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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 Feb 02, 2024 05:13 PM IST

In Class 12 Physics, Chapter 7 on alternating current is essential not only for school exams, but also for engineering and medical entrance exams. These class 12 physics chapter 7 notes cover important topics such as power in an AC circuit and transformer principles, which are required for exams such as JEE Main and NEET. Aside from exams, understanding concepts such as Capacitor Applied to AC Voltage and LCR Circuit Applied to AC Voltage is essential for practical applications in medical instrumentation and electrical engineering.

These Alternating Current class 12 notes, available online through platforms such as Careers360, provide students with a thorough understanding of alternating current, bridging theoretical knowledge with real-world application. By mastering these topics, students not only prepare for academic success but also gain the skills required to succeed in engineering and medicine. So, these cbse class 12 physics ch 7 notes serve as indispensable tools for aspiring professionals, guiding them toward achievement in both exams and future careers.

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NCERT Class 12 Physics Chapter 7 Notes

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

{"backgroundColor":"#ffffff","type":"$$","backgroundColorModified":false,"aid":null,"id":"2","code":"$$V\\,=\\,V_{0}\\sin \\omega t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,.......\\left(i\\right)$$","font":{"size":11,"color":"#000000","family":"Arial"},"ts":1636192242590,"cs":"mERDzGMk27Obe8cRFfkJAA==","size":{"width":226,"height":16}}

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

{"id":"3","backgroundColor":"#ffffff","backgroundColorModified":false,"code":"$$i\\,=\\,i_{0}\\sin \\omega t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,......\\left(ii\\right)$$","aid":null,"font":{"size":11,"color":"#000000","family":"Arial"},"type":"$$","ts":1636192221756,"cs":"bZFJ0y+7xxnI6sr96OCMzQ==","size":{"width":193,"height":16}}

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

{"aid":null,"font":{"family":"Arial","size":11,"color":"#000000"},"backgroundColor":"#ffffff","code":"$$T\\,=\\,\\frac{2\\pi}{\\omega}$$","id":"4","backgroundColorModified":false,"type":"$$","ts":1636192572589,"cs":"NoU8A8FOLZl+RS8Y61E+NA==","size":{"width":64,"height":34}}

  • 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 be T, then its frequency is

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{"backgroundColorModified":false,"backgroundColor":"#ffffff","id":"5","code":"\\begin{align*}\n{f\\,}&={\\,\\frac{1}{T}}\\\\\n{}&\\relempty{\\therefore f\\,=\\,\\frac{\\omega}{2\\pi}\\,}\\\\\n{\\,}&\\relempty{}\t\n\\end{align*}","font":{"color":"#000000","size":11,"family":"Arial"},"aid":null,"type":"align*","ts":1636193132248,"cs":"VJ52QohQE4XCTCLmvrm7IQ==","size":{"width":112,"height":70}}

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

{"backgroundColor":"#ffffff","code":"\\begin{lalign*}\n&{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,i_{m\\,}=\\,\\frac{2}{\\pi}i_{0}\\,=\\,0.637i_{0}}\\\\\n&{\\,\\,\\,\\,\\,}\t\n\\end{lalign*}","backgroundColorModified":false,"type":"lalign*","font":{"family":"Arial","color":"#000000","size":11},"id":"8","aid":null,"ts":1636202416842,"cs":"2FjMG20GMGW8l3cz+WYvtQ==","size":{"width":154,"height":34}}



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 you are the alternating current is given by

{"code":"$$i_{rms\\,}=\\,{\\sqrt[]{\\bar{i^{2}}}}\\,=\\,\\frac{i_{0}}{{\\sqrt[]{2}}\\,}=\\,0.707i_{0}$$","font":{"color":"#000000","size":11,"family":"Arial"},"backgroundColorModified":false,"id":"12","type":"$$","aid":null,"backgroundColor":"#ffffff","ts":1636197061833,"cs":"ShMjtC3YwHPfubBrccYMag==","size":{"width":225,"height":42}}

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

{"aid":null,"type":"$$","backgroundColor":"#ffffff","id":"13","font":{"family":"Arial","size":11,"color":"#000000"},"code":"$$V_{rms}\\,=\\,\\frac{V_{0}}{{\\sqrt[]{2}}}\\,=\\,0.707V_{0}$$","backgroundColorModified":false,"ts":1636197419352,"cs":"3Uq+gPkDxn3w9OqstOkYWQ==","size":{"width":174,"height":40}}

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

{"backgroundColor":"#ffffff","backgroundColorModified":null,"id":"14","code":"$V\\,=\\,V_{0}\\sin \\omega t$","aid":null,"font":{"color":"#000000","family":"Arial","size":11},"type":"$","ts":1636198510434,"cs":"keQX3z6NXLyQV5rpFwerSQ==","size":{"width":100,"height":13}}

{"backgroundColor":"#ffffff","backgroundColorModified":false,"aid":null,"id":"15","type":"$$","code":"$$i\\,=\\,i_{0}\\sin \\omega t$$","font":{"family":"Arial","size":11,"color":"#000000"},"ts":1636198539840,"cs":"l/EqJru5W07NNKDPnx5/yA==","size":{"width":89,"height":13}}

