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NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

Edited By Vishal kumar | Updated on Jul 09, 2025 04:27 PM IST

Ever wondered how electric current is generated without using a battery? That is exactly what you will explore in NCERT Notes Class 12 Physics Chapter 6: Electromagnetic Induction. This chapter helps you understand how magnetic fields and motion come together to produce electricity. These NCERT Notes is very important for students preparing for CBSE boards, JEE, or NEET, as it builds strong fundamentals in electromagnetic induction and boost your problem-solving skills.

In these Alternating current Class 12 Notes PDF, you will find clear explanations of Faraday’s laws, Lenz’s law, magnetic flux, inductance, mutual inductance and AC generator. These NCERT notes for class 12 include important formulas, important diagrams, and step-by-step solved few previous year questions to make your revision smooth and effective. Whether you are revising for an exam or learning the topic for the first time.

This Story also Contains
  1. NCERT Notes for Class 12 Physics Chapter 7: Download PDF
  2. NCERT Notes for Class 12 Physics Chapter 7 Alternating Current
  3. Alternating Current: Previous Year's Question and Answer
  4. NCERT Class 12 Notes Chapterwise
NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF
NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

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NCERT Notes for Class 12 Physics Chapter 7: Download PDF

Prepare effectively using the PDF version of Alternating Current Notes. By downloading it, you can view the PDF offline without needing the internet and revise anytime at your own convenience.

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NCERT Notes for Class 12 Physics Chapter 7 Alternating Current

The NCERT Notes of Alternating Current is useful for students to revise and gain a better understanding of the topics. Here, students will find simple explanations of the concepts along with clear diagrams to enhance learning.

Introduction

Alternating Voltage:-

The voltage whose magnitude and direction changes with time and attains the same magnitude and direction after a definite time interval is called an ‘ alternating voltage. ’ Taking the expression for the induced emf e=e0sinωt, the alternating voltage can be expressed as

$V=V_0 \sin \omega t \quad \ldots \ldots \ldots(i)$

Where V₀ is the maximum (or peak) value of voltage.

Alternating Current:-

Due to this alternating voltage, the current that flows through the circuit also changes continuously between zero and maximum value and flows in one direction in the first half rotation and in the opposite direction in the next half rotation. This type of current is called an ‘ alternating current’. The alternating current may be expressed as

$i=i_0 \sin \omega t \quad \ldots \ldots(i i)$

Where i₀ is the maximum (or peak) value of current.

We get a sine curve (as shown in fig) when we plot the graph between the alternating voltage V or alternating current i and time t.

1644568259228

Some Important Terms Regarding The Alternating Current

  • Maximum value or Peak value: In a magnetic field, a coil is rotating, which produces the alternating voltage or current is maximum at two positions of the coil. This maximum value of alternating current or alternating voltage is known as the peak value of voltage or current. As shown in the above fig. It is represented by V₀ or i₀, respectively.

  • Periodic Time: The time taken by the alternating current to complete one cycle is called the periodic time of the current. If in Eqn (i) or (ii), t is increased by 2. The value of V or i remains unchanged. The periodic time of alternating current is
    $T=\frac{2 \pi}{\omega}$

  • Frequency: The number of cycles completed by an alternating current in one second is called the frequency of the current. It is equal to the number of rotations completed by the coil in one second. If the periodic time of the alternating current is T, then its frequency is
    $\begin{aligned} f= & \frac{1}{T} \\ & \therefore f=\frac{\omega}{2 \pi}\end{aligned}$

Mean Value of An Alternating Current

An alternating current flows during a one-half cycle in one direction and during the other half-cycle in the opposite direction. So, the mean (or average) value of alternating current is zero for one complete cycle.

The mean value of alternating current over a half cycle is a finite quantity and quantity which is defined as the mean value of alternating. It is given by

$i_m=\frac{2}{\pi} i_0=0.637 i_0$

Thus, the mean value of a.c. for a half-cycle(t = 0 to = T/2) is 0.637 times, or 63.7% of the peak value.

Similarly, the mean value of ac for the other half-cycle will be (t = T/2 to T) will be -2i0/Π=-0.637i0. It is zero for the full cycle.

Root mean square value of alternating current: The root-mean-square value of an alternating current is defined as the square root of the average of i² during a complete cycle, where i is the instantaneous value of the alternating current.

