NCERT Solutions Class 12 Physics Chapter 7
NCERT Solutions Class 12 Physics Chapter 7 Alternating Current
The NCERT Solutions Class 12 Physics Chapter 7 will guide students to understand the concepts and solve questions appropriately. Class 12 Physics Chapter 7 NCERT Solutions are prepared by the subject experts at Extramarks. All answers are explained clearly in simple language while preparing the solutions, and all the guidelines given by CBSE have been kept in mind
CBSE or Central Board of Secondary Education is the most preferred educational board among schools. It covers all topics required to understand the concept clearly and thoroughly. The NCERT Solutions Class 12 Physics Chapter 7 act as a helping hand for students to learn the subject effectively.
Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 7
Chapter 7 Physics Class 12 explains the concept of Alternating current, its properties and use cases, a vital sub-topic under the Electromagnetic Induction unit. Students can expect questions on this topic in first term examinations. To score good marks, students may refer to NCERT Solutions Class 12 Physics Chapter 7.
Following are the topics covered in NCERT Solutions Class 12 Physics Chapter 7
|Topic Number||Topic Name|
|7.2||AC Voltage Applied to resistor|
|7.3||AC Current, voltage by rotating vectors|
|7.4||AC voltage to inductor|
|7.5||AC Voltage to capacitor|
|7.6||AC Voltage to Series LCR|
A brief of the topics covered in NCERT Solutions Class 12 Physics Chapter 7 is below.
Topic 7.1 Introduction
Alternating current (AC) is an electric powered current that periodically reverses direction and adjusts its magnitude constantly with time, in contrast to direct current (DC), which flows best in a single direction.
Topic 7.2 AC Voltage Is Applied to a Resistor
To have a sinusoidal alternating current, we want to have an AC voltage supply because the current is at once proportional to voltage. An AC generator or AC dynamo may be used as an AC voltage supply.
Voltage V(t) is applied across resistance R. V(t) is sinusoidal voltage with peak Vm and time period T.
Where f is frequency and ω is the angular frequency. This kind of circuit is a purely resistive circuit. According to Kirchhoff’s law –
Here voltage and current have equal frequency, and each is in a similar phase.
The average current value may be located out with the aid of summing over the whole change inside the voltage and dividing it using the number of instances we do the measurements.
Over a cycle, the average value of AC current is 0 because, in the 1st half of the time, the current is positive and in the second half, the current is negative.
Root Mean Square Value of Current, i.e. irms=im2
Topic 7.3 Representation of AC Current and Voltage through Rotating Vectors or Phasors
The projection of the phasor at the vertical axis represents the value of the quantity. For example, inside the case of a current or a vector phasor, the projection of the phasor at the vertical axis, given via means of vmsinωt and imsinωt, respectively, offers a current value or voltage value.
Topic 7.4 AC Voltage Applied to an Inductor
It implies an AC circuit composed only of an inductor of inductance L linked to an AC supply. The AC voltage throughout the supply is V = Vm sin(t). The converting current output of the AC supply offers a returned emf inside the coil of magnitude that’s given via the means of VL = L di/dt.
Topic 7.5 AC Voltage Applied to a Capacitor
When an alternating voltage is carried out throughout a capacitor, the current leads the voltage via a phase angle of ninety degrees. This approach oscillates 1 / 4 of the cycle before the voltage.
Topic 7.6 AC Voltage Applied to a Series LCR Circuit
A chain LCR circuit is connected to an AC supply. The voltage of an AC supply is V=Vmsinωt. Potential distinction throughout the ends of capacitor =qC and potential distinction throughout the resistance ends =IR.
Topic 7.7 Power in AC Circuit: The Power Factor
Power factor (PF) is the ratio of running power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA). Apparent power, additionally called demand, is the degree of the quantity of power used to run equipment and devices all through a certain period. It is observed with the aid of using multiplying (kVA = V x A).
Topic 7.8 LC Oscillations
The LC Oscillator employs a tank circuit (comprising an inductor and a capacitor) to offer tremendous vital comments to preserve oscillations in a circuit going.
Working of an LC Oscillator:
- When a completely powered capacitor is connected to a de-energised inductor, the capacitor gets all the circuit’s energy while the inductor receives none. Let’s name the energy saved inside the capacitor (electrical energy) UE and the magnetic energy saved inside the inductor (UB).
- Current flows from the capacitor to the inductor, energizing the inductor and discharging the capacitor. The inductor’s energy starts to evolve to rise simultaneously as the capacitor’s energy falls. The bars underneath the circuit diagram display that 1/2 of the energy saved in an inductor is equal to 1/2 of the energy saved in a capacitor, indicating that the capacitor has transferred 1/2 of its energy to the inductor.
