Alternating Current MCQ Chapter 7
Below are some of the very important NCERT Alternating Current MCQ Class 12 Physics Chapter 7 with answers. These Alternating Current MCQ have been prepared by expert teachers and subject experts based on the latest syllabus and pattern of CBSE Term 1 examination.
We have given these Alternating Current MCQ Class 12 Physics questions with answers to help students understand the concept.
MCQ Questions for Class 12 Physics are very important for the latest CBSE Term 1 and Term 2 pattern. These MCQs are very important for students who want to score high in CBSE Board, NEET and JEE exam.
We have put together these NCERT Alternating Current MCQ for Class 12 Physics Chapter 7 with answers for the practice on a regular basis to score high in exams. Refer to these MCQs questions with answers here along with a detailed explanation.
1. Alternating current cannot be measured by DC ammeter because
- AC is virtual
- AC changes its direction
- AC cannot pass through DC ammeter
- average value of complete cycle is zero
2. An alternating current of equivalent value of Io/√2 is
- RMS current
- DC current
- all of these
3. In an AC circuit I = 100sin200πt. The time required for current to achieve its peak value will be
- 1/200 second
- 1/400 second
- 1/100 second
- 1/300 second
4. The ratio of mean value over half cycle to RMS value of AC is
5. The peak value of an alternating EMF E given by E = Eocosωt is 10V and its frequency is 50 Hz. At that time t=1/600 s, the instantaneous EMF is
- 5√3 V
- 5 V
- 10 V
- 1 V
6. The frequency of an alternating voltage is 50 cps and its amplitude is 120 V. Then the RMS value of voltage is
- 56.5 V
- 70.7 V
- 101.3 V
- 84.8 V
7. An electric heater of 40 ohm is connected to a 200V, 50 Hz main supply. The peak value of electric current flowing in the circuit is approximately
- 10 A
- 5 A
- 7 A
- 2.5 A
8. In the case of an inductor
- voltage leads the current by π/4
- voltage leads the current by π/3
- voltage leads the current by π/2
- voltage lacks the current by π/2
9. A resistance of 20 ohm is connected to a source of an alternating potential V = 220sin(100πt). The time taken by the current to change from its peak value to RMS value is
- 2.5 x 10-3 s
- 25 x 10-3 s
- 0.25 s
- 0.2 s
10. The RMS value of an AC of 50 Hz is 10A. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be
- 1 x 10-2 s and 7.07 A
- 2 x 10-2 s and 14.14 A
- 5 x 10-2 s and 14.14 A
- 5 x 10-3 s and 7.07 A
11. Determine The RMS value of the EMF given by
E (in V) = 8 sin(ωt) + 6 sin(2ωt)
- 10√2 V
- 10 V
- 5√2 V
- 7√2 V
12. An alternating current of frequency f is flowing in a circuit containing a resistor of resistance R and a choke of inductance L in series. The impedance of the circuit is
- R + 2πfπL
- √(R2 + L2)
- √(R2 + 2fπL)
- √(R2 + 4π2f2L2)
13. A generator produces a voltage that is given by V = 240 sin 120t V, where t is in seconds. The frequency and RMS voltage and nearly
- 19 Hz and 120 V
- 19 Hz and 170 V
- 60 Hz and 240 V
- 754 Hz and 170 V
14. The instantaneous voltage through a device of impedance 20 ohm is e = 80sin100πt. The effective value of the current is
- 1.732 A
- 2.828 A
- 3 A
- 4 A
15. A 15μF capacitor is connected to 220V, 50Hz source. Find the capacitive reactance and the RMS current
- 212.1 Ω; 1.037 A
- 212.1 Ω; 2.037 A
- 412.1 Ω; 1.037 A
- 412.1 Ω; 2.037 A
Click Below To Learn Physics Term 1 Syllabus Chapter-Wise MCQs
16. In an AC circuit an alternating voltage V = 200√2 sin100t is connected to a capacitor of capacity 1μF. The RMS value of the current in the circuit is
17. In an LR circuit, the value of L is (0.4/π) and the value of R is 30Ω. If in the circuit, an alternating EMF of 200V at 50 cps is connected, the impedance of the circuit and current will be
- 50 Ω, 4 A
- 40.4 Ω, 5 A
- 30.7 Ω, 6.5 A
- 11.4 Ω, 17.5 A
18. In an AC circuit the voltage applied is E = Eosinωt. The resulting current in the circuit is I = Iosin(ωt – π/2). The power consumption in the circuit is given by
- P = EoIo / 2
- P = EoIo / √2
- P = √EoIo
- P = 0
19. In an LCR circuit AC circuit, the voltage across each of the components L, C and R is 50V. The voltage across the LC combination will be