The root mean square of the alternating current is given by

$i_{r m s}=\sqrt{\overline{i^2}}=\frac{i_0}{\sqrt{2}}=0.707 i_0$

Thus, the root means the square value of an alternating current is 0.707 times and 70.7% of the peak value.

Similarly, the root mean square value of an alternating voltage represented by V = V₀ sin, t is given by

$V_{r m s}=\frac{V_0}{\sqrt{2}}=0.707 V_0$

The phase difference between voltage and current i in an alternating current circuit

In some circuits, the current reaches its maximum value before the voltage becomes maximum. Then, it is said the current is leading the voltage in phase. In some other circuits, the current reaches its maximum value after the voltage has become maximum. In such cases, the current is said to be lagging behind the voltage in phase.

If the alternating current circuit the voltage and the currents are in the same phase is that when the voltage is maximum the current is also maximum and when the voltage is zero the current is also zero then the alternating voltage and the alternating current are represented by the following equations

$\begin{aligned} & V=V_0 \sin \omega t \\ & i=i_0 \sin \omega t\end{aligned}$

If in an alternating current circuit, the current leads the voltage V by a phase difference ϕ, then the expression for the voltage and the current will be written as follows

$\begin{aligned} & V=V_0 \sin \omega t \\ & i=i_0 \sin (\omega t+\phi)\end{aligned}$

The graphical representation is shown in Fig 2.

1644568261000

If in a circuit, the alternating current i lags behind the alternating voltage by a phase difference ϕ. Then,

$
\begin{aligned}
& V=V_0 \sin \omega t \\
& i=i_0 \sin (\omega t)
\end{aligned}
$

This graphical representation is shown in Fig 3.

1644568258856

Phasors and Phasor diagrams:- The (rotating) vectors representing current and voltage are called phasors.

Phasor diagrams show alternating current and alternating voltage with their phase angles as rotating vectors (phasors).

AC Voltage Applied to a Resistor

Let a.c. voltage applied is

$v=v_m \sin \omega t$ ------(I)

where $v_m$ is the maximum value of applied voltage (or amplitude of oscillating potential difference) and $\omega$ is the angular frequency. From Kirchhoff's loop rule, $v_m \sin \omega t=I R$

$I=\left(\frac{v_m}{R}\right) \sin \omega t=i_m \sin \omega t$ ------(ii)


where $I_m=\left(\frac{v_m}{R}\right)$ is the amplitude of current. From Eq. (i) & (ii), it is clear that voltage across the resistor and current through the resistor are in same phase.

Representation of AC Current and Voltage by Rotating Vectors — Phasors

To represent the phase relationship between voltage and current in a.c. circuit, phasor diagram is used. A phasor is a vector which rotates about the origin with angular speed $\omega$. The magnitudes of phasors $V$ and $I$ represent the amplitude or peak value $V_m$ and $I_m$ of oscillating quantities and the vertical components of $V$ and $I$ represent the sinusoidally varying quantities.

Figure shows the phasor diagram for voltage and current for an a.c. source connected to a resistor.


AC Voltage Applied to an Inductor

Let an alternating voltage is given by

$V=V_0 \sin \omega t$

1644569356561

The maximum current in the circuit,

$i=\frac{V_0}{\omega L} \sin \left(\omega t-\frac{\pi}{2}\right)$

Where i₀ = ( V₀/⍵L) is the peak value of current. Comparison of this equation with the voltage equation shows that in a pure inductor, the current lags behind the voltage by a phase angle. This means that when the voltage is at maximum, the current is zero, and vice versa. Fig 5 (a) shows this phase relationship graphically, and figure 5 (b) shows the phasor diagram AC circuit containing inductance only.

1644569283501

Inductive Reactance (XL) is the opposition offered by an inductor to the flow of alternating current, and it increases with frequency- like how a swing resists sudden pushes!

Formula:

$
X_L=2 \pi f L$

Where:
$X_L=$ Inductive Reactance (in ohms)
$f=$ Frequency of AC (in Hz )
$L=$ Inductance (in henrys)


1644569284246

$X_L=\omega L=2 \pi f L$

Where f is the frequency of the alternating current. Thus, the inductive reactance increases with the increasing frequency of the current

When L is in henry, F is in hertz, and XL is in ohm.

