**Single Phase AC Circuits MCQs ( With Explanatory Answers)**

**1.** In case of Inductive circuit, Frequency is ___________Proportional to the inductance (L) or inductive reactance (*X _{L}*

*).*

- Directly
- Inversely
- No Effect

### Show Explanatory Answer

### Show Explanatory Answer

**Answer: 1…Directly Proportional**

*Explanation:*X

_{L}=2πfL….. i.e.…. X

_{L}∞ f…… and also…..L ∞ f

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

- Directly
- Inversely
- No Effect

**3.** *In case of Inductive circuit, inductance (L) is ___________Proportional to the inductive reactance (**X _{L}*

*).*

- Directly
- Inversely
- No Effect

### Show Explanatory Answer

### Show Explanatory Answer

**Answer: 1…Directly**

**Explanation:**X

_{L}=2πfL….. i.e.…. X

_{L}∞L.

**4.** *In inductive circuit, when Inductance (L) or inductive reactance (X _{L}) increases, the circuit current decreases, but the circuit power factor ________?*

- Increases
- Also Decreases
- Remain Same
- None of the above

### Show Explanatory Answer

### Show Explanatory Answer

*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 = √ (R^{2}+X_{L}^{2}) = √ (10^{2} + 6.28^{2}) = 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 = √ (R^{2}+X_{L}^{2}) = √ (10^{2} + 12.56^{2}) = 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.
*

** 5.** *In inductive circuit, when Inductance (L) or inductive reactance (X _{L}) increases, the circuit current ________?*

- Also Increases
- Decreases
- Remain Same
- None of the above

### Show Explanatory Answer

### Show Explanatory Answer

**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 = √ (R^{2}+X_{L}^{2}) = √ (10^{2} + 6.28^{2}) = 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 = √ (R^{2}+X_{L}^{2}) = √ (10^{2} + 12.56^{2}) = 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.

**6.** *In case of Capacitive circuit, Frequency is ___________Proportional to the Capacitance (C) or Capacitive reactance (**X _{C}*

*).*

- Directly
- Inversely
- No Effect

### Show Explanatory Answer

### Show Explanatory Answer

*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)

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

- Directly
- Inversely
- No Effect

### Show Explanatory Answer

### Show Explanatory Answer

*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**

**8.** *In case of Capacitive circuit, Capacitance (C) is ___________ Proportional to the Capacitive reactance (**X _{C}*

*).*

- Directly
- Inversely
- No Effect

### Show Explanatory Answer

### Show Explanatory Answer

*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)

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

- Increases
- Decreases
- Remain Same
- None of the above

### Show Explanatory Answer

### Show Explanatory Answer

*Answer: 1. Increases.*

**Explanation: **

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

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

X_{C} = 1/2πfC = 1/(2 x 3.1415 x 50 x 5×10^{-04}F) = 6.37 Ω

Z = √ (R^{2}+X_{C}^{2}) = √ (10^{2} + 6.37^{2}) = 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^{-3}F,

V=220, R= 10 Ω, C=1000µF =1×10^{-3}F

X_{C} = 1/2πfC = 1/(2 x 3.1415 x 50 x 1×10^{-3}F) = 3.18 Ω

Z = √ (R^{2}+X_{C}^{2}) = √ (10^{2} + 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 ________?*

- Increases
- Decreases
- Remain Same
- None of the above

### Show Explanatory Answer

### Show Explanatory Answer

*Answer: 2. Decreases*

**Explanation:**

Suppose, when Capacitive reactance (Xc) = 6 Ω

V=220, R= 10 Ω, Xc = 6 Ω

Z = √ (R^{2}+X_{C}^{2}) = √ (10^{2} + 6^{2}) = 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 = √ (R^{2}+X_{C}^{2}) = √ (10^{2} + 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__________.*

- Infinite
- Maximum
- Normal
- Minimum
- Zero

### Show Explanatory Answer

### Show Explanatory Answer

*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

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

- Infinite
- Maximum
- Normal
- Minimum
- Zero

### Show Explanatory Answer

### Show Explanatory Answer

*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)

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

- Infinite
- Maximum
- Normal
- Minimum
- Zero

### Show Explanatory Answer

### Show Explanatory Answer

*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)

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

- Input = Output
- P
_{IN}= P_{OUT} - The circuit is resistive only
- The angle (θ) between Voltage and Current is Zero.