- All of the capacitor’s energy will now be transmitted to the inductor as quickly because the capacitor is drained. As a result, a vast quantity of electrical energy is converted to magnetic energy.
- Because the capacitor has been absolutely drained and the inductor has been absolutely energized, the inductor will now start to charge the capacitor in the same direction because of the current. The present state of the circuit is depicted in the fourth diagram. As a result, 1/2 of the inductor’s energy has been transferred to the capacitor.
- Finally, the capacitor may be absolutely charged once more, and the inductor may be activated. On the alternative hand, the capacitor will now have its polarity reversed. As a result, if the current inside the circuit begins to move once more from the capacitor, it’ll flow in the opposite direction. We might also say that the primary 1/2 of the AC cycle has ended, and the second 1/2 has started because the current inside the circuit now has an opposing flow of current.
- As a result, each capacitor and the inductor may be charged two times for the duration of the cycle.
Topic 7.9 Transformers
These are used to increase or decrease the voltage of alternating currents. A transformer includes coils of wire wound on a metallic core. An alternating voltage is carried out to at least one coil (the first coil). This causes a changing (alternating) magnetic area to be installed inside the core.
List of NCERT Solutions Class 12 Physics Chapter 7 Alternating current
Click on the below links to view NCERT Solutions Class 12 Physics Chapter 7
- NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.1
- NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.2
- NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.3
- NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.4
- NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.5
- NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.6
- NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.7
- NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.8
- NCERT Solutions Class 12 Physics Chapter 7: Exercise 7.9
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Key Features of NCERT Solutions Class 12 Physics Chapter 7
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- Detailed explanations for each topic under chapter 7 Physics class 12
- Curated by Subject Experts in respective fields
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Q.1 A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?
Q.2 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Q.3 A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Q.4 A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit.
Q.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer.
Q.6 Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?
Q.7 A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?
Q.8 Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
Q.9 A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?
Q.10 A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?
[Hint: For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.]
Q.11 Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L= 5.0 H, C = 80μF, R = 40 Ω
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.
Q.12 An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t =0.
(a) What is the total energy stored initially? Is it conserved during LC oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored?
(i) Completely electrical (i.e., stored in the capacitor)?
(ii) Completely magnetic (i.e., stored in the inductor)?
(d) At what times is the total energy shared equally between the inductor and the capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?
Q.13 A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply.
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?
Q.14 Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady state?
Q.15 A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.
(a) What is the maximum current in the circuit?
(b) What is the time lag between the current maximum and the voltage maximum?
Q.16 Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply? Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.
Q.17 Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.
Q.18 A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.
(a) Obtain the current amplitude and rms values.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]
Q.19 Suppose the circuit in Exercise 7.18 has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.
Q.20 A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
(d) What is the Q-factor of the given circuit?
Q.21 Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 μF, and R = 7.4 Ω. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.
Q.22 A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?
Q.23 Answer the following questions:
(a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?
(b) A capacitor is used in the primary circuit of an induction coil.
(c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.
(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.
(e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?
(a) Yes; the statement is not true for rms voltage.
It is true that in any ac circuit, the applied instantaneous voltage is equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit. But, this is not true for rms voltage because such voltages across different elements may not be in same phase.
(b) The capacitor is used in the primary circuit of an induction coil. This is because, when a circuit is broken, a high induced voltage is used to charge the capacitor to avoid sparking.
(c) The dc signal will appear across capacitor C because for dc signals, the impedance of an inductor (L) is very low while the impedance of a capacitor (C) is very high (almost infinite). The capacitor blocks the dc signal. Therefore, a dc signal appears across C. For an ac signal of high frequency, the impedance of L is high and that of C is negligible. The inductor blocks the ac signal of high frequency. Therefore, an ac signal of high frequency appears across L.
(d) On a dc line, choke offers no impedance. Therefore, the lamp shines brightly and the insertion of an iron core in the choke causes no change in the lamp’s brightness. However, on an ac line, both the choke coil and the iron core increase the impedance of the circuit. Therefore, if an iron core is inserted in the choke coil (which is in series with a lamp connected to the ac line), the lamp will glow dimly.
(e) A choke coil is needed in the use of fluorescent tubes with ac mains because it reduces ac across the tube without much loss of power. We cannot use an ordinary resistor instead of a choke coil for this purpose because it wastes power in the form of heat.
Q.24 At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m3s–1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms–2).
Q.26 A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step up transformer at the plant.
Q.27 Do the same exercise as above with the replacement of the earlier transformer by a 40,000-220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred?