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NCERT Exemplar Class 12 Physics Solutions Chapter 7 Alternating Current helps the students to understand how AC circuits work in real life. This chapter explains concepts like AC voltage, different types of reactance, impedance, resonance, different types of circuits and power in AC circuits. You will also learn about devices like transformers and AC generators, which are widely used in electricity distribution. With detailed solutions, students can easily grasp important formulas and problem-solving techniques.
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NCERT Exemplar Class 12 Physics Solutions Chapter 7 Alternating Current explains how AC voltage drives current and its importance in electrical devices. It covers concepts like power in ac circuit, nature of the circuit and the phasor diagram. These solutions help the students to learn the concept of complex problems and improve their understanding. Mastering this chapter is very important for scoring well in board exams and competitive tests.
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Also see - NCERT Solutions for Class 12 Other Subjects
Question:1
If the RMS current in a 50 Hz ac circuit is 5 A, the value of the current 1/300 seconds after its value becomes zero is
a. $5\sqrt{2}A$
b.$5\sqrt{\frac{3}{2A}}$
c.$\frac{5}{6A}$
d.$\frac{5}{\sqrt{2}A}$
Answer:
The answer is the option (b)Question:2
Answer:
For maximum power to be delivered from the generator to the loadQuestion:3
When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220V. This means
A. input voltage cannot be AC voltage, but a DC voltage.
B. maximum input voltage is 220V.
C. the meter reads not v but <v2> and is calibrated to read $\sqrt{< v^{2}> }$
D. the pointer of the meter is stuck by some mechanical defect.
Answer:
The answer is the option (c)Question:4
To reduce the resonant frequency in an LCR series circuit with a generator
A. the generator frequency should be reduced.
B. another capacitor should be added in parallel to the first.
C. the iron core of the inductor should be removed.
D. dielectric in the capacitor should be removed.
Answer:
The answer is the option (b)Question:5
Which of the following combinations should be selected for better tuning of an LCR circuit used for communication?
A. R = 20 Ω, L = 1.5 H, C = 35μF.
B. R = 25 Ω, L = 2.5 H, C = 45μF.
C. R = 15 Ω, L = 3.5 H, C = 30μF.
D. R = 25 Ω, L = 1.5 H, C = 45μF.
Answer:
The quality factorQuestion:6
An inductor of reactance 1Ω and a resistor of 2 Ω are connected in series to the terminals of a 6 V (rms) a.c. source. The power dissipated in the circuit is
A. 8 W.
B. 12 W.
C. 14.4 W.
D. 18 W.
Answer:
$X_L=1\Omega, R=2 \Omega \: \: E_{rms}=6V$
The average power dissipated in the circuit
$\\P_{av}=E_{rms}I_{rms}\cos\Phi \\ I_{rms}=\frac{E_{rms}}{Z}\\\\ Z=\sqrt{R^2+X_{L}^{2}}=\sqrt{4+1}=\sqrt{5}\\ \\ I_{rms}=\frac{6}{\sqrt{5}}A\\ \\ \cos \Phi \frac{R}{Z}=\frac{2}{\sqrt{5}}$
$\\P_{av}=6*\frac{6}{\sqrt{5}}*\frac{2}{\sqrt{5}}= \frac{72}{5}=14.4\: W$
Question:7
The output of a step-down transformer is measured to be 24 V when connected to a 12-watt light bulb. The value of the peak current is
$\\A.\: 1/\sqrt{2}A\\ B.\: \sqrt{2}A\\ C. \: 2A\\ D.\:2\sqrt{2}A$
Answer:
$\\ V_s=24 V P_s=12 W\\I_s V_s=12\\I_s=\frac{12}{V_s} =\frac{12}{24}=0.5 \: ampere\\ I_0=I_s \sqrt{2}=0.5\sqrt{2}=\frac{1}{\sqrt{2} }\: ampere$Question:8
As the frequency of an ac circuit increases, the current first increases and then decreases. What combination of circuit elements is most likely to comprise the circuit?
