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(i) Reactance of the capacitor will decrease, resulting in increase of the current in the circuit, Therefore, the bulb will glow brighter.

(ii) The bulb witt go dimmer.

(i) A

(ii) Capacitor

Average power over full cycle of the ac voltage source.

There is no effect of doubling the frequency of the a.c. source on R. Since X_L = 2πfL, the value of X_L becomes double on doubling the frequency of the a.c. source.
Further \(X_c = \frac{1}{2\pi fc}\)
∴ The value of X_c becomes half on doubling the frequency of the a.c. source.

1. Given E₀ = 300 V, E_av  = ?

Since E_v = \(\frac{E_0}{\sqrt{2}}\) = 0.707 × 300

or E_v = 212.1 V.

2. Given I_v = 10A, I₀ = ?

Since I₀ = \(\sqrt{2}\)E_v = 1.414 × 10

or I = 14.14 A.

Given V_rms = 220V, P = 100W; I_rms = ? ; V₀ = ? 

P = V² rms/R 

R = V² rms/P = (220)² /100 = 484 Ω 

Peak voltage = V₀ = V_rms x √2 = 220 x √2 = 311.126 V 

I_rms = V_rms/R = 220/484 = 0.45 A. 

(a) Expression for instantaneous value of alternating current = i = i_m sin ωt 

(b) Expression for instantaneous value of alternating voltage = v = v_m sin ωt 

(c) Expression for peak value of alternating current = i_m = v_m/x 

(d) Expression for peak value of alternating voltage = v_m = NABω 

(e) Expression for rms value of alternating current = effective current = I

= i_rms = i_m/√2 = 0.707 i_m 

(f) Expression for rms value of alternating voltage = v_rms = V = v_m/√2 = 0.707 v_m

Correct option is (A) Twice the ratio of the energy stored in the magnetic field to the rate of dissipation of energy in the resistance.

For an L–R circuit,

T (time constant) = \(\left(\frac L R\right)\) 

Now energy stored in magnetic field is \(\frac12\)LI² and rate of dissipation of energy is I²R.

Correct option is (B) 3:2 

T₁ (time constant) during build up = \(\left(\frac L{2R}\right)\)

T₂ during decay = \(\left(\frac L{3R}\right)\)

\(\therefore \frac{T_1}{T_2} = \frac32\)

Correct option is (C) 16 J/s

Energy stored per unit time = \(Li\frac{di}{dt}\)

= 2 (2) (4) = 16 J/s.

Correct option is (B) LI / 2R

In an L - R decay circuit, the initial current at t = 0 is 1. The total charge that has inductor has reduced to one-fourth of its initial value is LI / 2R.

Correct option is (C) \(\frac{2U}{P}\)

\(\frac1 2 LI^2 = U\)

\(I^2 R = P \)

\(T = \frac LR = \frac{2U}P\)

P_L=60W, I_L=0.54A

V_L=60/0.54=110V.

The transformer is step-down and have 1/2 input voltage. Hence

t_p=1/2xI₂=0.27A.

(c) R = 15 Ω, L = 3.5 H, C = 30μF.

(b) another capacitor should be added in parallel to the first. 

It is based up on the principle of electromagnetic induction.

(b)  5√3/2 A

The comparison of a.c. power supply and d.c. Power supply is given below:

Merits :

(i) The generation cost of a.c. is less than that of d.c. 

(ii) It can be made available in wide range of voltages using transformers. 

(iii) The a.c. devices such as motors are mechanically more robust and stout than the d.c. devices. 

(iv) The power loss in a.c. transmission is negligible as compared to that in d.c. transmission.

(v) The a.c. can be easily converted to d.c. The reverse is not true. 

(vi) For reducing the alternating current, we can use choke coils in which the loss of energy is much less than that in the rheostat used for reducing d.c.

Demerits : 

(i) The a.c. is more dangerous and fatal than d.c.

(ii) The 220 V a.c. supply has the peak value of about 311 V which can cause more severe shock to the persons coming in contact with it. 

(iii) The a.c. is transmitted mostly at the outer surface of the wire, and so the conductor needs to be in the form of several stranded wires. 

(iv) The a.c. contains higher harmonics in addition to the fundamental frequency.

(v) The a.c. cannot be used for electrolysis, electroplating, electro refining, electro-typing, etc. 

Current flowing in a circuit without any net dissipation of power, is called wattless current.

It is that steady current, which when passed through a resistance for a given time will produce the same amount of heat as the alternating current does in the same resistance and in the same time.

No power is developed across the inductor as heat.

Class 12 Physics MCQ Questions of Alternating Current with Answers with Answers are prepared as per the Latest Exam Pattern and syllabus. Students can solve these Alternating Current Class 12 MCQ Questions with Answers and assess their preparation level. These MCQ Questions for Class 12 with answers for a quick revision of the Chapter thereby helping you to enhance subject knowledge.