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

{"backgroundColor":"#ffffff","backgroundColorModified":false,"type":"$$","font":{"family":"Arial","color":"#000000","size":11},"id":"16-0","aid":null,"code":"$$V\\,=\\,V_{0}\\sin \\omega t$$","ts":1636198687131,"cs":"SNG+qRVu0RiKeLB7jsl/UQ==","size":{"width":100,"height":13}}

{"code":"$$i\\,=\\,i_{0}\\sin\\left(\\omega t\\,+\\,\\phi\\right)$$","id":"17-0","aid":null,"backgroundColor":"#ffffff","backgroundColorModified":false,"type":"$$","font":{"size":11,"family":"Arial","color":"#000000"},"ts":1636198760616,"cs":"S9vhwk5CgzNBFGRxVPBOLw==","size":{"width":138,"height":16}}

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,

{"backgroundColor":"#ffffff","backgroundColorModified":false,"type":"$$","font":{"family":"Arial","color":"#000000","size":11},"id":"16-1","aid":null,"code":"$$V\\,=\\,V_{0}\\sin \\omega t$$","ts":1636198687131,"cs":"zBvypbF5Dx33bkkjBCzZOw==","size":{"width":100,"height":13}}

{"aid":null,"font":{"size":11,"color":"#000000","family":"Arial"},"backgroundColor":"#ffffff","code":"$$i\\,=\\,i_{0}\\sin\\left(\\omega t\\,-\\,\\phi\\right)$$","backgroundColorModified":false,"id":"17","type":"$$","ts":1636198997352,"cs":"fl9KYyMSf28GQCojQiZpWw==","size":{"width":138,"height":16}}

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).

Different Types of A.C. Circuits

i. A circuit containing resistance only: let an alternating voltage V is given by

{"aid":null,"backgroundColorModified":false,"id":"16-2","type":"$$","font":{"family":"Arial","color":"#000000","size":11},"backgroundColor":"#ffffff","code":"$$V\\,=\\,V_{0}\\sin \\omega t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,.....\\left(i\\right)$$","ts":1636200246425,"cs":"SZaCkD4SSctpzo0VDXmAMw==","size":{"width":194,"height":16}}

And the peak value of the current in the circuit

1644569217491

{"backgroundColorModified":false,"id":"19","aid":null,"type":"$$","code":"$$\\therefore i\\,=\\,i_{0}\\sin \\omega t\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,......\\left(ii\\right)$$","font":{"size":11,"family":"Arial","color":"#000000"},"backgroundColor":"#ffffff","ts":1636200298232,"cs":"wvI0vB6o8Zgxmx+UPl21Cw==","size":{"width":228,"height":16}}

A comparison of equation (ii) with equation (i) shows that in a pure resistor the current I is always in phase with the applied voltage V . Figure (a) shows the phase relationship graphically and fig (b) is the phasor diagram for alternating voltage and current.

A comparison of equation (ii) with equation (i) shows that in a pure resistor the current I is always in phase with the applied voltage V . Figure (a) shows the phase relationship graphically and fig (b) is the phasor diagram for alternating voltage and current.

1644569250021

ii. A circuit containing inductance only: Let an alternating voltage is given by

{"backgroundColor":"#ffffff","backgroundColorModified":null,"id":"14","code":"$V\\,=\\,V_{0}\\sin \\omega t$","aid":null,"font":{"color":"#000000","family":"Arial","size":11},"type":"$","ts":1636198510434,"cs":"keQX3z6NXLyQV5rpFwerSQ==","size":{"width":100,"height":13}}

1644569356561

The maximum current in the circuit,

{"type":"lalign*","id":"24","backgroundColor":"#ffffff","font":{"color":"#000000","size":11,"family":"Arial"},"code":"\\begin{lalign*}\n&{i\\,=\\,\\frac{V_{0}}{\\omega L}\\sin\\left(\\omega t-\\,\\frac{\\pi}{2}\\right)}\t\n\\end{lalign*}","aid":null,"backgroundColorModified":false,"ts":1636203093154,"cs":"20YgYLX0T+wV/UJW8878aw==","size":{"width":164,"height":34}}


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 of ?/2. This means that when voltage is 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:- By applying Ohm’s law in the eqn of the peak value of the current in the coil i0=V0/?L we find that ?L has thedimension of resistance which is known as ‘ reactance of the coil’ or inductive reactance. It is denoted by XL.


1644569284246

{"font":{"family":"Arial","size":11,"color":"#000000"},"backgroundColorModified":false,"backgroundColor":"#ffffff","type":"$$","code":"$$X_{L}\\,=\\,\\omega L\\,=\\,2\\pi fL$$","id":"27","aid":null,"ts":1636206963060,"cs":"SbBpKW6Tx0jTbbu/r4OSCQ==","size":{"width":140,"height":14}}

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

When L is in henry and f is in hertz and XL is in ohm.

iii. Circuit containing Capacitance only:- The alternating Voltage is given by

{"backgroundColor":"#ffffff","backgroundColorModified":false,"type":"$$","font":{"family":"Arial","color":"#000000","size":11},"id":"16-0","aid":null,"code":"$$V\\,=\\,V_{0}\\sin \\omega t$$","ts":1636198687131,"cs":"SNG+qRVu0RiKeLB7jsl/UQ==","size":{"width":100,"height":13}}1644569480246







The maximum current in the circuit

{"id":"29","aid":null,"backgroundColor":"#ffffff","type":"$$","code":"$$i\\,=\\,i_{0}\\left(\\omega t\\,+\\frac{\\pi}{2}\\right)$$","font":{"color":"#000000","size":11,"family":"Arial"},"backgroundColorModified":false,"ts":1636208257558,"cs":"RkoU6vf1+wgZXaxIZ6BcMw==","size":{"width":121,"height":30}}



Where I0=V0/(1/?C) is the peak value of the current.

Comparing this equation with the voltage equation shows that in a perfect capacitor, the current leads by the voltage by a phase angle of ?/2 or we can say that the voltage lags behind the current by a phase angle of ?/2. This means that when the voltage is zero, the current is maximum and vice versa. Fig 6(a) shows the phase relationship graphically and fig6(b) is the phasor diagram of a.c. a circuit containing C only.

1644569480990


Capacitive Reactance:- On applying Ohm’s law in the eqn of the peak value of the current in the coil I0=V0/(1/?C) we find that 1/?C has the dimension of resistance which is known as reactance of the capacitor or capacitive reactance. It is denoted by XC

{"id":"31","code":"$$X_{C}\\,=\\,\\frac{1}{\\omega C}\\,=\\,\\frac{1}{2\\pi fC}$$","aid":null,"font":{"size":11,"color":"#000000","family":"Arial"},"backgroundColor":"#ffffff","backgroundColorModified":false,"type":"$$","ts":1636208961734,"cs":"eFDWgLCviBLIFkYsXmlsPw==","size":{"width":158,"height":37}}1644569658772

Where f is the frequency of the alternating current. Thus, the capacitive reactance decrease with increasing frequency of the current. The graph between XC and f is a rectangular hyperbola.

When C is in farad and f is in hertz then XC=1/2?fC is in ohm


iv. 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 the current in the circuit at any instant and VL and VR. The potential difference across L and R respectively at that instant. Then16445699058341644569905355

{"backgroundColor":"#ffffff","type":"$$","font":{"family":"Arial","color":"#000000","size":11},"id":"35","backgroundColorModified":false,"code":"$$V_{L\\,}\\,=\\,iX_{L}$$","aid":null,"ts":1636209962853,"cs":"ZvrKROY0pk3aEcCC+lVY9w==","size":{"width":77,"height":13}}

{"type":"$$","id":"36","backgroundColorModified":false,"backgroundColor":"#ffffff","aid":null,"font":{"family":"Arial","color":"#000000","size":11},"code":"$$V_{R}\\,=iR$$","ts":1636209986530,"cs":"/tKhI3t7Tjtsg8lWm/tYtA==","size":{"width":64,"height":13}}

Where XL is the inductive reactance now VR is in phase with the current i, while VLleads 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.

{"backgroundColor":"#ffffff","backgroundColorModified":false,"code":"\\begin{lalign*}\n&{V^{2\\,}\\,=\\,V_{R}^{2}\\,+\\,V_{L}^{2}}\\\\\n&{V^{2\\,}=\\,i^{2}\\left(R^{2}\\,+\\,X_{L}^{2}\\right)}\\\\\n&{i\\,=\\frac{V}{{\\sqrt[]{R^{2\\,}+\\,X_{L}^{2}}}}}\t\n\\end{lalign*}","font":{"size":11,"family":"Arial","color":"#000000"},"type":"lalign*","aid":null,"id":"42-0","ts":1636210622531,"cs":"NHPzyUaYefPMRyLTYfok4A==","size":{"width":144,"height":105}}

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

{"code":"$${\\sqrt[]{\\left(R^{2}+X_{L}^{2}\\right)}}$$","backgroundColor":"#ffffff","font":{"size":11,"family":"Arial","color":"#000000"},"aid":null,"backgroundColorModified":false,"id":"43-0-0","type":"$$","ts":1636210747538,"cs":"Ueny3/XX2fQyvGjSzYbJQw==","size":{"width":96,"height":30}}. We can decode it as Z, which is the impedance of the circuit. Hence, we have L - R circuit.

{"backgroundColor":"#ffffff","aid":null,"id":"43-1-0-0","code":"$$Z\\,=\\,{\\sqrt[]{\\left(R^{2}+X_{L}^{2}\\right)}}$$","type":"$$","backgroundColorModified":false,"font":{"color":"#000000","family":"Arial","size":11},"ts":1636210924472,"cs":"+bUMkdawexrw6hLjEs3fYA==","size":{"width":137,"height":30}}


But {"font":{"family":"Arial","color":"#000000","size":11},"type":"$$","backgroundColor":"#ffffff","backgroundColorModified":false,"id":"44-0","code":"$$X_{L\\,}\\,=\\,\\omega L$$","aid":null,"ts":1636210967843,"cs":"dLewGTO1BjSI1pH4OEUazw==","size":{"width":76,"height":13}}

