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Home » AC Fundamentals » Single Phase AC Circuits MCQs ( With Explanatory Answers)

Single Phase AC Circuits MCQs ( With Explanatory Answers)

Single Phase AC Circuits MCQs ( With Explanatory Answers)

1. In case of Inductive circuit, Frequency is ___________Proportional to the inductance (L) or inductive reactance (XL).

1. Directly
2. Inversely
3. No Effect

Explanation:
XL =2πfL….. i.e.…. XL∞ f…… and also…..L ∞ f

2. In case of Inductive circuit, Frequency is ___________ Proportional to the Current.

1. Directly
2. Inversely
3. No Effect

Explanation:

3. In case of Inductive circuit,  inductance (L) is ___________Proportional to the  inductive reactance (XL).

1. Directly
2. Inversely
3. No Effect

Explanation:
XL =2πfL….. i.e.…. XL∞L.

4. In inductive circuit, when Inductance (L) or inductive reactance (XL) increases, the circuit current decreases, but the circuit power factor ________?

1.   Increases
2.   Also Decreases
3.   Remain Same
4.   None of the above

Explanation:
Suppose, when Inductance (L) = 0.02H

V=220, R= 10 Ω, L=0.02 H, f=50Hz.

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

Z = √ (R2+XL2) = √ (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.

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

Z = √ (R2+XL2) = √ (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 XL increases, the circuit current decreases, but the circuit power factor also Decreases.

5. In inductive circuit, when Inductance (L) or inductive reactance (XL) increases, the circuit current  ________?

1.   Also Increases
2.   Decreases
3.   Remain Same
4.   None of the above

We know that, I = V / R,
but in inductive circuit, I = V/XL
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.

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

Z = √ (R2+XL2) = √ (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.

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

Z = √ (R2+XL2) = √ (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 XL increases, the circuit current decreases, and Vice Virsa.

6. In case of Capacitive circuit, Frequency is ___________Proportional to the Capacitance (C) or Capacitive reactance (XC).

1. Directly
2. Inversely
3. No Effect

Explanation:

In capacitive circuit,

XC= 1/2πfC, and

f = 1/2πXC

So here we can see that,

f = 1/ C …and also…f = 1/ XC.

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

7. In case of Capacitive circuit, Frequency is ___________ Proportional to the Current.

1. Directly
2. Inversely
3. No Effect

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

8.  In case of Capacitive circuit, Capacitance (C) is ___________ Proportional to the Capacitive reactance (XC).

1. Directly
2. Inversely
3. No Effect

Explanation:
In capacitive circuit,

XC = 1/2πfC, …i.e,

Xc  ∞ 1/C

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

9. In a Capacitive circuit, when Capacitance (C) increases, ( the circuit current also increases), then the circuit power factor ________?

1.   Increases
2.   Decreases
3.   Remain Same
4.   None of the above

Explanation:

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

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

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

Z = √ (R2+XC2) = √ (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

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

Z = √ (R2+XC2) = √ (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.

10.  In a Capacitive circuit, when Capacitive reactance increases, then the circuit power factor ________?

1.   Increases
2.   Decreases
3.   Remain Same
4.   None of the above

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

V=220, R= 10 Ω, Xc = 6 Ω

Z = √ (R2+XC2) = √ (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+XC2) = √ (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.

11. If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be__________.

1. Infinite
2. Maximum
3. Normal
4. Minimum
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

12. In pure inductive circuit, the power is __________?

1. Infinite
2. Maximum
3. Normal
4. Minimum
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)

13. In pure capacitive circuit, the power is __________?

1. Infinite
2. Maximum
3. Normal
4. Minimum
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)

14. If Power factor = Cos θ = 1, it means that _____________.

1. Input = Output
2. PIN = POUT
3. The circuit is resistive only
4. The  angle (θ) between Voltage and Current is Zero.

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.

15. Using P=VI Cos φ Formula, We Can Find_______.

1. Power of Single phase Circuit.
2. Voltage of Single Phase Circuit
3. Current of Single phase Circuit.
4. Power Factor of Single Phase Circuit
5. All of the above
6. None of the above

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

16. Reciprocal of Power Factor = _________?

1. Q Factor
2. Demand Factor
3. Diversity Factor
4. Utilization 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

17. Power Factor (Cos θ) =_________?

1. kW/kVA
2. R/Z
3. The Cosine of angle between Current and voltage
4. All of the above

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.

18. The relationship between Impedance (Z) and Admittance(Y) is ___________ ?

1. Z=1/Y
2. Z=1+Y
3. Z=1-Y
4. Z=Y2

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+XL2) …. In case of Inductive Circuit (*XL = Inductive Reactance)

Impedance = Z =√ ( R2+XC2) … in Case if Capacitive Circuit (*Xc = 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.

Y = 1/Z

= 1/ (V/I)

= I/V———> I=VY

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

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1. nice one

• Welcome Dear 🙂

3. Very good thanks

• نور الدين عبد الكريم ابراهيم الحمد<br />Welcome Dear.

4. question number 7 has been incorrectly interpreted here.<br />we know,<br />I=V/Xc<br />and Xc=1/(2*pi*f*C)<br />that makes f directly proportional to I.

5. any body can illustrate me question no 4 answer plz in my mind it is like this when inductance increased the power factor of the circuit decreases but here it is inversed……..<br />

• Dear..Did you check it properly…???? Try again….

6. electrical engineer

plz provide the answers of MCQs……I cant find them

• We are working on it. Thanks for positive feedback.

7. plz provide the answers of MCQs