Students are advised to practice the NCERT MCQ Questions for Class 12 Physics with Answers is available here. Solving the Class 12 Physics MCQ Questions of Alternating Current with Answers can be of extreme help as you will be aware of all the concepts.

Practice MCQ Question for Class 12 Physics chapter-wise 

1. An acceptor circuit is :

(a) series resonant circuit
(b) parallel resonant circuit
(c) LCR circuit
(d) None of these

2. In series resonant circuit:

(a) reactance is zero
(b) current is zero
(c) voltage is zero
(d) None of these

3. Transformer works on the principle of:

(a) convertor
(b) invertor
(c) mutual induction
(d) self-induction

4. The phase difference b/w the A.C. and e.m.f. π/2 Which of the following can not be the constituent of the circuit?

(a) LC
(b) L alone
(c) C alone
(d) RL

 5.  Faraday constant:

(a) depends on the amount of the electrolyte
(b) depends on the current in the electrolyte
(c) is a universal constant
(d) depends on the amount of charge passed through the electrolyte

6. A rejector circuit is:

(a) series resonant circuit
(b) parallel resonant circuit
(c) LCR circuit
(d) None of these

7. Alternating current cannot be measured by D.C. ammeter, because

(a) A. C. is virtual
(b) A. C. changes its direction
(c) A. C. cannot pass through D.C. ammeter
(d) average value of A. C for complete cycle is zero

8. A.C. power is transmitted from a power house at a high voltage as

(a) the rate of transmission is faster at high voltages
(b) it is more economical due to less power loss
(c) power cannot be transmitted at low voltages
(d) a precaution against theft of transmission lines

9. An A.C. source is connected to a resistive circuit. Which of the following is true?

(a) Current leads ahead of voltage in phase
(b) Current lags behind voltage in phase
(c) Current and voltage are in same phase
(d) Any of the above may be true depending upon the value of resistance.

10. With increase in frequency of an A.C. supply, the inductive reactance

(a) decreases
(b) increases directly with frequency
(c) increases as square of frequency
(d) decreases inversely with frequency

11. The capacitive reactance in an A.C. circuit is

(a) effective resistance due to capacity
(b) effective wattage
(c) effective voltage
(d) None of these

12. The frequency of A.C. mains in India is

(a) 30 Hz
(b) 50 Hz
(c) 60 Hz
(d) 120 Hz

13. The electric mains supply in our homes and offices is a voltage that varies like a sine function with time such a voltage is called ... A... and the current driven by it in a circuit is called the ... B... Here, A and B refer to

(a) DC voltage, AC current
(b) AC voltage, DC current
(c) AC voltage, DC voltage
(d) AC voltage, AC current

14. Alternating currents can be produced by a

(a) dynamo
(b) choke coil
(c) transformer
(d) electric motor

15. The parallel combination of inductor and capacitor is called as

(a) rectifier circuit
(b) tank circuit
(c) acceptor circuit
(d) filter circuit

16. The alternating current can be measured with the help of

(a) hot wire ammeter
(b) hot wire voltmeter
(c) moving magnet galvanometer
(d) suspended coil type galvanometer

17. The core of transformer is laminated to reduce

(a) flux leakage
(b) hysteresis
(c) copper loss
(d) eddy current

18. A transformer is based on the principle of

(a) mutual induction
(b) self induction
(c) Ampere’s law
(d) X-ray crystallography

19. Eddy currents in the core of transformer can't be developed by

(a) increasing the number of turns in secondary coil
(b) taking laminated transformer
(c) making step down transformer
(d) using a weak a.c. at high potential

20. A capacitor acts as an infinite resistance for

(a) DC
(b) AC
(c) DC as well as AC
(d) neither AC nor DC

21. A capacitor has capacitance C and reactance X, if capacitance and frequency become double, then reactance will be

(a) 4X
(b) X/2
(c) X/4
(d) 2X

22. In LCR circuit if resistance increases quality factor

(a) increases finitely
(b) decreases finitely
(c) remains constant
(d) None of these

23. The loss of energy in the form of heat in the iron core of a transformer is

(a) iron loss
(b) copper loss
(c) mechanical loss
(d) None of these

24. Which quantity is increased in a step-down transformer?

(a) Current
(b) Voltage
(c) Power
(d) Frequency

25. A coil of self-inductance L is connected in series with a bulb B and an ac source. Brightness of the bulb decreases when

(a) frequency of the ac source is decreased
(b) number of turns in the coil is reduced
(c) a capacitance of reactance X_c = X_L in included
(d) an iron rod is inserted in the coil

Answer :

1. Answer : (a) series resonant circuit

Explanation: Acceptor circuit provides the maximum response to currents at its resonant frequency. Series resonance circuit is known as acceptor circuit because the impedance at the resonance is at its minimum so as to accept the current easily such that the frequency of the accepted current is equal to the resonant frequency.