{"id":"43-1-1-0","type":"$$","backgroundColorModified":false,"font":{"color":"#000000","size":11,"family":"Arial"},"code":"$$\\therefore Z\\,=\\,{\\sqrt[]{\\left(R^{2}+\\left(\\omega L\\right)_{}^{2}\\right)}}$$","backgroundColor":"#ffffff","aid":null,"ts":1636211052460,"cs":"izrtNSX2ZWJ+XUx4eXEwbA==","size":{"width":178,"height":40}}

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 L - R circuit the applied voltage V leads the current i (or the current i, lags behind the voltage V) by a phase angle ϕ given by

{"font":{"family":"Arial","color":"#000000","size":11},"backgroundColor":"#ffffff","type":"lalign*","backgroundColorModified":false,"aid":null,"code":"\\begin{lalign*}\n&{\\tan \\phi=\\,\\frac{V_{L}}{V_{R}}\\,=\\,\\frac{iX_{L}}{iR}\\,=\\frac{X_{L}}{R}\\,}\\\\\n&{\\tan \\phi=\\,\\frac{\\omega L}{R}}\t\n\\end{lalign*}","id":"45","ts":1636211553217,"cs":"+6RLZSavVu7akj2GNoeTlw==","size":{"width":210,"height":77}}

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 represent VR and the vector OB represents VC the vector OD represents the resultant of VR and VCwhich is applied voltage V. Thus,


{"aid":null,"backgroundColor":"#ffffff","code":"\\begin{lalign*}\n&{V^{2\\,}\\,=\\,V_{R}^{2}\\,+\\,V_{C}^{2}}\\\\\n&{V^{2\\,}=\\,i^{2}\\left(R^{2}\\,+\\,X_{C}^{2}\\right)}\\\\\n&{i\\,=\\frac{V}{{\\sqrt[]{R^{2\\,}+\\,X_{C}^{2}}}}}\t\n\\end{lalign*}","type":"lalign*","backgroundColorModified":false,"font":{"family":"Arial","size":11,"color":"#000000"},"id":"42-1","ts":1636212383205,"cs":"oJi96vC/Mt4rGBVHBFQKPw==","size":{"width":145,"height":105}}

Applying Ohm's law we see that

{"type":"$$","backgroundColorModified":false,"id":"43-0","code":"$${\\sqrt[]{\\left(R^{2}+X_{C}^{2}\\right)}}$$","font":{"family":"Arial","color":"#000000","size":11},"aid":null,"backgroundColor":"#ffffff","ts":1636212420048,"cs":"wFrlzjajwf8vRlUSx1rLmg==","size":{"width":96,"height":30}}

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

{"backgroundColor":"#ffffff","backgroundColorModified":false,"type":"$$","font":{"size":11,"family":"Arial","color":"#000000"},"code":"$$Z\\,=\\,{\\sqrt[]{\\left(R^{2}+X_{C}^{2}\\right)}}$$","id":"43-1-0","aid":null,"ts":1636212444187,"cs":"WsWGWQss+hEBjxeOIaMwuA==","size":{"width":137,"height":30}}

But

{"type":"$$","backgroundColor":"#ffffff","font":{"size":11,"color":"#000000","family":"Arial"},"code":"$$X_{L\\,}\\,=\\,1/\\omega C$$","aid":null,"backgroundColorModified":false,"id":"44-1","ts":1636212463316,"cs":"cjysR4ub6eD2kvBeTH5bSg==","size":{"width":94,"height":16}}

{"aid":null,"backgroundColor":"#ffffff","code":"$$\\therefore Z\\,=\\,{\\sqrt[]{\\left(R^{2}+\\left(1/\\omega C\\right)_{}^{2}\\right)}}$$","font":{"family":"Arial","size":11,"color":"#000000"},"backgroundColorModified":false,"id":"43-1-1-1","type":"$$","ts":1636212482108,"cs":"VCb4jFgQRBGo89IdcxOoyQ==","size":{"width":196,"height":40}}

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

{"code":"\\begin{lalign*}\n&{\\tan \\phi=\\,\\frac{V_{C}}{V_{R}}\\,=\\,\\frac{X_{C}}{R}}\\\\\n&{\\tan \\phi=\\,\\frac{1/\\omega C}{R}}\t\n\\end{lalign*}","aid":null,"id":"53","backgroundColorModified":false,"type":"lalign*","font":{"color":"#000000","size":11,"family":"Arial"},"backgroundColor":"#ffffff","ts":1636212708841,"cs":"K2BIFsp/YIkw7bNPeWmSSw==","size":{"width":149,"height":80}}

vi. A circuit containing capacitance and Inductance (L-C Circuit):- In this case, the potential difference across L will lead the current i in the face by 90° while the potential difference {"font":{"family":"Arial","color":"#000000","size":11},"aid":null,"code":"$$V_{c}$$","id":"50","type":"$$","backgroundColor":"#ffffff","backgroundColorModified":false,"ts":1636212233329,"cs":"94Cccy6tv2pcRaX6au01qQ==","size":{"width":13,"height":13}} across C will lag behind the current i in phase by 90° the phasor diagram is drawn in the figure.