- 0 V
- 50 V
- 50√2 V
- 100 V
20. Find the capacitive reactance of a 10μF capacitor, when it is part of a circuit whose frequency is 100 Hertz.
- 159.2 Ω
- 412.1 Ω
- 612.1 Ω
- 812.1 Ω
21. The resonant frequency of a circuit is f. If the capacitance is made 4 times the initial values, than the resonant frequency will become
22. A coil of 10Ω and 10mH is connected in parallel to a capacitor of 0.1μF. The impedance of the circuit at resonance is
23. Which of the following curves correctly represent the variation of capacitive reactance (Xc) with frequency (f)?
24. How does the current in an RC circuit vary when the charge on the capacitor builds up?
- it decreases linearly
- it increases linearly
- it decreases exponentially
- it increases exponentially
25. The impedance in a circuit containing a resistance of 1Ω and an inductance of 0.1 H in series for AC of 50 Hz is
- √10 Ω
- 10√10 Ω
- 100 Ω
- 100√10 Ω
26. An AC circuit contains a resistance R, capacitance C and inductance L in series with the source of EMF e=eosin(ωt+f). The current through the circuit is maximum when
- ω2 = LC
- ωL = 1/ωC
- R = L = C
- ω = LCR
27. A charged 30μF capacitor is connected to a 27 mH inductor. The angular frequency of free oscillations of the circuit is
- 1.1 x 103 rad s-1
- 2.1 x 103 rad s-1
- 3.1 x 103 rad s-1
- 4.1 x 103 rad s-1
28. The frequency of the output signal becomes ________ times by doubling the value of the capacitance in the LC oscillator circuit.
29. In an LCR circuit, the sharpness of resonance depends on
- all of these
30. The average power dissipation in a pure capacitor in AC circuit is
31. In a series resonant circuit, having L, C and R as its elements, the resonant current is ‘i’. The power dissipated in circuit at resonance is
32. An AC supply gives 30 VRMS which passes through 10Ω. The power dissipated in it is
- 45√2 W
- 90√2 W
- 45 W
- 90 W
33. In a series LCR circuit alternating EMF(e) and current(i) are given by equation v=vosin(ωt), i=iosin(wt+π/3). The average power dissipated in the circuit over a cycle of AC is
34. In an AC circuit, the current flowing in inductance is I = 5sin(100t-π/2)A and the potential difference V = 200sin(100t)V. The power consumption is equal to
- 20 W
- 40 W
- 1000 W
35. The power factor in an AC series LR circuit is
- √(R2 + L2ω2)
- R√(R2 + L2ω2)
- R/(√R2 + L2ω2)
36. A transformer is employed in
- Convert DC into AC
- Convert AC into DC
- Obtain a suitable DC voltage
- Obtain a suitable AC voltage
37. The loss of energy in the form of heat in the iron core of a transformer is
- Copper loss
- Iron loss
- Mechanical loss
- None of these
38. The core of any transformer is laminated so as to
- Make it light weight
- Make it robust and strong
- Increase the secondary voltage
- Reduce the energy loss due to eddy currents
39. A step up transformer has a transformation ratio 5:3. What is the voltage in secondary if voltage in primary is 60 V?
- 60 V
- 180 V
- 20 V
- 100 V
40. A transformer has 50 turns in the primary and 100 in the secondary. If the primary is connected to a 220 V DC supply, what will be the voltage across the secondary?
- 19 V
- 30 V
- 62 V
- 0 V
41. The primary of a transformer has 400 turns while the secondary has 2000 turns. If the power output from the secondary at 1000 V is 12kW, what is the primary voltage?
- 200 V
- 400 V
- 300 V
- 500 V
42. A step down transformer is used on a 1000V line to deliver 20A at 120V at the secondary coil. If the efficiency of the transformer is 80% the current drawn from the line is
- 0.3 A
- 3 A
- 30 A
- 24 A
43. If the RMS current in a 50 Hz AC circuit is 5 A, the value of the current 1/300 s after its value becomes zero is
- 5√2 A
- 5√(3/2) A
- ⅚ A
- 5√2 A
44. An alternating current generator has an internal resistance Rg and an internal reactance Xg. It is used to supply power to a passive load consisting of a resistance Rg and a reactance XL. For maximum power to be delivered from the generator to the laid, the value of XL is equal to
