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Imagine a day without electricity...
No lights, no fans, no internet—even your phone battery would be useless! All of this is possible because of the Alternating Current (AC), which powers almost everything around us. In this chapter, you’ll learn how AC works, why it changes direction, and how it helps in sending electricity over long distances. It’s a super important and scoring Chapter that connects Physics to real life!
Alternating Current is an important and scoring chapter in the Class 12 Physics syllabus. This NCERT solutions page gives you clear, step-by-step answers prepared by Physics experts from Careers360.
The electricity we use at home is alternating current (AC). But do you know the voltage and frequency of the current that comes to our homes? You’ll find answers to such questions in this chapter. The NCERT questions covered in Class 12 Physics Chapter 7 are based on single-phase AC, and these solutions will help you understand the concepts easily.
NCERT Solutions for Class 12 Alternating Current are super helpful for understanding the basics needed for your CBSE board exam. By going through these solutions, you’ll get clear answers to all your questions about alternating current. In Chapter 7 of Class 12 Physics, you’ll also learn about phasor diagrams and how to solve problems related to them simply and easily.
Download the NCERT Class 12 Physics Chapter 7 (Alternating Current) Exercise Solutions PDF for free.
Perfect for quick revision and scoring well in the CBSE board exam! These expert-written solutions make tough concepts easy and help you practice important questions with confidence.
Q7.1 (a) A
a)What is the RMS value of current?
Answer:
Given,
RMS voltage in the circuit
Resistance in the circuit
Now,
RMS current in the circuit:
Hence, the RMS value of the current is 2.2A.
What is the net power consumed over a full cycle?
Answer:
Given,
Supplied RMS Voltage
Supplied RMS Current
The net power consumed over a full cycle:
Hence net power consumed is 484W.
Q7.2 (a) The peak voltage of an ac supply is
Answer:
Given
Peak Value of ac supply:
Now as we know in any sinusoidal function
Since our ac voltage supply is also sinusoidal
Hence RMS value of voltage os 212.13V.
Q7.2 (b) The RMS value of current in an ac circuit is
Answer:
Given,
RMS value of current
Since Current is also sinusoidal (because only resistance is present in the circuit, not the capacitor and inductor)
Hence the peak value of current is 14.1A.
Q7.3 A
Answer:
Given
Supply Voltage
Supply Frequency
The inductance of the inductor connected
Now
Inductive Reactance
RMS Value of the current :
Hence the RMS Value of current is 15.92A.
Q7.4 A
Answer:
Given,
Supply Voltage
Supply Frequency
The capacitance of the connected capacitor
Now,
Capacitive Reactance
RMS Value of current
Hence the RMS Value of current is 2.49A.
Answer:
As we know,
Power absorbed
Where
In both cases (inductive and capacitive), the power absorbed by the circuit is zero because in both cases the phase difference between current and voltage is 90 degree.
This can be seen as The elements(Inductor and Capacitor) are not absorbing the power, rather storing it. The capacitor is storing energy in electrostatic form and Inductor is storing the energy in magnetic form.
Answer:
Given
Capacitance
Inductance
Now,
Angular Frequency
Hence Angular Frequency is
Answer:
Given,
Resistance
Inductance
Capacitance
Voltage supply
At resonance, the supply frequency is equal to the natural frequency, and at the natural frequency, the total impedance of the circuit is equal to the resistance of the circuit
as inductive and capacitive reactance cancel each other. in other words,
As
Now,
Current in the circuit
Average Power transferred in the circuit :
Hence average power transferred is 2000W.
(a) Determine the source frequency which drives the circuit in resonance.
Answer:
Given,
Variable frequency supply voltage
Inductance
Capacitance
Resistance
a) Resonance angular frequency in this circuit is given by :
Hence this circuit will be in resonance when the supply frequency is 50 rad/sec.
Q7.8 (b) Figure shows a series LCR circuit connected to a variable frequency
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
Answer:
Given,
Variable frequency supply voltage
Inductance
Capacitance
Resistance
Now,
The impedance of the circuit is
At Resonance Condition
Hence, the Impedance at resonance is 40
Now, at resonance condition, impedance is minimum which means the current is maximum which will happen when we have a peak voltage, so
Current in the Resonance circuit is Given by
Hence amplitude of the current at resonance is 8.13A.