AC Voltage Applied to a Capacitor

The alternating Voltage is given by $V=V_0 \sin \omega t$

1644569480246

The maximum current in the circuit

$i=i_0\left(\omega t+\frac{\pi}{2}\right)$

Phasor Diagram:


It is clear that the current through a capacitor is ahead of voltage by $\frac{\pi}{2}$.

Capacitive Reactance (XC) is the opposition offered by a capacitor to the flow of alternating current, and it decreases as the frequency increases, like how a sponge passes water faster when the flow is rapid!

1644569658772

$
X_C=\frac{1}{2 \pi f C}
$

Where:
$X_C=$ Capacitive Reactance (in ohms)
$f=$ Frequency of AC (in Hz )
$C=$ Capacitance (in farads)

Circuit containing Inductance and Resistance in series(L-R series circuit).

Let an alternating voltage V=V0sin ,t be applied to a circuit containing an inductance L and a non-inductive resistance R are in series as shown in fig. The same current will flow both in L and R.

Let i be the current in the circuit at any instant and VL and VR. The potential difference across L and R., respectively at that instant. Then

16445699058341644569905355

$\begin{aligned} & V_L=i X_L \\ & V_R=i R\end{aligned}$

Where XL is the inductive reactance, VR is in phase with the current i, while VL leads i by 90°. Thus, VR and VL are mutually at right angles.

The phasor diagram is drawn in the figure. In this diagram, the vector represents VR while OB represents VL. The vector OD represents the resultant of VR and VL, which is applied voltage V.

$\begin{aligned} & V^2=V_R^2+V_L^2 \\ & V^2=i^2\left(R^2+X_L^2\right) \\ & i=\frac{V}{\sqrt{R^2+X_L^2}}\end{aligned}$

According to Ohm's law, a circuit's effective resistance is

$\sqrt{\left(R^2+X_L^2\right)}$

We can decode it as Z, which is the impedance of the circuit. Hence, we have the L - R circuit.

$Z=\sqrt{\left(R^2+X_L^2\right)}$

But,

$\begin{aligned} & X_L=\omega L \\ & \therefore Z=\sqrt{\left(R^2+(\omega L)^2\right)}\end{aligned}$

The quantity Z is measured in ohm is called impedance because it impedes the flow of alternating current in the circuit. The reciprocal of the impedance is called the admittance of the AC circuit. It is measured in mho or ohm-¹ (Ω-¹) or siemens (S).

The phasor diagram shows in the L - R circuit that the applied voltage V leads the current i (or the current i lags behind the voltage V) by a phase angle ϕ given by

$\begin{aligned} & \tan \phi=\frac{V_L}{V_R}=\frac{i X_L}{i R}=\frac{X_L}{R} \\ & \tan \phi=\frac{\omega L}{R}\end{aligned}$

It is clear from the situation that if L = 0, then ϕ = 0( the voltage and the current will be in the same phase): If R = 0, then ϕ = 90° (the voltage will lead the current by 90° ).

v. A circuit containing capacitance and resistance in series:- in this case, the instantaneous potential difference across C and R are given by VC=iXC and VR=iR where XC is the capacitive reactance and i is the instantaneous current.

16445699008771644569901714

Now, VR is in the phase with i while VC lags behind i by 90°. The phasor diagram is drawn in the figure in which the vector OA represents VR, and the vector OB represents VC. The vector OD represents the resultant of VR and VC, which is applied voltage V. Thus,

$\begin{aligned} & V^2=V_R^2+V_C^2 \\ & V^2=i^2\left(R^2+X_C^2\right) \\ & i=\frac{V}{\sqrt{R^2+X_C^2}}\end{aligned}$

Applying Ohm's law, we see that

$\sqrt{\left(R^2+X_C^2\right)}$

is the effective resistance of the circuit. It is called the impedance of the circuit and is represented by Z. Thus, in a C - R circuit, we have

$Z=\sqrt{\left(R^2+X_C^2\right)}$

But

$\begin{aligned} & X_L=1 / \omega C \\ & \therefore Z=\sqrt{\left(R^2+(1 / \omega C)^2\right)}\end{aligned}$

The phasor diagram shows that in the C R circuit, the applied voltage V lags behind the current i ( or the current i leads the voltage V) by a phase angle ϕ, given by

$\begin{aligned} \tan \phi & =\frac{V_C}{V_R}=\frac{X_C}{R} \\ \tan \phi & =\frac{1 / \omega C}{R}\end{aligned}$

AC Voltage Applied to a Series LCR Circuit

Let an alternating voltage V=V0sinωt be applied to a circuit containing an inductance L, a capacitance C, and resistance R all joined in series. As shown in fig. The same current i will flow in all three and the vector sum of the potential difference across them will be equal to the applied voltage.

