Voltage across AB, is V_Z = 9V
Voltage drop across R_s = 15 - 9 = 6V
Therefore current through the resistor R_s,
I = \(\frac { 6 }{ 1\times { 10 }^{ 3 } } \) = 6 mA
Voltage across the load resistor, = V_AB = 9V
Current through load resistor,
\({ I }_{ L }=\frac { { V }_{ AB } }{ { R }_{ L } } =\frac { 9 }{ 2\times { 10 }^{ 3 } } =4.5mA\)
The current through the Zener diode,
I_Z = I - I_L = 6 mA - 4.5mA = 1.5 mA

(i) In beta decay, a radioactive nucleus emits either electron or positron. If electron (e^-) is emitted, it is called β^- decay and if positron (e⁺) is emitted, it is called β^- decay
(ii) The positron is an anti-particle of an electron whose mass is same as that of electron and charge is opposite to that of electron - that is, +e. Both positron and electron are referred to as beta particles.
β^- decay:
(iii) β^- decay: In β^- decay, the atomic number of the nucleus increases by one but mass number remains the same. This decay is represented by
\(_{ Z }^{ A }{ X }\rightarrow _{ Z+1 }^{ A }{ Y+ }{ e }^{ - }+\bar { v } \) ...(1)
(iv) It implies that the element X becomes Y by giving out an electron and antineutrino (⊽). 
(v) In other words, In each β^- decay, one neutron (n) in the nucleus of X is converted into a proton(p) by emitting an electron (e^-) and antineutrino(⊽). It is given by
\(n\rightarrow p+{ e }^{ - }+\bar { v } \) 
Example :
\(_{ 6 }^{ 14 }{ C }\rightarrow _{ 7 }^{ 14 }{ N+ }{ e }^{ - }+\overline { v } \)
β⁺ decay:
(vi) In β⁺ decay, the atomic number is decreased by one and the mass number remains the same. This decay is represented by
\(_{ Z }^{ A }{ X }\rightarrow _{ Z-1 }^{ A }{ Y+ }{ e }^{ + }+v\)
(vii) It implies that the element X becomes Y by giving out an positron (e⁺) and neutrino (v),
In otherwords, in each β⁺ decay, one proton(p) in the nucleus of X is converted into a neutron by emitting a positron (e⁺) and a neutrino. It is given by
\(p\rightarrow n+{ e }^{ +}+{ v } \)
Example: 
\(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }{ Ne }+{ e }^{ + }+v\)
(viii) However a single proton (not inside any nucleus) cannot have β⁺ decay due to energy conservation, because neutron mass is larger than proton mass.
(ix) But a single neutron (not inside any nucleus) can have  β^- decay.
(x) It is important to note that the electron or positron which comes out from nuclei during beta decay never present inside the nuclei rather they are produced during the conversion of neutron into proton or proton into neutron inside the nucleus.

Magnetic field lines can be entirely confined within the core of a toroid since the toroid has no ends. θ solenoid is open ended and the field lines inside it which are parallel to the length of the solenoid cannot form closed curves inside the solenoid.

(i) The potential energy difference per unit charge is given by
\(\frac { \Delta U }{ q' } =\frac { q'\int _{ R }^{ P }{ (-\overset { \rightarrow }{ E } ) } .d\overset { \rightarrow }{ r } }{ q' } =\int _{ R }^{ P }{ \overset { \rightarrow }{ E } } .d\overset { \rightarrow }{ r } \quad ...(1)\)
(ii) The above equation (1) is independent of q'. The quantity \(\frac { \Delta U }{ q' } =\int _{ R }^{ P }{ \overset { \rightarrow }{ E } } .d\overset { \rightarrow }{ r } \) is called electric potential difference between P and R and is denoted as V_P - V_R = ∆V.
(iii) In other words the electric potential difference is also defined as the work done by an external force to bring unit positive charge from point R to point P.
\({ V }_{ p }-{ V }_{ R }=\Delta V=\int _{ R }^{ P }{ \overset { \rightarrow }{ E } } .d\overset { \rightarrow }{ r } \)
(iv) The electric potential energy difference can be written as ∆U = q' ∆V.

Qualitative treatment:
The electromagnetic oscillations of LC system can be compared with the mechanical oscillations of a spring-mass system.
There are two forms of energy involved in LC oscillations. One is electrical energy of the charged capacitor, the other magnetic energy of the inductor carrying current.
Table: Energy in two oscillatory systems:

Likewise, the mechanical energy of the spring-mass system exists in two forms; the potential energy of the compressed or extended spring and the kinetic energy of the mass. The Table lists these two pairs of energy.
By examining the table, the analogies between the various quantities can be understood and these correspondences are given in the Table.
The angular frequency of oscillations of a spring-mass is given by,
\(\omega= \sqrt \frac{k}{m}\)
From Table, k→ 1/C and m → L. Therefore, the angular frequency of LC oscillations is given by,
\(\omega= \frac{1} {\sqrt {LC}}\)

Let the length MN = y and the point P is on its perpendicular bisector. Let O be the point on the conductor as shown in figure. Therefore,
\(OM=ON=\frac { y }{ 2 } ,then\)
\(cos\varphi _{ 1 }=\frac { \frac { y }{ 2 } }{ \sqrt { \frac { { y }^{ 2 } }{ 4 } +{ d }^{ 2 } } } =\frac { adjacent \ length }{ hypotenuse \ length } \)
\(=\frac { ON }{ PH } =-\frac { \frac { y }{ 2 } }{ \sqrt { \frac { { y }^{ 2 } }{ 4 } +{ a }^{ 2 } } } =-\frac { y }{ \sqrt { { y }^{ 2 }+{ 4a }^{ 2 } } } \)
\(cos\varphi _{ 1 }=\frac { adjacent \ length }{ hypotenuse \ length } =\frac { OM }{ PM } \)
\(=-\frac { \frac { y }{ 2 } }{ \sqrt { \frac { { y }^{ 2 } }{ 4 } +{ a }^{ 2 } } } =-\frac { y }{ \sqrt { { y }^{ 2 }+{ 4a }^{ 2 } } } \)
Hence,
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 4\pi a\sqrt { { y }^{ 2 }+{ 4a }^{ 2 } } } \hat { n } \)
For long straight wire, Y\(\rightarrow \infty ,\)
\(\overset { \rightarrow }{ B } =\frac { { \mu }_{ ° }I }{ 2\pi a } \hat { n } \)
The result obtained is the same as we obtained in equation (3.39).

Energy stored in the capacitor
i) Capacitor not only stores the charge but also it stores energy. When a battery is connected to the capacitor, electrons of total charge - Q are transferred from one plate to the other plate. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor.
ii) To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
dW = V dQ
Where \(V=\frac { Q }{ C } \)  .....(1)
iii) The total work done to charge a capacitor is
\(W=\int _{ 0 }^{ Q }{ \frac { Q }{ C } } dQ=\frac { { Q }^{ 2 } }{ 2C } \quad \quad ....(2)\)
This work done is stored as electrostatic potential energy (U_E) in the capacitor.
\({ U }_{ E }=\frac { { Q }^{ 2 } }{ 2C } =\frac { 1 }{ 2 } { CV }^{ 2 },\quad (\therefore Q=CV)\quad ....(3)\)
(iv) This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor.Substituting  \(C=\frac { { \varepsilon }_{ 0 }A }{ d } \) and V = Ed. 
\(U=\frac { 1 }{ 2 } \left( \frac { { \varepsilon }_{ 0 }A }{ d } \right) { (Ed) }^{ 2 }=\frac { 1 }{ 2 } { \varepsilon }_{ 0 }(Ad){ E }^{ 2 }\quad \quad \quad \quad \quad ...(4)\)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy density \({ u }_{ E }=\frac { U }{ Volume } \)       
Equation (4) ⇒ \({ u }_{ E }=\frac{1}{2}{ \varepsilon }_{ 0 }{ E }^{ 2 }\).....(5)
(v) From equation (5),
(a) We infer that the energy is stored in the electric field existing between the plates of the capacitor. Once the capacitor is allowed to discharge, the energy is retrieved.
(b) The energy density depends only on the electric field and not on the size of the plates of the capacitor. 
(c) This is true for the electric field due to any type of charge configuration.

A = 5 x 10^-2 m²; B = 0.2 T
(i) θ = 0°;
\({ \Phi }_{ B }=BAcos\theta =0.2\times 5\times { 10 }^{ -2 }\times { cos }0^{ o }\)
\({ \Phi }_{ B }=1\times { 10 }^{ -2 }Wb\)
(ii) θ = 90° – 60° = 30°;
\(\Phi_B\) = BAcosθ  = 0.2 x 5 x 10^-2 x cos 30^o
\({ \Phi }_{ B }=1\times { 10 }^{ -2 }\times \frac { \sqrt { 3 } }{ 2 } =8.66\times { 10 }^{ -3 }Wb\)
(iii) θ = 90°;
\({ \Phi }_{ B }\) = BA cos90^o = 0

Given magnetic moment 0.5 J T^-1 and distance r = 0.1 m
(a) When the point lies on the axial line of the bar magnet, the magnetic field for short magnet is given by
\({ { \vec B }_{ axial } } =\frac { { \mu }_{ ° } }{ 4\pi } \left( \frac { 2{ p }_{ m } }{ { r }^{ 3 } } \right) \hat { i } \)
\({ { \vec B }_{ axial } } =1{ 0 }^{ -7 }\times \left( \frac { 2\times 0.5 }{ { \left( 0.1 \right) }^{ 3 } } \right) =1\times { 10 }^{ -4 }\hat { i } \ T\)
Hence, the magnitude of the magnetic field along axial is B_axial = 1 x 10^-4 T and direction is towards South to North.
(b) When the point lies on the normal bisector (equatorial) line of the bar magnet, the magnetic field for short magnet is given by
\({ {\vec B }_{ equatorial } } =-\frac { { \mu }_{ ° } }{ 4\pi } \frac { { p }_{ m } }{ { r }^{ 3 } } \hat { i } \)
\({ {\vec B }_{ equatorial } } =-1{ 0 }^{ -7 }\left( \frac { 0.5 }{ { \left( 0.1 \right) }^{ 3 } } \right) \hat { i } =-0.5\times 1{ 0 }^{ -4 }\hat { i } \ T \)
Hence, the magnitude of the magnetic field along axial is B_equatorial = 0.5 x 10^-4 T and direction is towards North to South.
Note that magnitude of B_axial is twice that of magnitude of B_equatorial and the direction of B_axial and B_equatorial are opposite.