Q7.8 (c)Figure shows a series LCR circuit connected to a variable frequency
Answer:
Potential difference across any element =
Now
The potential difference across the capacitor:
The potential difference across the inductor
The potential difference across Resistor
=40 Irms=230V
The potential difference across LC combination
Hence at resonating, the frequency potential difference across LC combination is zero.
What is the total energy stored initially? Is it conserved during
Answer :
Given,
The inductance of the inductor:
The capacitance of the capacitor :
The initial charge on the capacitor:
Total energy present at the initial moment:
Hence initial energy in the circuit is 1J. Since we don't have any power-consuming elements like resistance in the circuit, the energy will be conserved
What is the natural frequency of the circuit?
Answer:
Given,
The inductance of the inductor:
The capacitance of the capacitor :
The initial charge on the capacitor:
The natural angular frequency of the circuit:
Hence the natural angular frequency of the circuit is
The natural frequency of the circuit:
Hence the natural frequency of the circuit is 159Hz.
Q1. (c-i) An
Answer:
at any instant, the charge on the capacitor is:
Where time period :
Now, when the total energy is purely electrical, we can say that
this is possible when
Hence Total energy will be purely electrical(stored in a capacitor) at
Q1.(c-ii)An
Let the instant the circuit is closed be
Answer:
The stored energy will be purely magnetic when the pure electrical stored is zero. i.e. when the charge on the capacitor is zero, all energy will be stored in the inductor.
So, t for which charge on the capacitor is zero is
Hence at these times, the total energy will be purely magnetic.
Q1. (d) An
Answer:
The energy will be shared equally when the energy in the capacitor is half of the maximum energy it can store. i.e.
From here, we got
So now, we know the charge on the capacitor, we can calculate the time for which
From here,
Hence for these times, the total energy will be shared equally between capacitor and inductor.
Q1. (e) An
Answer:
If the resistance is added to the circuit, the whole energy will dissipate as heat eventually. energy will keep moving between the capacitor and inductor with reducing in magnitude in each cycle and eventually all energy will be dissipated.
Answer:
Given,
The inductance of the coil
The resistance of the coil
Supply voltage
Supply voltage frequency
Now, as we know peak voltage =
Peak voltage
The impedance of the circuit :
Now peak current in the circuit :
Hence peak current is 1.82A in the circuit.
Answer:
Let the voltage in the circuit be
Current in the circuit be
Where
V is maximum At
t = 0
Hence, the time lag between voltage maximum and the current maximum is
For phase difference
Hence time lag between the maximum voltage and the maximum current is
Answer:
Given,
The inductance of the coil
The resistance of the coil
Supply voltage
Supply voltage frequency
a)
Now, as we know peak voltage =
Peak voltage
Now,
The impedance of the circuit :
Now peak current in the circuit :
Hence peak current is
The current in the circuit is very small, which is one of the indications of inductor working as a nearly open circuit in the case of high frequency.
b)
For phase difference
Now
Hence time lag between the maximum voltage and the maximum current is
In the DC circuit, after attaining the steady state, inductor behaves line short circuit as
Answer:
Given,
The capacitance of the capacitor
The resistance of the circuit
Voltage supply
Frequency of voltage supply
The maximum current in the circuit
Hence maximum current in the circuit is 3.24A.
Answer:
In the case of a capacitor, we have
So,
So the time lag between max voltage and the max current is :
Answer:
Given,
The capacitance of the capacitor
The resistance of the circuit
Voltage supply
Frequency of voltage supply
The maximum current in the circuit
Hence maximum current in the circuit is 3.9A.
b)
In the case of capacitor, we have
So,
So the time lag between max voltage and max current is :
At high frequencies,
In the DC circuit, after a steady state is achieved, Capacitor acts like an open circuit.
Answer:
As we know, in the case of a parallel RLC circuit:
The current will be minimal when
Which is also the condition of natural frequency. Hence the total current is minimum when source frequency is equal to the natural frequency.
RMS value of current in R
RMS value in Inductor
RMS value in capacitor
Capacitor current and inductor current will cancel out each other so the current flowing in the circuit is 5.75A.