1644570405906

Let i be the current in the circuit at any instant of time and VL, VC and VR are the potential difference across L, C and R, respectively at that instant. Then

$V_L=i X_L, V_C=i X_C$ and $V_R=i R$

Now XL and XC are the inductive and capacitive reactance, respectively.

Now VR is in phase with i but VL leads by 90°. The phasor diagram is drawn as a figure. In this diagram, the vector OA represents VR and the vector OB represents VL and the vector OC represents VC. VLand VC is opposite to each other. if VL>VC then their results will be VL-VCwhich is represented by OD. Finally, the vector OF represents the resultant of VR and VL-VC, ie., the resultant of all the three, which is applied voltage V. Thus

$\begin{aligned} & V^2=V_R^2+\left(V_L-V_C\right)^2 \\ & V^2=i^2\left(R^2+\left(X_L-X_C\right)\right) \\ & i=\frac{V}{\sqrt{R^2+\left(X_L-X_C\right)^2}}\end{aligned}$

Applying Ohm's law, we see that
$\sqrt{R^2+\left(X_L-X_C\right)^2}$ is the effective resistance of the circuit and is called the impedance Z of the circuit. Thus, in the L-C-R circuit, we have

$\begin{aligned} & Z=\sqrt{R^2+\left(X_L-X_C\right)^2} \\ & X_L=\omega L \text { and } \\ & X_C=\frac{1}{\omega C}\end{aligned}$

$\therefore \quad Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}$

The phasor diagram (fig) shows that in the L-C-R circuit the applied voltage V leads the current i was a phase angle ϕ given by

$\begin{aligned} & \tan \phi=\frac{V_L-V_C}{V_R}=\frac{X_L-X_C}{R} \\ & \tan \phi=\frac{\omega L-\frac{1}{\omega C}}{R}\end{aligned}$

The following three cases arise

When $ω_L>1/ω_C$ then tanΦ is positive. In this case, the voltage V leads the current i.

When $ω_L$ < 1/ $ω_C$ then tanΦ negative. In this case, the voltage V lags behind the current i.

When $ω_L$=1/$ω_C$ then tanΦ=0 in this case the voltage V and the current i are in the phase.

Again when $ω_L$=1/$ω_C$,

$\therefore \quad Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}=R$

Which is the minimum value Z can have. Thus, in case, impedance is minimum and hence, the current is maximum. This is the case of electrical resonance. Hence, at resonance

$\begin{aligned} & \omega_L=\frac{1}{\omega C} \\ & \omega=\frac{1}{\sqrt{L C}}\end{aligned}$

But ω=2πf, where f is the frequency of the applied voltage. Therefore

$
f=\frac{1}{2 \pi \sqrt{L C}}=f_0
$

Where

$
f_0\left(=\frac{1}{2 \pi \sqrt{L C}}\right)
$
is the natural frequency of the circuit when the resistance is zero.

The condition for resonance is the frequency of the applied voltage should be equal to the natural frequency of the circuit when the resistance of the circuit is zero.

1644570408696

Impedance triangle:- The impedance of an L-C-R a.c. The circuit is given by

$
Z=\sqrt{R^2+\left(X_L-X_C\right)^2}
$

and the phase relationship is given by

$
\tan \phi=\frac{X_L-X_C}{R}
$

This triangle is called the impedance triangle.

Power in AC Circuit: The Power Factor

$
V=V_m \sin \omega t
$


$
I=\frac{V_m}{Z} \sin (\omega t+\phi)=I_m \sin (\omega t+\phi) \text { assuming } V_C>V_L
$


$
\begin{aligned}
P & =V \times I=V_m \cdot I_m \cdot \sin \omega t \cdot \sin (\omega t+\phi) \\
& =\frac{V_m \cdot I_m}{2} \cdot[\cos (\omega t-\omega t+\phi)-\cos (\omega t+\omega t+\phi)]
\end{aligned}
$


$
\frac{V_m \cdot I_m}{2} \cdot[\cos \phi-\cos (2 \omega t+\phi)]
$


$
\bar{P}=V_{\text {rms }} I_{\text {rms }} \cdot \cos \phi
$


As $\cos (2 \omega t+\phi)$ for one complete cycle, is ZERO

$
\begin{aligned}
& \bar{P}=\frac{V_{\mathrm{rms}}^2}{Z} \cdot \cos \phi \\
& \bar{P}=I_{\mathrm{rms}}^2 \times Z \times \cos \phi=I_{\mathrm{rms}}^2 \times Z \times \frac{R}{Z}=I_{\mathrm{rms}}^2 \times R
\end{aligned}
$

(From impedance triangle $\cos \phi=\frac{R}{Z}$ )
The terms cos$\phi$ is known as power factor as it determines the power consumed in the circuit.

