Single Phase AC Circuits MCQs ( With Explanatory Answers)

Answer: 2…Also Decreases
Explanation:  
Suppose, when Inductance (L) = 0.02H

V=220, R= 10 Ω, L=0.02 H, f=50Hz.

X_L = 2πfL = 2 x 3.1415 x 50 x 0.02 = 6.28 Ω

Z = √ (R²+X_L²) = √ (10² + 6.28²) = 11.8 Ω

I = V/Z = 220/11.8 = 18.64 A

Cos θ = R/Z = 10/11.05 = 0.85

Now we increases Inductance (L) form 0.02 H to 0.04 H,

V=220, R= 10 Ω, L=0.04 H, f=50Hz.

X_L = 2πfL= 2 x 3.1415 x 50 x 0.04 = 12.56 Ω

Z = √ (R²+X_L²) = √ (10² + 12.56²) = 16.05 Ω

I = V/Z = 220 / 16.05 = 13.70 A

Cos θ = R/Z = 10/16.05 = 0.75

Conclusion:

We can see that, When inductance (L) was 0.02, then circuit current were 18.64 A, and Circuit power factor was (Cos θ) = 0.85.

But, when Circuit inductance increased from 0.02H to 0.04 H, then current decreased from13.70 A to 18.64A, also Power Factor (Cos θ) decreased from 0.85 to 0.75.

Hence proved,

In inductive circuit, when inductive reactance X_L increases, the circuit current decreases, but the circuit power factor also Decreases.

Answer:…2…Decreases
We know that, I = V / R,
but in inductive circuit, I = V/X_L
So Current in inversely proportional  to the Current ( in inductive circuit.
Let’s check with an example.. 

Suppose, when Inductance (L) = 0.02H

V=220, R= 10 Ω, L=0.02 H, f=50Hz.

X_L = 2πfL = 2 x 3.1415 x 50 x 0.02 = 6.28 Ω

Z = √ (R²+X_L²) = √ (10² + 6.28²) = 11.8 Ω

I = V/Z = 220/11.8 = 18.64 A

Now we increases Inductance (L) form 0.02 H to 0.04 H,

V=220, R= 10 Ω, L=0.04 H, f=50Hz.

X_L = 2πfL= 2 x 3.1415 x 50 x 0.04 = 12.56 Ω

Z = √ (R²+X_L²) = √ (10² + 12.56²) = 16.05 Ω

I = V/Z = 220 / 16.05 = 13.70 A

Conclusion:

We can see that, When inductance (L) was 0.02, then circuit current were 18.64 A,

But, when Circuit inductance increased from 0.02H to 0.04 H, then current decreased from13.70 A to 18.64A.

Hence proved,

In inductive circuit, when inductive reactance X_L increases, the circuit current decreases, and Vice Virsa.

Answer:     2.    Inversely 
Explanation:

In capacitive circuit,

X_C= 1/2πfC, and

f = 1/2πX_C C

So here we can see that,

f = 1/ C …and also…f = 1/ X_C. 

So, in a capacitive circuit, frequency is inversely proportional to the Capacitance (C) and Capacitive reactance (Xc)

Answer:    1     Directly

Explanation:
We know that,

I = V/R

but in capacitive circuit

I = V/Xc……(1)

But we also know that

Xc = 1/2πfC ….(2)….. i.e ….. Xc ∞ 1/f
Puttint (2) into (1)
I = V/ (1/2πfC)…i.e ..I = V x 2πfC

Hence Proved, I ∞ f

Answer:     2.    Inversely
Explanation:
In capacitive circuit,

X_{C =} 1/2πfC, …i.e,

Xc  ∞ 1/C

So, in a capacitive circuit, Capacitance (C) is inversely proportional to the Capacitive reactance (Xc)

Answer:      1.    Increases.
Explanation:

Suppose, when Capacitance (C) = 500µF = or 5×10^-04F

V=220, R= 10 Ω, C=500µF = (5×10^-04F), f=50Hz.

X_C = 1/2πfC = 1/(2 x 3.1415 x 50 x 5×10^-04F) = 6.37 Ω

Z = √ (R²+X_C²) = √ (10² + 6.37²) = 11.85 Ω

I = V/Z = 220/11.8 = 18.56 A

Cos θ = R/Z = 10/11.85 = 0.84

Now we increased Capacitance (C) = 1000µF = or 1×10^-3F,

V=220, R= 10 Ω, C=1000µF =1×10^-3F

X_C = 1/2πfC = 1/(2 x 3.1415 x 50 x 1×10^-3F) = 3.18 Ω

Z = √ (R²+X_C²) = √ (10² + 3.18 ²) = 10.49 Ω

I = V/Z = 220/11.8 = 20.97A = 21A

Cos θ = R/Z = 10/11.85 = 0.95

Conclusion:

We can see that, When Capacitance (C) was 500µF, then circuit current were 18.56 A, and Circuit power factor was (Cos θ) = 0.84.

But, when we increased Circuit Capacitance from 500µF to 1000µF, then current also increased from18.56 A to 21A, also Power Factor (Cos θ) increased from 0.84 to 0.95.

Hence proved,

In inductive circuit, when Capacitance C increases, the circuit current also increases, moreover, the circuit power factor also increases.

Answer:   2.   Decreases

Explanation:
Suppose, when Capacitive reactance (Xc) = 6 Ω

V=220, R= 10 Ω, Xc = 6 Ω

Z = √ (R²+X_C²) = √ (10² + 6²) = 11.66 Ω

Cos θ = R/Z = 10/11.66 = 0.85

Now we increased Capacitive reactance = 10 Ω

V=220, R= 10 Ω, Xc = 10 Ω

Z = √ (R²+X_C²) = √ (10² + 10 ²) = 14.14 Ω

Cos θ = R/Z = 10/14.14 = 0.70

Conclusion:

We can see that, When Capacitive reactance (Xc) = 6 Ω, then circuit power factor was (Cos θ) = 0.85.

But, when we increased Capacitive reactance from 6 Ωto 10 Ω, then Power Factor (Cos θ) decreased from 0.85 to 0.70.

Hence proved,

In Capacitive circuit, when Capacitive reactance (Xc) increases, then the circuit power factor also increases.

Answer:          5.       Zero
Explanation:If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be zero. The reason is that,
We know that Power in AC Circuit
P= V I Cos φ
if angle between current and Voltage are 90 ( φ = 90) Degree. then
Power P = V I Cos ( 90) = 0
[ Note that Cos (90) = 0] So if you put Cos 90 = 0→Then Power will be Zero

Answer            5.        Zero
Explanation: We know that in Pure inductive circuit, current is lagging by 90 degree from voltage ( in other words, Voltage is leading 90 Degree from current) i.e the pahse difference between current and voltage is 90 degree.
So  If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be zero. The reason is that,
We know that Power in AC Circuit
P= V I Cos φ
if angle between current and Voltage are 90 ( φ = 90) Degree. then
Power P = V I Cos ( 90) = 0
[ Note that Cos (90) = 0] So if you put Cos 90 = 0→Then Power will be Zero (In pure Inductive circuit)

Answer:              5.      Zero
Explanation:
We know that in Pure capacitive circuit, current is leading by 90 degree from voltage ( in other words, Voltage is lagging 90 Degree from current) i.e the phase difference between current and voltage is 90 degree.
So  If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be zero. The reason is that,
We know that Power in AC Circuit
P= V I Cos φ
if angle between current and Voltage are 90 ( φ = 90) Degree. then
Power P = V I Cos ( 90) = 0
[ Note that Cos (90) = 0] So if you put Cos 90 = 0→Then Power will be Zero (In pure capacitive circuit)

Answer:      4.     Theangle θ  between Voltage and Current is Zero
Explanation: We know that Power factor = Cos θ 
Given value of Power factor is = 1.
But, this is only possible when θ = 0 ( in case of Power factor = Cos θ).
I.e, Cosθ = Cos (0) = 1.

Answer:          5.       All of the above
Explanation: As we know that it depends of the given values or data. but generally we can find all these quantity this way by this formula.

For Power: P=VI Cos φ

For Voltage = V = P / (I Cos φ)

For Current = I = P / (V Cos φ)

For Power Factor =  Cos φ = P / VI

Answer:     1.  Q Factor

Explanation:

Opposite of Power factor is called the Q-Factor or Quality Factor of a Coil or its figure of merit.

Q Factor = 1/ Power Factor=1/Cosθ= Z/R    …    (Where Power Factor Cosθ = R/Z)

If R is too small with respect to Reactance

Then Q factor = Z/R = ωL/R = 2πfL / R    …    (ωL/R = 2πf)

Also Q = 2π (Maximum Energy Stored/Energy dissipate per Cycle) in the coil.

For More Detail : Q Factor in Electrical and Electronics Engineering

Answer.     4.     All of the above.

Explanation:As we know that power in single phase AC Circuits = P = VI Cos θ. Therefore Cos θ = P / V I ===> Cos θ = P (in Watts) / V I (in Volt- Ampere) ===> Cos θ = W/VI .
And Cos θ = R/Z = the ratio between Resistance and Impedance = Resistance / Impedance = R / Z
Also Cos θ = The Cosine of angle between Current and voltage = P = V I Cos θ.

Answer:   1.    Z=1/Y

Explanation:

Impedance: The overall resistance in AC Circuit is Called Impedance. It is represented by Z and the unit of impedance is same like resistance i.e. = Ω (Ohm) where is

Impedance = Z =√ ( R²+X_L²) …. In case of Inductive Circuit (*X_L = Inductive Reactance)

Impedance = Z =√ ( R²+X_C²) … in Case if Capacitive Circuit (*X_c = Capacitive Reactance)

Admittance: The Admittance is defined as the reciprocal of impedance just as conductance is the reciprocal of resistance i.e. it is represented by Y.

Admittance,

Y = 1/Z

= 1/ (V/I)

= I/V———> I=VY

The unit of admittance is Siemens and its unit symbol is S.