Answer:
The inductance of the inductor
The capacitance of the capacitor
Voltage supply
Frequency of voltage supply
Here, we have
Impedance
Now,
Current in the circuit will be
where,
The negative sign is just a matter of the direction of current.so,
here
But, since the value of R is zero(since our circuit have only L and C)
Hence
Now,
RMS value of this current:
Answer:
As we know,
RMS potential drop across an element with impedance Z:
SO,
RMS potential difference across inductor:
RMS potential drop across capacitor
(c) What is the average power transferred to the inductor?
Answer:
Since
Current flowing in the circuit is sinusoidal and hence average power will be zero as the average of sin function is zero.in other words, the inductor will store energy in the positive half cycle of the sin (0 degrees to 180 degrees) and will release that energy in the negative half cycle(180 degrees to 360 degrees), and hence average power is zero.
Answer:
As we know,
Average power
Since in the circuit, phase difference
Answer:
Since the phase difference between voltage and current is 90 degree, even the total power absorbed by the circuit is zero. This is an ideal circuit, we can not have any circuit in practical that consumes no power, that is because practically resistance of any circuit is never zero. Here only inductor and capacitor are present and none of them consumes energy, they just store it and transfer it like they are doing it in this case.
Answer:
The inductance of the inductor
The capacitance of the capacitor
The resistance of a
Voltage supply
Frequency of voltage supply
As we know,
Impedance
Current flowing in the circuit :
Now,
Average power transferred to the resistor:
Average power transferred to the inductor = 0
Average power transferred to the capacitor = 0:
Total power absorbed by circuit :
Hence circuit absorbs 788.44W.
Q9. (a) A series LCR circuit with
What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
Answer:
The inductance of the inductor
The capacitance of the capacitor
The resistance of the resistor
Voltage supply
Frequency of voltage supply
As we know,
the current amplitude is maximum at the natural frequency of oscillation, which is
Also, at this frequency,
SO,
The maximum current in the circuit :
Hence maximum current is 14.14A.
Q9. (b) A series LCR circuit with
Answer:
Since the resistor is the only element in the circuit which consumes the power, the maximum absorbed power by circuit will be maximum when power absorbed by the resistor will be maximum. power absorbed by the resistor will be maximum at when current is maximum which is the natural frequency case,
Hence when source frequency will be equal to the natural frequency, the power absorbed will be maximum.
Hence frequency
Maximum Power Absorbed
Answer:
The value of maximum angular frequency is calculated in the first part of the question and whose magnitude is 4166.67
Q-factor of any circuit is given by
Hence Q-factor for the circuit is 21.74.
Answer:
As
Power
Power
As,
At half powerpoint :
here,
On putting values, we get, two values of
And they are:
Also,
The current amplitude at these frequencies
Answer:
The inductance of the inductor
The capacitance of the capacitor
The resistance of the resistor
Now,
Resonant frequency
Q-Factor of the circuit
Now, to improve the sharpness of resonance by reducing its full width at half maximum, by a factor of 2 without changing
we have to change the resistance of the resistor to half of its value, that is
Answer:
Yes, at any instant the applied voltage will be distributed among all element and the sum of the instantaneous voltage of all elements will be equal to the applied. But this is not the case in RMS because all elements are varying differently and they may not be in the phase.
Q11.(b) Answer the following questions: A capacitor is used in the primary circuit of an induction coil.
Answer:
Yes, we use capacitors in the primary circuit of an induction coil to avoid sparking. when the circuit breaks, a large emf is induced and the capacitor gets charged from this avoiding the case of sparking and short circuit.
Q11. (c) Answer the following questions:
An applied voltage signal consists of a superposition of a
Answer:
For a high frequency, the inductive reactance and capacitive reactance:
Hence the capacitor does not offer resistance to a higher frequency, so the ac voltage appears across L.
Similarly
For DC, the inductive reactance and capacitive reactance:
Hence DC signal appears across Capacitor only.
Q11.(d) Answer the following questions:
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.
Answer:
For a steady state DC, the increasing inductance value by inserting iron core in the choke, has no effect on the brightness of the connected lamp, whereas, for ac when the iron core is inserted, the light of the lamp will shine less brightly because of increase in inductive impedance.