2. Answer : (d) None of these

Explanation: In RLC series circuit, when the circuit current is in phase with the voltage, the circuit is said to be in Series Resonance. The resonance condition in the circuit arises when the inductive reactance is equal to the capacitive reactance. At this condition, the circuit draws the maximum current.

3. Answer : (c) mutual induction

Explanation: A transformer works on the principles of "electromagnetic induction" as a mutual induction. Mutual induction is the process by which a coil of wire magnetically induces a voltage in another closely located coil.

4. Answer : (d) RL

Explanation: If the phase difference between

V and I is π/2

∴ Power = V⋅I = VI cos π/2 = 0

Since Power is 0 it should only consists of reactive elements (L & C), Resistance R cannot be present.

 5. Answer : (c) is a universal constant

Explanation: The constant is named after the British physicist Michael Faraday. It is a universal constant. It is the amount of electric charge carried by 1 mole of electrons. Represented by F & its measuring unit is Coulombs per mole (C).

6. Answer : (b) parallel resonant circuit

Explanation: A circuit comprising a capacitor and an inductor connected in parallel, having values chosen such that the combination offers a very high impedance to signals of a particular frequency.

7. Answer : (d) average value of A. C for complete cycle is zero

Explanation: Average value of A.C. for complete cycle is zero. Hence A.C. can not be measured by D.C. ammeter.

8. Answer : (b) it is more economical due to less power loss

Explanation: In long distance transmission of electricity, long wires are used which have very high resistance, R∝ l, so direct multiplication of R into a high number will increase power wasted. So current should be low, voltage can be high because resistance is denominator, so it will reduce the power wasted as heat. So less power is wasted therefore it is more economical way.

9. Answer : (c) Current and voltage are in same phase

Explanation: When resistance is connected to A.C source, then current & voltage are in same phase.

10. Answer : (b) increases directly with frequency

Explanation:\(X_L=\omega L\rightarrow X_L\propto\omega\)

11. Answer : (a) effective resistance due to capacity

Explanation: Capacitive reactance in an A.C. circuit is \(X_C=\frac{1}{\omega C}ohm\) where C is the capacitance of capacitor & \(\omega=2\pi n\) (n is the frequency of A.C source)

12. Answer : (b) 50 Hz

Explanation: The standard frequency of alternating current in India is 50 Hz.

13. Answer : (d) AC voltage, AC current

Explanation: The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. Such a voltage is called alternating voltage and the current driven by it in a circuit is called the alternating current.

14. Answer : (a) dynamo

Explanation: Dynamos and alternators are the two main types of electromagnetic generators. A commutator is used in dynamos to produce pulsating direct current. Alternators are devices that produce alternating current.

15. Answer : (b) tank circuit

Explanation: The parallel combination of inductor and capacitor is called as tank circuit.

16. Answer : (a) hot wire ammeter

Explanation: Alternating current can be measured by hot wire ammeter whereas moving coil galvanometer and tangent galvanometer are used to measure dc current.

17. Answer : (d) eddy current

Explanation: Since the core forms a closed path, current gets induced (Eddy Current). So, the core has some resistance due to which losses are produced which decreases the efficiency of the transformer. Thus, the core of a transformer is laminated to reduce the Eddy Current.

18. Answer : (a) mutual induction

Explanation: A transformer works on the principles of "electromagnetic induction" as a mutual induction. Mutual induction is the process by which a coil of wire magnetically induces a voltage in another closely located coil.

19. Answer : (b) taking laminated transformer

Explanation: Eddy currents are locally generated current loops in the transformer core. This happens because transformer core is in close proximity of the varying magnetic field of the primary.Eddy current is a loss. To reduce this loss, core is made from bundle of thin laminated sheets.

20. Answer : (a) DC

Explanation: A fully charged capacitor acts as an open circuit. As a result, the capacitor now acts as an open circuit and thus, there is no more flow of charge in this circuit. In other words, we can say that a fully charged capacitor acts as an infinite resistance for DC.

21. Answer : (c) X/4

Explanation: \(X=\frac{1}{C\omega}\)

\(X'=\frac{1}{4C\omega}\)

\(\therefore X'=\frac{X}{4}\)

22. Answer : (b) decreases finitely

Explanation: Here, \(Q=\frac{\omega^2L}{R}\)

if R resistance increases, quality factor Q decreases finitely.

23. Answer : (a) iron loss

Explanation: Iron loss is the energy loss in the form of heat due to the formation of eddy currents in the iron core of the transformer.