16445703471521644570346681


It meets the phase difference between VL and VC is shown 180 ° i.e, they are opposite in phase to each other. Hence the resultant potential difference and LC circuit is

{"type":"$$","aid":null,"backgroundColorModified":null,"backgroundColor":"#ffffff","id":"54","font":{"family":"Arial","color":"#000000","size":11},"code":"$$V\\,=\\,V_{L\\,}\\approx V_{C}$$","ts":1636269087465,"cs":"a9x6P8h80zSzUylAxGdgag==","size":{"width":104,"height":13}}

and the impedance of the circuit is

{"id":"55","font":{"size":11,"color":"#000000","family":"Arial"},"aid":null,"code":"$$Z\\,=\\,X_{L}\\,\\approxat{}\\,X_{C}$$","backgroundColor":"#ffffff","type":"$$","backgroundColorModified":false,"ts":1636269214453,"cs":"k8nNfcV11xOT14lGzBe5aw==","size":{"width":114,"height":13}}

if XL=XC then the impedance Z = 0 in this situation the amplitude of the current in the circuit would be infinite. It is the condition of electrical resonance. Thus, in the condition electrical resonance

{"code":"\\begin{lalign*}\n&{X_{L\\,}=\\,X_{C}}\\\\\n&{\\omega L\\,=\\,\\frac{1}{\\omega C}}\\\\\n&{2\\pi fL\\,=\\,\\frac{1}{2\\pi fC}}\\\\\n&{f^{2}\\,=\\,\\frac{1}{4\\pi^{2}LC}}\\\\\n&{f\\,=\\,\\frac{1}{2\\pi}{\\sqrt[]{\\frac{1}{LC}}}}\t\n\\end{lalign*}","type":"lalign*","id":"57","backgroundColorModified":false,"font":{"size":11,"family":"Arial","color":"#000000"},"aid":null,"backgroundColor":"#ffffff","ts":1636269565035,"cs":"l7bqftPvK2KvEjnsSjUqdQ==","size":{"width":113,"height":188}}

This is called a resonant frequency of the circuit.

vii. A circuit containing inductance capacitance and resistance in a series (LCR series 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

{"backgroundColorModified":false,"backgroundColor":"#ffffff","font":{"size":11,"color":"#000000","family":"Arial"},"type":"$$","code":"$$V_{L}\\,=\\,iX_{L}$$","id":"62","aid":null,"ts":1636270047589,"cs":"53bEtTL1jYcy9NCCKmntjA==","size":{"width":74,"height":13}}, {"backgroundColorModified":false,"aid":null,"type":"$$","id":"63","backgroundColor":"#ffffff","font":{"family":"Arial","size":11,"color":"#000000"},"code":"$$V_{C}\\,=\\,iX_{C}$$","ts":1636270085454,"cs":"ToGZ+1/nKuYkDbnuEWKq/w==","size":{"width":76,"height":13}} and {"backgroundColor":"#ffffff","backgroundColorModified":false,"aid":null,"id":"64","type":"$$","code":"$$V_{R}\\,=\\,iR$$","font":{"family":"Arial","color":"#000000","size":11},"ts":1636270105056,"cs":"TfyW0FuzJD0QzW4UbdRWCw==","size":{"width":66,"height":13}}


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

{"id":"78","backgroundColor":"#ffffff","backgroundColorModified":false,"type":"lalign*","aid":null,"font":{"family":"Arial","color":"#000000","size":11},"code":"\\begin{lalign*}\n&{V^{2}\\,=\\,V_{R}^{2}\\,+\\,\\left(V_{L}\\,-\\,V_{C}\\right)^{2}}\\\\\n&{V^{2}\\,=\\,i^{2}\\left(R^{2}\\,+\\,\\left(X_{L}\\,-\\,X_{C}\\right)\\right)}\\\\\n&{i\\,=\\,\\frac{V}{{\\sqrt[]{R^{2}\\,+\\,\\left(X_{L}\\,-\\,X_{C}\\right)^{2}}}}}\t\n\\end{lalign*}","ts":1636271300421,"cs":"w8sAR0W5j0vi5TI4jSutLw==","size":{"width":209,"height":106}}

Applying ohm's law we see that


{"code":"$${\\sqrt[]{R^{2}\\,+\\,\\left(X_{L}\\,-\\,X_{C}\\right)^{2}}}$$","backgroundColor":"#ffffff","id":"79","font":{"color":"#000000","family":"Arial","size":11},"aid":null,"type":"$$","backgroundColorModified":false,"ts":1636271335994,"cs":"OU2n56ARWSBoIVlCFbv31g==","size":{"width":156,"height":30}} 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

{"aid":null,"font":{"size":11,"family":"Arial","color":"#000000"},"id":"80","code":"$$Z\\,=\\,{\\sqrt[]{R^{2}\\,+\\,\\left(X_{L}\\,-\\,X_{C}\\right)^{2}}}$$","backgroundColorModified":false,"backgroundColor":"#ffffff","type":"$$","ts":1636271416373,"cs":"FawCDxnx/6acK8iFB6TGBA==","size":{"width":198,"height":30}}