Case 1: Circuit containing pure resistance only:-

$\begin{aligned} & \bar{P}=\frac{1}{2} V_0 i_0=\frac{V_0}{\sqrt{2}} \frac{i_0}{\sqrt{2}} \\ & \bar{P}=V_{r m s} \times i_{r m s}\end{aligned}$

Case-2 : Purely Inductive or Capacitive Circuits: The phase difference $\phi=\frac{\pi}{2}$. Thus power factor $\cos \phi=0$. Thus power consumed is ZERO. Although current flows through the circuit but power consumed is ZERO. Such a current is known as wattless current.
Case-3 : In an series LCR circuit, power dissipated is given by

$
\bar{P}=v_{r m s} \cdot I_{r m s} \cdot \cos \phi
$

where $\phi=\tan ^{-1} \frac{\left(X_C-X_L\right)}{R}$ so $\phi$ may be non-zero in a $R L$ circuit or $R C$ or $R C L$ circuit. Even in such cases, power is dissipated in the resistor only.
Case-4 : Power dissipated at resonance in LCR circuit. At resonance $X_C-X_L=0, \phi=0$, therefore $\cos \phi=1$ and $P=I^2 R$. That is maximum power is dissipated in a circuit (through $R$ ) at resonance.

Transformers

It works on the principle of mutual induction. The transformer is a device that is used for converting a large alternating current at low voltage into a small current at high voltage and vice versa. The transformers that convert low voltage into higher ones are called step-up transformers, while those that convert high voltages into lower ones are called step-down transformers

.1644570447124

Construction: A simple transformer consists of two coils called the primary and the secondary, insulated from each other and bound on a common soft iron laminated core. One of the two coils has a smaller number of turns of thick insulated copper wire cover while the other hand a large number of turns of thin insulated copper wire. In a step-up transformer, the coil of the copper wire has a smaller number of turns in a primary coil and the coil of wire has a large number of turns in the secondary coil (fig a) in the step-down transformer the order is reversed.

Theory:- The given source of EMF says AC mains is always connected to the primary coil. When the alternating current flows through the primary coil, then in each cycle of the current, the core is magnetised once in one direction and once in the opposite direction. Moreover, since the secondary coil is also wound on the same core, the magnetic flux passing through it is subjected to continuous changes as the core is magnetised and de-magnetised again and again. Consequently, alternating EMFs at the same frequency are induced in the secondary coil through mutual induction. Depending on the ratio of turns in the two coils, the secondary coil induces its own EMF.

Let Np and Ns be the number of turns in the primary and secondary coil, respectively. Let us assume that there is no leakage of magnetic flux so that the same flux passes through each turn of the primary and secondary. Let ΦB be the magnetic flux linked with each turn of either coil at any instant. Then by Faraday's law of electromagnetic induction, the EMF induced in the primary coil is given by

$e_p=-N_p \frac{\Delta \phi_B}{\Delta t}$

The EMF induced in the secondary coil is given by

$\begin{aligned} & e_s=-N_s \frac{\Delta \phi_B}{\Delta t} \\ & \therefore \quad \frac{e_s}{e_p}=\frac{N_s}{N_p}\end{aligned}$

The induced EMF ep in the primary coil will be nearly equal to the applied voltage Vp across its ends if there is no resistance in the primary circuit and no loss of energy in it. Likewise, if the secondary circuit is open, then the voltage

VS across its and will equal the EMF es induced in it. Under this condition, we have

$\frac{V_s}{V_p}=\frac{e_s}{e_p}=\frac{N_s}{N_p}=r$

Where r is called the transformer ratio. In a step-up transformer, r is more than 1, whereas in a step-down transformer r is less than 1. Thus,

