power 

V = Voltage

I* = Conjugate of current

P = Active power

Q = Reactive power

Calculation

Given, V_s = 100 V

R = 3Ω 

X_L = 4Ω 

The current is given by:

\(I={100\over 3+j4}\)

\(I={100\over 5\angle 53.13}\)

\(I=20\angle -53.13\)

\(I^*=20\angle +53.13\)

\(S=(100)(20\angle53.13)\)

\(S=2000\angle53.13\)

P = 2000 cos(53.13°)

P = 2000 × 0.6

P = 1200 W

3 phase connections

1.) Delta connection

The phase and line voltages are the same.

\(V_L=V_P\)

The line current is √3 times of phase current and lags the phase current by 30°.

I_L = √3 I_P ∠-30° 

2.) Star connection

The phase and line currents are the same.

\(I_L=I_P\)

The line voltage is √3 times the phase voltage and leads the phase voltage by 30°.

V_L = √3 VP ∠30° 

Hence, the correct answer is option 1.

Relationship between line and phase voltages and currents in star-connected and delta-connected networks:

The correct answer is option 1): 1.11

The form factor is defined as the ratio of the RMS value to the average value of an alternating quantity.

F.F. (Form factor) = \(R.M.S\: Value\over Average\: Value\)

Crest Factor ‘or’ Peak Factor is defined as the ratio of the maximum value to the R.M.S value of an alternating quantity.

C.F. ‘or’ P.F. =\( Maximum\:Value\over R.M.S\:Value\)

 For a sinusoidal waveform:

Form Factor = 1.11

Crest Factor = 1.414

Concept:

In AC circuits, the power factor is defined as the ratio of the real power flowing to the load to the apparent power in the circuit.

Power factor = cos ϕ

Where ϕ is the angle between voltage and current.

If the current lags the voltage, the power factor will be lagging.

If the current leads the voltage, the power factor will be leading.

Power factor = cos ϕ = R/Z

\(Z=\sqrt{{{R}^{2}}+{{X}^{2}}}\)

Where,

R = resistance

Z = impedance

X = reactance

The power factor is unity when the circuit is purely resistive.

Calculation:

Power factor = R/Z = 10/10 = 1 (unity)

Concept:

Reactive power (Q) = V I sinϕ

Where,

V = Voltage in volt

I = Current in ampere

ϕ = phase difference between voltage & current

Calculation:

Voltage (V) = 230 V

Current (I) = 5 A

Phase difference (ϕ) = 30°

Q = 230 × 5 × sin30°

The correct option is 1

Concept:

Voltage v = 100 sin (90 πt - \(\frac {π} {6}\))

Current i = 100 sin (90 πt - \(\frac {\pi} {6}\))

The phase difference between voltage and current is 0 only in the resistive part. there is no inductive or capacitive part.

w = angular frequency = \(2\times \Pi \times f\) = \(90\times \Pi \)

 f =\(\frac{90\times \Pi }{2\times \Pi }\)  = 45 HZ, resistive

Additional point

At a resonance 

\(I = \frac{V}{Z}=\frac{V}{R}\)

In a series resonance circuit current vs frequency graph for the series resonance circuit

• Line voltage (line-to-line voltage) in a poly-phase system is the voltage between any two given phases.
• While the phase voltage is the voltage between any given phase and neutral.
• In delta connected circuits,

1. Line voltage is equal to phase voltage
2. Line current is equal to √3 times the phase current with a lag of 30°.

• In star connected circuits,

1. Line voltage is equal to √3 times the phase voltage with a lead of 30°.
2. Line current is equal to phase current.

• Note that neutral is available in star connection but not in delta connection.

• Inductive reactance (X_L), capacitive reactance (X_C), impedance (Z) & resistance (R) are calculated/rated in ohm (Ω).
• Reactive power (Q) is calculated/rated in VAR.
• Current (i) is calculated/rated in ampere.
• Active power (P) is calculated in watt.

In pure inductive circuit, current lags the voltage by 90°.

In pure capacitive circuit, current leads the voltage by 90°.

In R - L circuits, current lags the voltage but it is not exactly 90°.

In R - C circuits, current leads the voltage but it is not exactly 90°.

Concept:

For a sinusoidal alternating waveform

• Peak to peak value of voltage (V_p-p) = 2 (Peak value)
• The peak value of voltage (V_p) = √2 V_rms
• RMS value of voltage \(\left( {{V_{rms}}} \right) = \frac{{{V_p}}}{{\surd 2}}\)
• The average value of voltage \(\left( {{V_{avg}}} \right) = \frac{{2{V_p}}}{\pi }\)

Calculation:

V_rms = 250 V

V_p = √2 × 250 V

V_p-p = 2 × √2 × 250

∴ V_p-p = 707 V

For a balanced system the phase difference between all three phases is 120°.

For delta connection,

Line voltage is equal to phase voltage, i.e. V_L = V_ph

Line current is equal to the phasor difference of the current in two phases connected to the line, i.e.

Current in line R = I_R – I_B

Current in line Y = I_Y – I_R

Current in line B = I_B – I_Y

Where, I_R, I_Y, and I_B are the phase current in delta connection.

For a balanced system, the phasor diagram is given below.

Where,

V_R, V_Y, and V_B are the phase voltages

I_R, I_Y, and I_B are the phase currents

I_R, I_Y, and I_B are the line currents

• Line voltage lags behind phase current by an angle equal to the power factor angle φ.
• The angle between the line voltage and line current is (30° - φ)
• The angle between line current and phase current is 30°

The correct answer is option 1):Root Mean square Value

Concept:

• RMS value of voltage is said to be s the square root of the mean square of instantaneous values of the voltage signal
• AC voltages (and currents) are always given as RMS values because this allows a sensible comparison to be made with steady DC voltages
• RMS is Root Mean square Value 
• The RMS voltage (RMS) of a sinusoidal waveform is obtained by multiplying the peak voltage value by 0.7071

Vrms =\( V_m\over \sqrt {2}\)

where

Vm is the maximum voltage