You may also read about Admittance here

What is Admittance and how does calculate it?

Ans:      ( 2 )

For a sinusoidal alternating signal, the average value is equal to the steady state value of that electrical parameter which transfers the same change across the circuit as its alternating counterpart. It is determined by averaging all the instant values of the electrical parameter. Since the sinusoidal wave is symmetrical, the average value in the positive cycle would be equal to the average values in the negative cycle. Hence it is zero for a full cycle.

Ans.      ( 3 )

Consider the sinusoidal alternating current signal given below.

This is a sinusoidal current waveform, where the value of current at any point is given as:

i = I_m sin⁡θ

For a small segment of thickness dθ, the area would be:  idθ

Integrating for a half cycle, the area would be:

The average value would be given as :

Ans:      ( 3 )

The form factor for a sinusoidal waveform is the ratio of its RMS value and its average value. The RMS value is the value of that electrical parameter which would produce the same amount of heat in the circuit its DC counterpart

Mathematically, Form Factor, 

Ans:       ( 1.14 )

Peak Factoris the ratio of maximum value to RMS value. It indicates the maximum value.

Mathematically, 

Answer:      ( 1 )

Phasor Diagram is the representation of a sine wave using a single line rotating in counter-clockwise direction with a constant velocity. While the length of the line represents the magnitude, i.e. the maximum value of the alternating parameter, its orientation with the X-axis represents the frequency. Two sine waves of same frequency would have represented by rotating phasors which maintain a fixed position relative to each other.

Ans.      ( 2 )

For the above circuit, the inductive reactance is given as:

The Resistance, R = 100 Ohms

Substituting the values, f = 50 Hz, L = 0.1H, we get  = 31.4 Ohms approx.

The impedance is given as:

Substituting the corresponding values, we get, Z = 104.8 Ohms approx.

The value of current, I = V/Z = 1.9 Amperes

The phase difference is given as:

Calculating we get, ∅  = 17.43 Degrees

Hence Power Factor, cos∅ = 0.95

Answer:      ( 2 )

Average power consumed is the power consumed in the resistor, since an ideal inductor does not consumes power. Thus power consumed is given as:

P = VIcos∅

V = Supply Voltage = 200 Volts

Now, from above, I = 1.9 Amperes

Power factor, cos∅ = 0.95

Hence, P = 361 Watts

Ans:      ( 2 )

We have the following parameters given

Non-Inductive Resistance, R = 100 Ohms

Inductance of Coil, L = 0.1 H

Resistance of Coil, R_L = 10 Ohms

Frequency, f = 50 Hz

Supply Voltage, V = 200 Volts

Inductive Reactance, X_L = 2_{πfl =} 31.4 Ohms

Therefore, Impedance of Coil,  = 

Total Impedance, Z = Z_L+ R = 132.95 Ohms

Total current passing through the circuit, I = V/Z = 1.5 Amperes

Now, power factor for only the coil, cosθ = R_L/Z_L = 0.3

Voltage across coil, V_L = IZ_L = 49.425 Volts

Power Consumed by the coil, P_L = V_LIcosθ = 22.24 Watts

Power consumed by the non-inductive resistance, P_R = I²R = 225 Watts

Total power consumed or real power, P_rl = P_L + P_R = 247.4 Watts

Total input power, P_A = VI = 300 Watts

Hence, total power factor, p_f = P_rl/P_A = 0.8

Answer:      ( 2 )

Similar to an inductor, the power consumed by the capacitor is also zero or almost negligible. Hence for the circuit, the power consumed is the power consumed by the resistance only.

Therefore, power consumed = I²R

Now, I = V/Z

V = Supply Voltage, = 200 Volts

Z = Impedance given as:

Now, current, I = 1.84 Amperes

Therefore, Power consumed = 338.56 Watts = 339 Watts approx

Maximum charge in the capacitor, Q_max = CV_m= CV√2 = 0.02 Columbs

Maximum energy, E_max = 0.5CV_m² = 3 Joules

Answer:

For a straight conductor, the alternating current produces a magnetic flux, which is more at the centre compared to the surface. This causes inductance to be more at the centre than at the surface. Thus current is more at the outer surface where the inductance is less. Hence useful cross section of the conductor is lesser than the actual one and the effective resistance is higher.

Answer:      ( 1 )

When current across a series RLC circuit is in phase with the supply voltage, it is said to be in resonance. Consider the below series RLC circuit.

Capacitive Reactance = X_C = 1/2πfC

Inductive Reactance = X_L = 2πfL

Since current is in phase with supply voltage, the total phase difference would be zero and hence X_C = X_L.

Therefore, impedance, Z = R and current, I = V/Z = V/R

Since opposition to flow of current is only due to the resistor, maximum current flows across the circuit and thus, voltage drop across the capacitive and inductive components is maximum. Thus series resonance is avoided in power systems.

Answer:      ( 1 )

Parallel AC systems are used frequently than series AC systems because all the electrical appliances or devices are operated with the same supply voltage and also each one needs to be operated independently using its corresponding switch.