A. Inductor and capacitor.
B. Resistor and inductor.
C. Resistor and capacitor.
D. Resistor, inductor and capacitor.
Answer:
For I to be maximum in a circuit, the reactance should be minimum. In the LCR circuit, if the frequency is increased, the inductance will increase while the capacitance decreases.Question:9
In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?
A. Only resistor.
B. Resistor and an inductor.
C. Resistor and a capacitor.
D. Only a capacitor.
Answer:
The correct answer is the options (c, d)Question:10
Electrical energy is transmitted over large distances at high alternating voltages. Which of the following statements is (are) correct?
A. For a given power level, there is a lower current.
B. Lower current implies less power loss.
C. Transmission lines can be made thinner.
D. It is easy to reduce the voltage at the receiving end using step-down transformers.
Answer:
Since the power is to be transmitted over the large distances at high alternating voltages, the current flowing through the wires will below.Question:11
A. Here, the power factor cos$\phi \geq 0, P \geq 0.$
B. The driving force can give no energy to the oscillator (P = 0) in some cases.
C. The driving force cannot syphon out (P < 0) the energy out of oscillator.
D. The driving force can take away energy out of the oscillator.
Answer:
The answers are the options (a, b, c)Question:12
When an AC voltage of 220 V is applied to the capacitor C
A. the maximum voltage between plates is 220 V.
B. the current is in phase with the applied voltage.
C. the charge on the plates is in phase with the applied voltage.
D. power delivered to the capacitor is zero.
Answer:
The correct answer are the options (c, d) On connecting the capacitor to ac supply the plate of the capacitor connected to the +ve terminal is at higher potential compared to the plate connected to negative terminal.Question:13
The line that draws power supply to your house from street has
A. zero average current.
B. 220 V average voltage.
C. voltage and current out of phase by 90°.
D. voltage and current possibly differing in phase $\phi$ such that $\left | \phi \right |< \frac{\pi}{2}$
Answer:
The correct answer are the options (a,d)Question:14
Answer:
If a L-C circuit is considered analogous to a harmonically oscillating spring block system, energy due to motion of charge particle i.e., magnetic energy is analogous to kinetic energy and electrostatic energy due to charging of capacitor is analogous to potential energy.Question:15
Answer:
Inductive reactanceQuestion:16
Study the circuits (a) and (b) shown in Fig 7.2 and answer the following questions.
(a) Under which conditions would the rms currents in the two circuits be the same?
(b) Can the rms current in circuit (b) be larger than that in (a)?
Answer:
$I_{rms} \: in (a)=\frac{V_{rms}}{R}$Question:17
Answer:
The instantaneous power output of an AC source can be negative because there are chances for power to get absorbed at some point of time. However, the average output over a cycle cannot be negative at any instant. It can only be either positive or zero as in any LC circuit.Question:18
Answer:
Question:19
Answer:
Question:20
Answer:
Phase angle $\theta$ by which Voltage leads current is given by,
$\tan \theta= \frac{X_L-X_C}{R}=\frac{\omega L-\frac{1}{\omega C}}{R}$
For low frequencies, $X_L<X_C\Rightarrow \tan\theta <0$
For resonance
$X_L<X_C\Rightarrow \tan\theta =0\left (\nu =\nu _0=\frac{1}{2\pi\sqrt{2C}} \right )$
Question:21
A device ‘X’ is connected to an a .c source. The variation of voltage, current and power in one complete cycle is shown in Fig 7.5.
(a) Which curve shows power consumption over a full cycle?
(b) What is the average power consumption over a cycle?
(c) Identify the device ‘X’.