Q11. (e) Answer the following questions:
Why is a 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?
Answer:
We need choke coil in the use of fluorescent tubes with ac mains to reduce the voltage across the tube without wasting much power. If we use simply resistor for this purpose, there will be more power loss, hence we do not prefer it.
Answer:
Given,
Input voltage:
Number of turns in the primary coil
Output voltage:
Now,
Let the number of turns in secondary be
Now as we know, in a transformer,
Hence the number of turns in secondary winding is 400.
Answer:
Given,
Height of the water pressure head
The volume of the water flow per second
Turbine generator efficiency
Mass of water flowing per second
The potential energy stored in the fall for 1 second
Hence input power
Now as we know,
Hence output power is 176.4 MW.
Answer:
Power required
The total resistance of the two-wire line
Input Voltage
Output Voltage:
RMS Current in the wireline
Now,
Power loss in the line
Hence, power loss in line is 600kW.
How much power must the plant supply, assuming there is negligible power loss due to leakage?
Answer:
Power required
The total resistance of the two-wire line
Input voltage
Output voltage:
RMS current in the wireline
Now,
Total power delivered by plant = line power loss + required electric power = 800 + 600 = 1400kW.
Answer:
Power required
The total resistance of the two-wire line
Input Voltage
Output Voltage:
RMS Current in the wireline
Now,
Voltage drop in the power line =
Total voltage transmitted from the plant = 3000+4000=7000
as power is generated at 440V, The rating of the power plant is 440V-7000V.
Answer:
Power required
The total resistance of the two-wire line
Input Voltage
Output Voltage:
RMS current in the wireline
Now,
a) power loss in the line
b)
Power supplied by plant = 800 kW + 6 kW = 806 kW.
c)
Voltage drop in the power line =
Total voltage transmitted from the plant = 300+40000=40300
as power is generated at 440V, The rating of the power plant is 440V-40300V.
We prefer high voltage transmission because power loss is a lot less than low voltage transmission.
Q1:
The power factor of R-L circuit is
Answer:
Power factor -
And For RL circuit
where
So
Power factor
on Solving we get
Q2:
In the above circuit,
Answer:
Phase difference -
- wherein
R = resistance
Phase difference -
For current
For
Q3:
An inductance of
Answer:
As we learn
Inductive susceptance
Inductive susceptance
Q4:
Figure shows a series LCR circuit connected to a variable frequency 230 V source.
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.
Answer:
The rms potential drop across R is given by
The rms potential drop across
The rms potential drop across C is given by
The rms potential drop across L-C is given by
Q5:
In the LCR circuit current resonant frequency is 600Hz, and half power points are at 650 and 550 Hz. The quality factor is-
Answer:
In a series LCR circuit, the quality factor is a measure of the sharpness of the resonance and is defined as the ratio of the resonant frequency
Given the following values:
Resonant frequency
Lower half-power frequency
Upper half-power frequency
The bandwidth (
Now, calculating the Q-factor:
Therefore, the Q-factor of the circuit is 6.
Understand basic terminology –
Alternating current (AC): Current that reverses direction periodically
Peak value, RMS value, frequency, and time period
Get basic equations –
Instantaneous current:
RMS current:
RMS voltage:
Know AC vs DC differences – Direction, value over time, frequency, and use
Understand impedance in AC circuits –
Apply power formulas in AC –
Here are the exercise-wise solutions of the NCERT Class 12 physics book:
In the previous year question papers, questions related to identifying the circuit elements are repeatedly asked. Along with this simple numerical questions are also asked from Alternating Current. For practising numerical, solve NCERT exercise questions, NCERT exemplar questions and previous year papers.
One question for JEE main can be expected from the Class 12 chapter Alternating Current.
One or two questions may be asked from NCERT chapter Alternating Current for NEET exam.
In India, the AC supply has a frequency of 50 Hz, which means it changes direction 50 times in one second.
A phasor diagram is a visual way to show the phase difference between voltage and current in an AC circuit using rotating vectors.
AC is easier to transmit over long distances and can be easily stepped up or down using transformers.
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As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
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