- Introduction
- |
- Alternating current and Alternating EMF
- |
- Average or mean current
- |
- Root Mean square value of AC
- |
- Phasor diagram
- |
- A.C through pure resistor
- |
- A.C through pure inductor
- |
- AC through pure capacitor
- |
- Circuit containing inductance and resistance in series
- |
- Circuit containing capacitance and resistance in series
- |
- LCR series circuit

- Figure below shows a circuit containing a capacitor ,resistor and inductor connected in series through an alternating voltage source

- Same amount of current will flow in all the three circuit components and vector sum of potential drop across each component would be equal to the applied voltage

- If i be the amount of current in the circuit at any time and V
_{L},V_{C}and V_{R}the potential drop across L,C and R respectively then

V_{R}=iR ⇒ Voltage is in phase with i

V_{L}=iωL ⇒ Voltage is leading i by 90^{0}

V_{C}=i/ωC ⇒ Voltage is lagging behind i by 90^{0}

- Since V
_{L}is ahead of i by 90 and V_{C}is behind by 90 so that phase difference between V_{L}and V_{C}is 180 and they are in direct opposition to each other as shown in the figure 12(b)

- In figure 12(b) we have assumed that V
_{L}is greater than V_{C}which makes i lags behind V.If V_{C}> V_{L}then i lead V

- In this phasors diagram OA represent V
_{R},AD represent V_{C}and AC represent V_{L}.So in this case as we have assumed that V_{L}> V_{C},there resultant will be (V_{L}-V_{C}) represented by vector AD

- Vector OB represent resultant of vectors V
_{R}and (V_{L}-V_{C}) and this vector OB is the resultant of all the three ,which is equal to applied voltage V,thus

is called impedance of the circuit

- From phasors diagram 12(b),current i lag behind resultant voltage V by an phase angle given by,

- From equation (20) three cases arises

**(i)**When ωL > 1/ωC then tanφ is positive i.e. φ is positive and voltage leads the current i

**(ii)**When ωL < 1/ωC,then tanφ is negative i.e. φ is negative and voltage lags behind the current i

**(iii)**When ωL = 1/ωC ,then tanφ is zero i.e. φ is zero and voltage and current are in phase

- Again considering case (iii) where ωL = 1/ωC,we have

- This is the case where X
_{L}=X_{C},the circuit is said to be in electric resonance where the impedance is purely resistive and minimum and currents has its maximum value

- Hence at resonance

ωL = 1/ωC

or ω=1/√LC ---(21)

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

f_{0}=1/2π√LC ---(22)

This frequency is called resonant frequency of the circuit and peak current in this case is

i_{0}=V_{0}/R

and reactance is zero

- We will now define resonance curves which shows the variation in circuit current (peak current i
_{0}) with change in frequency of the applied voltage

- Figure below shows the shape of resonance curve for various values of resistance R

- for small value of R,the resonance is sharp which means that if applied frequency is lesser to resonant frequency f
_{0},the current is high otherwise

- For large values of R,the curve is broad sided which means that those is limited change in current for resonance and non -resonance conditions

Class 12 Maths Class 12 Physics

- NCERT Solutions: Physics 12th
- NCERT Exemplar Problems: Solutions Physics Class 12
- Concepts of Physics - Vol. 2 HC Verma
- Dinesh New Millennium Physics Class -XII (Set of 2 Vols) (Free Complete Solutions to NCERT Textbook Problems & NCERT Exemplar Problems in Physics-XII)
- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- CBSE Chapterwise Questions-Solutions Physics, Class 12
- CBSE 15 Sample Question Paper: Physics for Class 12th

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