{"aid":null,"backgroundColor":"#ffffff","font":{"family":"Arial","size":11,"color":"#000000"},"id":"81","type":"$$","backgroundColorModified":false,"code":"$$X_{L}\\,=\\,\\omega L$$","ts":1636271442399,"cs":"FkxAnPBKdKm/lWtgOVorXg==","size":{"width":72,"height":13}} and

{"font":{"size":11,"color":"#000000","family":"Arial"},"id":"82","type":"$$","code":"$$X_{C}\\,=\\,\\frac{1}{\\omega C}$$","backgroundColorModified":false,"aid":null,"backgroundColor":"#ffffff","ts":1636271468950,"cs":"gNyu6EsAk6rDf3oLfEDEBA==","size":{"width":81,"height":34}}

{"type":"$$","backgroundColorModified":false,"font":{"family":"Arial","color":"#000000","size":11},"id":"83-0","aid":null,"backgroundColor":"#ffffff","code":"$$\\therefore\\,\\,\\,\\,Z\\,=\\,{\\sqrt[]{R^{2}\\,+\\,\\left(\\omega L\\,-\\,\\frac{1}{\\omega C}\\right)^{2}}}$$","ts":1636271525771,"cs":"PrJPYBpU210+svL901UZxg==","size":{"width":244,"height":50}}


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

{"id":"84","type":"lalign*","code":"\\begin{lalign*}\n&{\\tan \\phi=\\,\\frac{V_{L}-\\,V_{C}}{V_{R}}\\,=\\,\\frac{X_{L}\\,-\\,X_{C}}{R}}\\\\\n&{\\tan \\phi=\\,\\frac{\\omega L-\\,\\frac{1}{\\omega C}}{R}}\t\n\\end{lalign*}","aid":null,"backgroundColorModified":false,"backgroundColor":"#ffffff","font":{"family":"Arial","size":11,"color":"#000000"},"ts":1636271723254,"cs":"pvSRtES+Eoz1ckB0hqbFzA==","size":{"width":242,"height":84}}

The following three cases arise

When ωL>1/ωC then tanΦ is positive. In this case, the voltage V leads the current i.

When ωL<1/ωC then tanΦ negative. In this case, the voltage V lags behind the current i.

When ωL=1/ωC then tanΦ=0 in this case the voltage V and the current i are in the phase.

Again when ωL=1/ωC,

{"type":"$$","backgroundColor":"#ffffff","code":"$$\\therefore\\,\\,\\,\\,Z\\,=\\,{\\sqrt[]{R^{2}\\,+\\,\\left(\\omega L\\,-\\,\\frac{1}{\\omega C}\\right)^{2}}}\\,=\\,R$$","aid":null,"font":{"color":"#000000","family":"Arial","size":11},"id":"83-1","backgroundColorModified":false,"ts":1636272245968,"cs":"6INL0gPw3ZG5PtRG8IAAQw==","size":{"width":286,"height":50}}

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

{"aid":null,"type":"lalign*","backgroundColor":"#ffffff","id":"90","font":{"color":"#000000","size":11,"family":"Arial"},"backgroundColorModified":false,"code":"\\begin{lalign*}\n&{\\omega L\\,=\\,\\frac{1}{\\omega C}}\\\\\n&{\\omega\\,=\\,\\frac{1}{{\\sqrt[]{LC}}}}\t\n\\end{lalign*}","ts":1636272435318,"cs":"RlVJhDKtQHvjHs7+Ehss7w==","size":{"width":84,"height":80}}

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

{"backgroundColor":"#ffffff","code":"$$f\\,=\\,\\frac{1}{2\\pi{\\sqrt[]{LC}}}\\,=\\,f_{0}$$","backgroundColorModified":false,"aid":null,"type":"$$","font":{"family":"Arial","size":11,"color":"#000000"},"id":"92","ts":1636272541289,"cs":"snpXehPvWt2pOg1ns8ydnA==","size":{"width":144,"height":40}}

Where

{"id":"93","backgroundColor":"#ffffff","aid":null,"font":{"color":"#000000","size":11,"family":"Arial"},"type":"$$","backgroundColorModified":false,"code":"$$f_{0}\\,\\left(=\\frac{1}{2\\pi{\\sqrt[]{LC}}}\\right)$$","ts":1636272586780,"cs":"+hLCFRLhS4TWh/q2Ougc3w==","size":{"width":121,"height":41}}

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

{"aid":null,"font":{"size":11,"family":"Arial","color":"#000000"},"id":"80","code":"$$Z\\,=\\,{\\sqrt[]{R^{2}\\,+\\,\\left(X_{L}\\,-\\,X_{C}\\right)^{2}}}$$","backgroundColorModified":false,"backgroundColor":"#ffffff","type":"$$","ts":1636271416373,"cs":"FawCDxnx/6acK8iFB6TGBA==","size":{"width":198,"height":30}}

and the phase relationship is given by

{"backgroundColor":"#ffffff","id":"94","font":{"size":11,"family":"Arial","color":"#000000"},"backgroundColorModified":false,"aid":null,"code":"$$\\tan \\phi=\\,\\frac{X_{L}-X_{C}}{R}$$","type":"$$","ts":1636272789998,"cs":"udeMBFWIj8MkYbgIiW+5lg==","size":{"width":136,"height":34}}

this triangle is called the impedance triangle.