Answer:
(a) $\text {Power (P)}=\text {VI}$Question:22
Answer:
In a DC circuit, 1 Ampere is defined as 1 Coulomb per second, but for AC circuit, the direction of current keeps changing. Instead, Joule’s heating is used to define the rms value of AC current. This is done because Joule’s heating is independent of the direction of the current. RMS value of AC is equal to the value of DC required to generate the same amount of heat through a given resistor in a given time.Question:23
Answer:
$Inductance (L)=0.01 \; H$Question:24
Answer:
$P_{s}=60\; W$Question:25
Answer:
Under the effect of AC current, capacitor plates charge and discharge alternately. Current is a direct result of this charging and discharging. On increasing the frequency of current, the capacitor will pass more current through it.Question:26
Answer:
Under the flow of current, Inductor develops a back emf. On decreasing the current, the induced emf will be such that it tries to maintain the existing flow of current. Induced emf is directly proportional to the rate of change of current.Question:27
Answer:
Impedance,Question:28
1MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if
(i) power is transmitted at 220V. Comment on the feasibility of doing this.
(ii) a step-up transformer is used to boost the voltage to 11000 V, power transmitted, then a step-down transformer is used to bring voltage to 220 V.
$(\rho _{cu}=1.7\times10^{-8} SI \; unit)$
Answer:
$L=20\; km=20000\; m$Question:29
Answer:
Taking Resistor andAlternating EMF circuit, $i=i_{1}+i_{2}$Question:30
For an LCR circuit driven at frequency $\omega$, the equation reads
$L\frac{di}{dt}+Ri+\frac{q}{C}=v_{1}=v_{m}\; \sin \omega t$
(i) Multiply the equation by i and simplify where possible.
(ii) Interpret each term physically.
(iii) Cast the equation in the form of a conservation of energy statement.
(iv) Integrate the equation over one cycle to find that the phase difference between v and i must be acute.
Answer:
$V=V_{m}\sin \omega t$Question:31
In the LCR circuit shown in Fig 7.7, the ac driving voltage is $v=v_{m}\sin \omega t$
(i) Write down the equation of motion for q (t).
(ii) At $t=t_{0}$, the voltage source stops and R is short circuited. Now write down how much energy is stored in each of L and C.
(iii) Describe subsequent motion of charges.
Answer:
a) $V=V_{m}\sin \omega t$Introduction
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AC voltages are preferred over DC because they can be efficiently converted to a voltage of choice through transformers, which can step up or step down the voltage. Class 12 Physics NCERT Exemplar solutions chapter 7 explains the graphical representation of AC current and voltage, principles behind an AC generator, defines angular frequency, time period, factors and phasors, dives deep into current and potential relations and explores Resistors, Capacitors, and Inductors, involves a study of Phasor algebra, circuit series and Power in an AC circuit.
Also, check NCERT Solutions for Class 12 Physics
1. AC Voltage & Current Representation
2. Impedance & Reactance
$
Z=\sqrt{R^2+\left(X_L-X_C\right)^2}
$
3. Phasor Representation
4. Power in AC Circuits
5. Resonance in LCR Circuit
$
f_r=\frac{1}{2 \pi \sqrt{L C}}
$
6. LC Oscillations
* The energy oscillates between inductor and capacitor with frequency:
$
\omega=\frac{1}{\sqrt{L C}}
$
7. Transformers
Chapter 7 Alternating Current |
Frequently Asked Questions (FAQs)
For Class 12 students, AC is a crucial chapter from boards, entrance exam and higher education point of view. This is one of the major chapters in electrical engineering.
One can use these NCERT exemplar Class 12 Physics solutions chapter 7 for better understanding of the topics and how the questions will be formed in the board exam.
Chapter covers topics like AC circuit, capacitor, resonance, LC oscillations, transformers, LCR circuit, etc.
Our esteemed physics teachers and team have solved each and every question as per the CBSE pattern and marking scheme in NCERT exemplar Class 12 Physics chapter 7 solutions
On Question asked by student community
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Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.
From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .
If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.
The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.
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