24. Answer : (a) Current

Explanation: In step -down transformer, voltage decreases and corresponding current increases.

25. Answer :(d) an iron rod is inserted in the coil

Explanation: Brightness of the bulb decreases when inductive reactance increases, i.e. when iron rod is inserted in the coil, its inductance L increases, hence current decreases.

Click here to practice  MCQ Question for Alternating Current Class 12

The alternating current easily convert high potential to low potential and reduced the energy loss easily.

L = 1.5 H
C = 35 µF = 35 x 10^-6 F
R = 20q
E_v = 200V
When the frequency of the supply equals the natural frequency of the circuit, the impedance of the circuit is equal to the resistance of the circuit. Therefore, rms value of the current in the circuit
\(I_V=\frac{E_v}{R}=\frac{200}{20}= 10A\)
Therefore, average power transferred to the circuit in one complete cycle,
P = EI = 200 × 10 = 2000 W.

(i) When a dielectric slab is introduced between the plates of the capacitor, its capacitance will increase. Hence, the potential drop across the capacitor will decrease (V= Q/C  ) thereby decreasing the potential drop across the bulb because, both the bulb and capacitor is connected in series. So, brightness of the bulb will increase.

(ii) When resistance (R) is increased keeping the capacitance same, the potential drop across the resistor will increases. Therefore, the potential drop across the bulb will decrease because both are connected in series. So, brightness of the bulb will decrease. 

P₀ = 80 W (given)

Prms = P₀/2 = 40W

Energy consumed = P × t = 40 × 100 = 4000J = 4.0KJ

E = 3 × 10⁶V/m, A = 20cm², d = 0.1mm

Potential diff. across the capacitor = Ed = 3 × 10⁶ × 0.1 × 10^–3 = 300V

Max. rms Voltage = V/√2 = 300/√2 = 212V

The AC current through pure L and C circuit is called wattles current. This is because power factor of pure L and C circuits is zero. 

r.m.s. value of current. 

The working of a.c. ammeter is based on heating effect of electric current and heat produced is directly proportional to l₂ (and not l). Therefore, an a.c. ammeter has non-linear scale. 

The frequency of direct current is zero.

The power dissipated in the circuit is (c) 14.4 W.

The value of the peak current is (a) 1/√2 A.

No nothing is perfect. It is impossible to make a perpetual machine.

The current lags behind the voltage by phase angle π/2.

Capacitive reactance, \(X_C = \frac1{\omega C} = \frac1{2\pi fc}\)

(i) By using laminated iron core, we minimize loss of energy due to eddy current. 

(ii) By selecting a suitable materials for the core of a transformer, the hysteresis loss can be minimized.

1. Step down transformer.

2. Power loss = 200w -10w = 190w

3. The possible energy losses in a transformer: 

• Eddy current loss 
• Copper loss 
• Hysteries loss 
• Flux leakage

4. In this case input power = VI = 48 × 1 = 48 W.

If transformer does not waste energy, input power = out put power.

Hence maximum output power 48 W. But bulb requires 100 w. Hence the bulb does not glow with this low input power.

1. Step down transformer.

2. Since the transformer is ideal

V_pI_p = V_s I_s = 960 W, V_p = 220 v, V_s = 110 v
V_pI_p =960

I_s = \(\frac{960}{110}\)= 8.72 A

3. R =  \(\frac{V^2}{P}=\frac{110^2}{960}\) = 12.60Ω

Step down transformer.

1. Self-induction

2. \(\frac{d\phi}{dt}\) = \(\frac{1}{2}\) = = 0.5V.

3. When AC is connected the brightness of bulb will be decreased. This is due to the back emf in the circuit.

1. 200 v.

2. V = V₀ sin ωt 

3. V_max = \(\frac{V_0}{\sqrt2}\) = \(\frac{200}{\sqrt{2}}\) = 141.8v.

Any two of the following (or any other correct) reasons:

(i) AC can be transmitted with much lower energy losses as compared to dc.

(ii) AC voltage can be adjusted (stepped-up or stepped down) as per requirement.

(iii) AC current in a circuit can be controlled using (almost) wattless devices like choke coil.

(iv) AC is easier to generate.

The capacitor.

When a choke coil in series with a lamp is connected to a d.c. line, L has no effect on the steady value of the current. Therefore, brightness of the lamp is not affected on the insertion of iron core in the choke. On a.c. line, the lamp will shine dimly due to the impedance of the choke coil. The brightness of the lamp will further go dim on the insertion of iron core, which increases the impedance of the choke coil.

(a), (d).

(a) zero average current.

(d) voltage and current possibly differing in phase φ such that |φ|<π/2.

In a resistance coil, when an alternating current is flown, there will be a magnetic field generated across the coil and so there will be an inductance induced into the coil. Hence it will have more impedance compared to the one withDC current.