$\frac{\text { voltage obtained across } \sec \text { ondary }}{\text { voltage applied across primary }}=\frac{\text { no. of turns in } \sec \text { ondary }}{\text { no. of turns in primary }}$

If ip and isbe the current in the primary and the secondary at any instant and the energy losses be zero, then

Power in the secondary = power in the primary

$\begin{aligned} & V_s \times i_s=V_p \times i_p \\ & \frac{i_p}{i_s}=\frac{V_s}{V_p}=\frac{N_s}{N_p}=r\end{aligned}$

When voltage is stepped-up, the current is correspondingly reduced in the same ratio and vice-versa. Thus, the energy obtained from the secondary coil is equal to the energy given to the primary coil.

Alternating Current: Previous Year's Question and Answer

Q1: 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:

$X_L=1 \Omega, R=2 \Omega \quad E_{r m s}=6 \mathrm{~V}$

The average power dissipated in the circuit

$\begin{aligned}
& P_{a v}=E_{r m s} I_{r m s} \cos \Phi \\
& I_{r m s}=\frac{E_{r m s}}{Z} \\
& Z=\sqrt{R^2+X_L^2}=\sqrt{4+1}=\sqrt{5} \\
& I_{r m s}=\frac{6}{\sqrt{5}} A \\
& \cos \Phi= \frac{R}{Z}=\frac{2}{\sqrt{5}} \\
& P_{a v}=6 * \frac{6}{\sqrt{5}} * \frac{2}{\sqrt{5}}=\frac{72}{5}=14.4 \mathrm{~W}
\end{aligned}$

Hence, the answer is the option 3.

Q2: 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) $5 \sqrt{\frac{3}{2}} A$
(C) $\frac{5}{6} A$
(D) $\frac{5}{\sqrt{2}} A$

Answer :

$
I=I_0 \sin \omega t
$
It is given that $I_0=\sqrt{2} * 5=5 \sqrt{2} A, \nu=50 \mathrm{~Hz}, t=\frac{1}{300 \mathrm{~s}}$

$
\begin{aligned}
& I=5 \sqrt{2} \sin 2 \pi \nu t \\
& =5 \sqrt{2} \sin 2 \pi * 50 * \frac{1}{300} \\
& =5 \sqrt{2} \sin \left(\frac{\pi}{3}\right)=5 \sqrt{2} \sin 60^{\circ} \\
& =5 \sqrt{2} * \frac{\sqrt{3}}{2}=5 \sqrt{\frac{3}{2}} A
\end{aligned}
$
Hence, The answer is the option (2).

Q3: When an AC voltage of 220 V is applied to the capacitor C, then
(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 power applied to the circuit is given by

$
\begin{aligned}
& \quad P_{a v}=V_{r m s} I_{r m s} \cos \phi \\
& \phi=90^{\circ} \text { for pure capacitor circuit } \\
& P_{a v}=V_{r m s} I_{r m s} \cos 90^{\circ}=0
\end{aligned}
$

Hence, the answer is the option (4).

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Frequently Asked Questions (FAQs)

1. Is Alternating Current class 12th notes important for JEE?

Yes, understanding Alternating Current (AC) class 12th notes is important for JEE preparation as it aligns with the syllabus and provides essential knowledge for tackling physics questions in the exam.

2. What are the main derivations included in the NCERT Class 12 Physics chapter 7?

The main derivations covered in the NCERT book are Circuit containing Capacitance and resistance, Circuit containing inductance and capacitance, and Circuit containing Inductance, Capacitance, and resistance in series

3. In what way does the chapter contribute to the CBSE board exam?

It is one of the important chapters, can expect 4 to 6 marks questions from the chapter alternating Current.

4. What are the main topics covered in alternating current Class 12 notes?

There are some major topics covered in alternating current Class 12 notes, including The Application of AC Voltage to a Resistor, Representating AC Current, and Voltage by Rotating Vectors - Phasors, The Application of AC Voltage to an Inductor, The Application of AC Voltage to a Capacitor, The Application of AC Voltage to a Series LCR Circuit, The Application of AC Voltage to an LCR Circuit, and Power in AC Circuit: The Power Factor

5. Define the transformer according to Alternating Current Class 12 Physics chapter 7 notes.

The transformer is a device that is used for converting a large alternating current at low voltage into a small current at high voltage and vice versa.

These topics can also be downloaded from Alternating Current Class 12 notes pdf download.

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Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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