45. When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220 V. This means
- Input voltage cannot be AC voltage, but a DC voltage
- Maximum input voltage is 220 V
- The meter reads not v but <v2> and is calibrated to read √<v2>
- The pointer of the meter is stuck by some mechanical defect
46. To reduce the resonant frequency in an LCR series circuit with a generator
- The generator frequency should be reduced
- Another capacitor should be added in parallel to the first
- The iron core of the inductor should be removed
- Dielectric in the capacitor should be removed
47. Which of the following combinations should be selected for better tuning of an LCR circuit used for communication?
- R = 20Ω, L = 1.5 H, C = 35 μF
- R = 25Ω, L = 2.5 H, C = 45 μF
- R = 15Ω, L = 3.5 H, C = 30 μF
- R = 25Ω, L = 1.5 H, C = 45 μF
48. An inductor of reactance 1Ω and a resistor of 2Ω are connected in series to the terminals of a 6V(rms) AC source. The power dissipated in the circuit is
- 8 W
- 12 W
- 14.4 W
- 18 W
49. The selectivity of a series LCR AC circuit is large when
- L is large, R is large
- L is small, T is small
- L is large , R is small
- L = R
50. The phase difference between the current and the voltage in series LCR circuit at resonance is
1. (4) 2. (1) 3. (2) 4. (3) 5. (1) 6. (4) 7. (3) 8. (3) 9. (1) 10. (3) 11. (3) 12. (4) 13. (2) 14. (2) 15. (1) 16. (2) 17. (1) 18. (4) 19. (1) 20. (1) 21. (1) 22. (4) 23. (2) 24. (3) 25. (2) 26. (2) 27. (1) 28. (4) 29. (4) 30. (4) 31. (2) 32. (4) 33. (3) 34. (1) 35. (4) 36. (4) 37. (2) 38. (4) 39. (4) 40. (4) 41. (1) 42. (2) 43. (2) 44. (3) 45. (3) 46. (2) 47. (3) 48. (3) 49. (3) 50. (4)
Assertion-Reasoning Based MCQ
- Both assertion and reason are true and reason is the correct explanation of assertion.
- Both assertion and reason are true but reason is not the correct explanation of assertion.
- Assertion is true but reason is false.
- Assertion is false but reason is true.
1. Assertion AC is more dangerous in use than DC
Reason It is because the peak value of AC is greater than indicated value
2. Assertion Average value of AC over a complete cycle is always zero
Reason average value of AC is always defined over half cycle
3. Assertion The alternating current lags behind the EMF by a phase angle of when AC flows through and inductor
Reason The inductive reactance increases as the frequency of AC source decreases
4. Assertion Capacitor serves as a block for DC and offers an easy path to AC
Reason Capacitive reactance is inversely proportional to frequency
5. Assertion In series LCR resonant circuit the impedance is equal to the ohmic resistance
Reason At resonance the inductive reactance exceeds the capacitive reactance
6. Assertion An alternating current shows magnetic effect
Reason Alternating current varies with time
7. Assertion In series LCR circuit resonance can take place
Reason Resonance takes place in inductance and capacitive reactance are equal and opposite
8. Assertion Power factor correction is must in heavy machinery
Reason A low power factor implies larger power loss in transmission
9. Assertion Choke coil is preferred over a registered to adjust current in an AC circuit
Reason Power factor for inductance is zero
10. Assertion When AC circuit containing resistor only its power is minimum
Reason Power of a circuit is independent of phase angle
11. Assertion A transformer cannot work on DC supply
Reason DC change is neither in magnitude nor in direction
12. Assertion A laminated core is used in transformer to increase eddy currents
Reason The efficiency of a transformer increases with increase in eddy currents
13. Assertion Soft iron is used as a core of transformer
Reason Area of hysteresis loop for soft iron is small
14. Assertion An AC generator is based on the phenomenon of electromagnetic induction
Reason In single coil we consider self induction only
Assertion-Reasoning Based MCQ Answers
AC is more dangerous in use than DC. It is because the peak value of AC is greater than the indicated value.
The mean or average value of alternating current or EMF during half cycle is given by
Im = 0.636 Io
Em = 0.6363 Eo
During the next half cycle, the mean value of AC will be equal in magnitude but opposite in direction. For this reason the average value of AC over a complete cycle is always zero. So the average value is always defined over a half cycle of AC.
When AC flows through an inductor current lags behind the EMF, by phase of π/2 inductive reactance
XL = ωL = 2πfL
So, when frequency increases correspondingly inductive reactance also increases.
The capacitive reactance of capacitor is given by
XC = 1/ ωC = 1/2πfC
So this is infinite for DC and has a very small value for AC. Hence, a capacitor blocks DC.
In series resonance circuit inductive reactance is equal to capacitive reactance.
ωL = 1/ωC
Like direct current, an AC also produces magnetic field. But the magnitude and direction of the field goes on changing continuously with time.
At resonant frequency,
XL = XC, Z = R (minimum)
A heavy machinery requires a large power.
The average power is given by,
Pav = ErmsIrmscosΦ
The required power can be supplied to the heavy machinery either by supplying larger current or by improving power factor. The first method is costly. Hence, the second one is used.
We can use a capacitor of suitable capacitance as a choke coil, because average power consumed per cycle in an ideal capacitor is zero. Therefore, like a choke coil a condenser can reduce AC without power dissipation.