Power in A.C. Circuit:-

(i) Circuit containing pure resistance only:-

{"font":{"size":11,"color":"#000000","family":"Arial"},"id":"95-0","type":"$$","backgroundColor":"#ffffff","aid":null,"code":"$$\\bar{P}=\\,\\frac{1}{2}V_{0}i_{0}\\,=\\,\\frac{V_{0}}{{\\sqrt[]{2}}}\\,\\frac{i_{0}}{{\\sqrt[]{2}}}$$","backgroundColorModified":false,"ts":1636274873685,"cs":"7VHXU216F1fQNgYcEGiwSA==","size":{"width":176,"height":40}}


{"aid":null,"code":"$$\\bar{P}\\,=\\,V_{rms}\\,\\times\\,i_{rms}$$","type":"$$","id":"96","backgroundColorModified":false,"font":{"size":11,"family":"Arial","color":"#000000"},"backgroundColor":"#ffffff","ts":1636274938977,"cs":"VCEZ4Os8vZDZf3uhqKMQyQ==","size":{"width":128,"height":16}}

(ii) Circuit containing both inductance and resistance (L-R) circuit:-


{"code":"$$\\bar{P}=\\,\\frac{1}{2}V_{0}i_{0}\\,\\cos \\phi=\\,\\frac{V_{0}}{{\\sqrt[]{2}}}\\,\\frac{i_{0}}{{\\sqrt[]{2}}}\\,\\cos \\phi$$","font":{"family":"Arial","color":"#000000","size":11},"type":"$$","backgroundColor":"#ffffff","id":"95-1","backgroundColorModified":false,"aid":null,"ts":1636275237023,"cs":"q+WVvpplh5bXpsoUZGjAVA==","size":{"width":257,"height":40}}


{"aid":null,"id":"97","code":"$$\\bar{P}\\,=\\,V_{rms}\\times i_{rms}\\times\\,\\cos \\phi$$","backgroundColorModified":false,"font":{"size":11,"family":"Arial","color":"#000000"},"type":"$$","backgroundColor":"#ffffff","ts":1636275314288,"cs":"HjCJZtss6L5+xKaCOK23kQ==","size":{"width":185,"height":16}}


cosΦ is known as the power factor of the circuit. Its value depends on the nature of the circuit.

{"backgroundColor":"#ffffff","id":"99","type":"$$","aid":null,"backgroundColorModified":false,"code":"$$\\cos \\phi=\\,\\frac{R}{{\\sqrt[]{R^{2}\\,+\\,\\omega^{2}L^{2}}}}$$","font":{"size":11,"color":"#000000","family":"Arial"},"ts":1636275680782,"cs":"caxUxXVn+XwT7ZScDJb+sQ==","size":{"width":164,"height":40}}

Since the value of cosΦ is always less than 1 the power dissipation in the L-R circuit is less than that in the pure-resistance circuit for the same voltage and the current.

Wattless Current:- If If the resistance in the a.c. the circuit is zero, although current flows in the circuit, yet the average power remains zero, that is, there is no energy dissipation in the circuit. The current in such a circuit is called wattless current.

Electrical Oscillations in L-C circuit

Electrical Oscillation takes place when a charged capacitor is discharged through an inductor of negligible ohmic resistance. These are known as L-C oscillations. As shown in the figure a capacitor C and resistance-free induction coil L be connected to a battery. Let the switch S be first connected to a to charge the capacitor then to b to discharge the capacitor through the inductance coil L.

When it is fully charged, the charge on the capacitor is q0 then the amount of energy q02/2Cis stored as electrical energy in an electrical field between its plates. At this moment the current in the circuit is zero. (fig a)


1644570446538

When the capacitor is connected to the inductance coil L and starts discharging, an anticlockwise current starts flowing in the circuit(fig b). As the current rises from zero, a magnetic field starts developing around and the inductance L, and when the current reaches its peak value i0 the energy stored as a magnetic field is Li02/2 and the capacitor is entirely discharged. Thus, the electrical energy stored between the plates of the capacitor is transferred to the magnetic field around an inductance. After the capacitor is discharged the current i0 does not become zero suddenly but due to the inductance L continuously flowing in the same direction for the same time while the magnetic field dies down. The current continues to flow in the same direction and the capacitor is again charged in the reverse direction ( fig c). After the capacitor is charged the current becomes zero. The energy stored in the form of a magnetic field is stored in the form of an electric field between the plates of the capacitor.