### Show Explanatory Answer

### Show Explanatory Answer

**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_______.*

- Power of Single phase Circuit.
- Voltage of Single Phase Circuit
- Current of Single phase Circuit.
- Power Factor of Single Phase Circuit
- All of the above
- None of the above

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### Show Explanatory Answer

*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 = _________?*

- Q Factor
- Demand Factor
- Diversity Factor
- Utilization Factor

### Show Explanatory Answer

### Show Explanatory Answer

**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**

*17**. Power Factor (Cos θ) =_________? *

- kW/kVA
- R/Z
- The Cosine of angle between Current and voltage
- All of the above

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### Show Explanatory Answer

*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 ___________ ?*

- Z=1/Y
- Z=1+Y
- Z=1-Y
- Z=Y
^{2}

### Show Explanatory Answer

### Show Explanatory Answer

**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 =√ ( R^{2}+X_{L}^{2}) …. In case of Inductive Circuit (*X_{L }= Inductive Reactance)

Impedance = Z =√ ( R^{2}+X_{C}^{2}) … 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

**19. Average value of a sinusoidal alternating signal is ————-for a full cycle.**

- Maximum
- Zero
- Finite Value
- Infinite

### Show Explanatory Answer

### Show Explanatory Answer

**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.

**20. The average value of a sinusoidal alternating signal is ————–**

- Equal to the maximum value
- Half of the maximum value
- 637 times the maximum value
- None of the above

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### Show Explanatory Answer

**21. Form Factor for a sinusoidal waveform is :**

- 1.21
- 0.5
- 1.11
- 0

### Show Explanatory Answer

### Show Explanatory Answer

**22. Peak Factor for a sinusoidal waveform is:**

- 1.3
- 1.02
- 1.14
- 0.5

### Show Explanatory Answer

### Show Explanatory Answer

**23. Sine Waves of the same frequency can be represented using same phasor**

- True
- False

### Show Explanatory Answer

### Show Explanatory Answer

**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.

**24. For the circuit given below, the approximate value of reactance, impedance, current and power factor is———-respectively.**

- 35 Ohms, 90 Ohms, 1 Amperes, 1
- 4 Ohms, 104.8 Ohms, 1.9 Amperes, 0.95
- 40 Ohms, 100 Ohms, 5 Amperes, 0.5
- 50 Ohms, 200 Ohms, 2 Amperes, 0.7

### Show Explanatory Answer

### Show Explanatory Answer

**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

**25. For the circuit given in previous question, the power consumed is—-**

- 200 Watts
- 361 Watts
- 300 Watts
- 500 Watts

### Show Explanatory Answer

### Show Explanatory Answer

**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

**26. Power Factor for the circuit given below is——–**

- 1
- 0.8
- 0.9
- 0.5

### Show Explanatory Answer

### Show Explanatory Answer

**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_{L}Icosθ = 22.24 Watts

Power consumed by the non-inductive resistance, P_{R} = I^{2}R = 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

**27. For the circuit given in question 6, assume the inductor is replaced by a capacitor of 75 microfarads. The power consumed, maximum charge and maximum energy in the circuit is————respectively**

- 300 Watts, 0.1 C, 4 Joules
- 339 Watts, 0.2 C, 3 Joules
- 100 Watts, 0.01C, 12 Joules
- 150 Watts, 0.4 C, 2 Joules

### Show Explanatory Answer

### Show Explanatory Answer

**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 = I^{2}R

Now, I = V/Z

V = Supply Voltage, = 200 Volts

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_{m}^{2} = 3 Joules

**28. Alternating current across a conductor is ———**

- Uniform throughout
- More at the outer surface than at the core
- More at the core than at the outer surface
- Zero at the core

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### Show Explanatory Answer

**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.

**29. For a series RLC circuit at resonance, the voltage drop across capacitive and inductive components is ——-**

- Large
- Small
- Zero
- Infinite

### Show Explanatory Answer

### Show Explanatory Answer

**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.

**30. Parallel AC systems are used frequently than series AC systems**

- True
- False

### Show Explanatory Answer

### Show Explanatory Answer

**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.

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nice one

Thanks Dear

Its very Helpfull… Thanks

Welcome Dear 🙂

Very good thanks

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

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.

We have correct it. Thanks for review.

Sir please provide other mcqs..

Sir please provide other mcqs like this… other topics

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….

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

We are working on it. Thanks for positive feedback.

plz provide the answers of MCQs