The power of an AC circuit is given by,
P = EIcosΦ
Where cosΦ is a power factor and is Φ phase angle. In case of circuit containing resistance only, phase angle is zero and power factor is equal to 1. Therefore power is maximum in case of circuit containing resistor only.
Transformer works on AC only AC changes in magnitude as well as in direction and induced EMF.
Large eddy currents are produced in non laminated iron core of the transformer by induced EMF, as the resistance of bulk iron core is very small. By using thin iron sheets are score the resistance is increased. Laminating the core substantially reduces the eddy currents. Eddy currents heat up the core of the transformer. More the eddy current greater the loss of energy and efficiency goes down.
Hysteresis loss in the core of transformer is directly proportional to the hysteresis loop area of the core material. Since soft iron has narrow hysteresis loop area, that is why soft iron core is used in transformer.
According to electromagnetic induction, whenever the magnetic flux changes and EMF will be induced in the coil.
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Case-Study Based MCQ
1. The figure shows a series LCR circuit.
For such a citcuit, the impedance Z is given by Z = √R2 + (XL – XC)2 where XL and XC are inductive and capacitive resistances respectively. As the frequency of AC is increased, at a particular frequency, XL becomes equal to XC. For that frequency, maximum current occurs. This is because the impedcane becomes equal to the least value which is R.
Current through the circuit is I = V/R. This circuit behaves like a pure resistive circuit and current and voltage will be in phase. This is called resonance. Frequency of AC at which resonance occurs is called resonant frequency. If frequency is less than the resonant frequency, then the capacitive reactance will be more. The circuit will be capacitive in nature.
If frequency is more than the resonant frequency, inductive reactance will be more. Circuit is inductive in nature and the current lags behind the voltage by a phase of π/2.
An LCR circuit which has a resistance 50 ohm has a resonant angular frequency 2 x 103 rad/s. At resonance, the voltage across the resistance and inductance are 25 V and 20 V respectively.
(i) The value of inductance is
(ii) The value of capacitive reactance is
(d) 12.5 μF
(iii) The impedance at resonance is
(a) 50 ohm
(b) 16 ohm
(c) 64 ohm
(d) 25 ohm
(iv) Which of the following angular frequency of AC will see the circuit as inductive in nature?
(a) 1.5 x 103 rad/s
(b) 103 rad/s
(c) 2 x 103 rad/s
(d) 5 x 103 rad/s
(v) At angular frequency 103 rad/s, the nature of circuit is
(d) none of these
2. A series LCR circuit consist of series combination of a resistance, an inductor and a capacitance. A similar series LCR circuit is shown in figure. The given series LCR circuit is connected across a 200 V 60 Hz line consisting of capacitive reactance 30 ohm a non-inductive resistor of 44 ohm and a coil of inductive reactance 90 ohm and resistance 36 ohm.
(i) Calculate the total impedance of the circuit.
(a) 1000 ohm
(b) 100 ohm
(c) 3600 ohm
(d) 4900 ohm
(ii) Calculate the current flowing in the circuit.
(a) 1 A
(b) 5 A
(c) 2 A
(d) 10 A
(iii) What is the impedance of the coil?
(a) 97 ohm
(b) 87 ohm
(c) 100 ohm
(d) 110 ohm
(iv) what is the potential difference across the coil?
(a) 194 V
(b) 186 V
(c) 180 V
(d) 190 V
(v) Calculate the power dissipated in the coil.
(a) 100 W
(b) 122 W
(c) 130 W
(d) 144 W
Case-Study Based MCQ Answers
1. (i) (a) XL = VL/I
I = VR/R = 25/50 = 1/2
XL = 20/(1/2) = 40 ohm
XL = ωL ; L = 40 / (2 x 103) = 20 x 10-3 = 20mH
(ii) (d) ω2 = 1/LC ; C = 1/( ω2L) = 1 / ((2 x 103)2 x 20 x 10-3) = 12.5 μF
(iii) (a) At resonance, the impedance equal just resistance.
(iv) (d) For inductive nature, ω > ωr
(v) (b) If ω < ωr, the circuit will be capacitive in nature.
2. (i) (b) Z = √((R1 + R2)2 + (XL – XC)2) = √((44 + 36)2 + (90 – 30)2) = 100 ohm
(ii) (c) Current, I = V/Z = 200/100 = 2A
(iii) (a) Impedance of the coil, ZL = √(R22 + XL2) = √((36)2 + (90)2) = 97 ohm
(iv) (a) Potential difference across the coil, VL = IZL = 2 x 97 = 194 V
(v) (d) Power dissipated in the inductive coil, P = I2R2 = (2)2 x 36 = 144 W
Click Below To Learn Physics Term 1 Syllabus Chapter-Wise MCQs
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