Now the capacitor discharge in the opposite direction and the current starts in the opposite direction and the current starts increasing from zero in the reverse direction. This process of charging and discharging the capacitor is repeated again and again. According to the current in the L-C-R circuit first rises from zero to a maximum value in one direction, reduced to zero, then rises to the same maximum value in the reverse direction and again reduces to zero. The electrical oscillation occurs in the circuit. The frequency of these oscillations depends upon the capacitance C of the capacitor and the inductance L. It can be shown by calculation that if the resistance of the circuit be negligible, the frequency of the electrical oscillation in the circuit is given by

{"id":"105","backgroundColor":"#ffffff","backgroundColorModified":false,"type":"$$","aid":null,"font":{"color":"#000000","family":"Arial","size":11},"code":"$$f\\,=\\,\\frac{1}{2\\pi}{\\sqrt[]{\\frac{1}{LC}}}\\,=\\,\\frac{1}{2\\pi}{\\sqrt[]{\\frac{1/C}{L}}}$$","ts":1636362834962,"cs":"UvD6hIw0lotb4NTEPu3TjA==","size":{"width":222,"height":41}}

Normally it runs in lacs. These oscillations do not continue for infinite time because there is always some resistance in the circuit due to which some energy is lost in the form of energy.

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 magnetized 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 magnetized and de-magnetized 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

{"backgroundColor":"#ffffff","aid":null,"font":{"color":"#000000","family":"Arial","size":11},"code":"$$e_{p}\\,=\\,-\\,N_{p}\\frac{\\Delta \\phi_{B}}{\\Delta t}$$","id":"109","backgroundColorModified":false,"type":"$$","ts":1636364541089,"cs":"TonsOgQYlpjqoIZK0Owmsw==","size":{"width":128,"height":36}}

And the EMF induced in the secondary coil is given by

{"font":{"color":"#000000","family":"Arial","size":11},"aid":null,"code":"$$e_{s}\\,=\\,-\\,N_{s}\\frac{\\Delta \\phi_{B}}{\\Delta t}$$","id":"110","type":"$$","backgroundColorModified":false,"backgroundColor":"#ffffff","ts":1636364635130,"cs":"jYXXo157dAGS+4VGAKsgMQ==","size":{"width":126,"height":36}}

{"code":"$$\\therefore\\,\\,\\,\\,\\,\\frac{e_{s}}{e_{p}}\\,=\\,\\frac{N_{s}}{N_{p}}$$","backgroundColor":"#ffffff","type":"$$","font":{"size":11,"color":"#000000","family":"Arial"},"id":"111","aid":null,"backgroundColorModified":false,"ts":1636364694024,"cs":"ViFdY/9jfF7xZ7WZowIg6g==","size":{"width":106,"height":40}}

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

{"backgroundColorModified":false,"id":"116","font":{"family":"Arial","color":"#000000","size":11},"aid":null,"type":"$$","backgroundColor":"#ffffff","code":"$$\\frac{V_{s}}{V_{p}}\\,=\\,\\frac{e_{s}}{e_{p}}\\,=\\,\\frac{N_{s}}{N_{p}}\\,=\\,r,$$","ts":1636365040499,"cs":"O9Rv7ecEl4hfh4rMBM3tcg==","size":{"width":170,"height":40}}

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

{"font":{"family":"Arial","color":"#000000","size":11},"type":"$$","backgroundColor":"#ffffff","id":"117","code":"$$\\frac{voltage\\,\\,obtained\\,across\\,\\sec ondary}{voltage\\,applied\\,across\\,primary}\\,=\\,\\frac{no.\\,of\\,turns\\,in\\sec ondary}{no.\\,of\\,turns\\,in\\,primary}$$","aid":null,"backgroundColorModified":false,"ts":1636365401818,"cs":"Kf9HgFxHMEd5Bw60SLe5XQ==","size":{"width":489,"height":38}}

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

{"code":"\\begin{lalign*}\n&{V_{s}\\times\\,i_{s}\\,=\\,V_{p}\\,\\times\\,i_{p}}\\\\\n&{\\frac{i_{p}}{i_{s}}\\,=\\,\\frac{V_{s}}{V_{p}}\\,=\\,\\frac{N_{s}}{N_{p}}\\,=\\,r}\t\n\\end{lalign*}","font":{"size":11,"color":"#000000","family":"Arial"},"id":"120","type":"lalign*","backgroundColorModified":null,"backgroundColor":"#ffffff","aid":null,"ts":1636365769042,"cs":"Np7nHfC7ANflR2+iyKvROg==","size":{"width":165,"height":61}}

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.


Significance of NCERT class 12 physics chapter 7 notes

Comprehensive Revision: NCERT Alternating Current class 12 notes are an excellent resource for thoroughly reviewing the chapter.

Understanding Main Topics: These notes help you understand the primary topics covered in the chapter, such as voltage applied to a resistor, resistor representation of AC current, and transformer principles.

CBSE Syllabus Coverage: The class 12 physics chapter 7 notes make it easier to cover the essential topics outlined in the CBSE Class 12 Physics Syllabus, which are closely aligned with the curriculum.

Preparation for Competitive Exam: These cbse class 12 physics ch 7 notes not only help students with school exams, but also prepare them for various competitive tests such as VITEEE, BITSAT, JEE MAIN, NEET, and so on.

Offline Preparation Convenience: By downloading the physics class 12 chapter 7 notes pdf format, students can study whenever and wherever they want, according to their schedule.

NCERT Class 12 Notes Chapterwise

Subject Wise NCERT Exemplar Solutions

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NCERT Books and Syllabus

Frequently Asked Question (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.

Articles

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

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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