Since the resistances are connected in parallel, therefore, the equivalent resistance in the circuit is
\(\frac { 1 }{ { R }_{ p } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } =\frac { 1 }{ 4 } +\frac { 1 }{ 6 } \)
\(\frac { 1 }{ { R }_{ p } } =\frac { 5 }{ 12 } \Omega \quad or\quad { R }_{ p }=\frac { 12 }{ 5 } \Omega \)
The resistors are connected in parallel, the potential difference (voltage) across them is the same.
\({ I }_{ 1 }=\frac { V }{ { R }_{ 1 } } =\frac { 24V }{ 4\Omega } =6A\)
\({ I }_{ 2 }=\frac { V }{ { R }_{ 2 } } =\frac { 24 }{ 6 } =4A\)
The current I is the sum of the currents in the two branches. Then,
I = I₁ + I₂ = 6 A + 4 A = 10 A

The proton and the electron attract each other. The magnitude of the electrostatic force between these two particles is given by
\(F_e=\frac { ke^{ 2 } }{ { r }^{ 2 } } =\frac { 9\times 10^{ 9 }\times (1.6\times 10^{ -19 })^{ 2 } }{ (5.3\times 10^{ -11 })^{ 2 } } \)
=\(\frac { 9\times 2.56 }{ 28.09 } \) x 10^-7 = 8.2 x 10^-8 N
The gravitational force between the proton and the electron is attractive. The magnitude of the gravitational force between these particles is
F_G = \(\frac { G{ m }_{ e }{ m }_{ p } }{ { r }^{ 2 } } \)
= \(\frac { 6.67\times 10^{ -11 }\times 9.1\times 10^{ -31 }\times 1.6\times 10^{ -27 } }{ (5.3\times 10^{ -11 })^{ 2 } } \)
= \(\frac { 97.11 }{ 28.09 } \) x 10^-47 = 3.4 x 10^-47N
The ratio of the two forces \(\frac { { F }_{ e } }{ F_{ G } } =\frac { 8.2\times 10^{ -8 } }{ 3.4\times 10^{ -47 } } \)
= 2.41 x 10³⁹
Note that F_e ≈ 10³⁹ F_G
The electrostatic force between a proton and an electron is enormously greater than the gravitational force between them. Thus the gravitational force is negligible when compared with the electrostatic force in many situations such as for small size objects and in the atomic domain. This is the reason why a charged comb attracts an uncharged piece of paper with greater force even though the piece of paper is attracted downward by the Earth. This given figure is shown in below.

Electrostatic attraction between a comb and pieces of papers

Since, the prism is equilateral, A = 60^o;
Given, i₁ = 30^o;i₂ = 75^o
Equation for angle of deviation, d = i₁ + i₂ – A
Substituting the values, d = 30°+ 75°– 60°= 45°
The angle of deviation produced d = 45^o

Given, n_go = 1.25 and n_g = 1.5
Refractive index of glass with respect to oil,
\({ n }_{ go }=\cfrac { { n }_{ g } }{ { n }_{ 0 } } \) 
Rewriting for refractive index of oil,
\({ n }_{ p }=\cfrac { { n }_{ g } }{ { n }_{ go } } =\cfrac { 1.5 }{ 1.25 } =1.2\) 
The refractive index of oil is, n_o = 1.2

L_air  = 4.8 x 10^-3H
L_iron = 1.8H
L_air = \(\mu_{o}\)n²Al = 4.8 x 10^-3H
L_iron = \(\mu_{o}\)n²Al = \(\mu_{o}\mu_r\)n²Al = 1.8H
\(\therefore { \mu }_{ r }=\frac { { L }_{ iron } }{ { L }_{ air } } =\frac { 1.8 }{ 4.8\times { 10 }^{ -3 } } =375\)

The maximum torque experienced by the dipole is when it is aligned perpendicular to the applied field.
ፒ_max = pE sin 90⁰ = 3.4 x 10^-30 x 3 x 10⁴ Nm
ፒ_max = 10.2 x 10^-26 Nm.

By using superposition principle, the net electric field at point A is
\(\vec { E_{ A } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1A }^{ 2 } } \hat { { r }_{ 1A } } +\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 2A }^{ 2 } } \hat { { r }_{ 2A } } \)
where r_1A and r_2A are the distances of point A from the two charges respectively
\(\vec { E_{ A } } =\frac { 9\times 10^{ 4 }\times 1\times { 10 }^{ -6 } }{ (2\times { 10 }^{ -3 })^{ 2 } } (\hat { j } )+\frac { 9\times 10^{ 9 }\times 1\times { 10 }^{ -6 } }{ (2\times 10^{ -3 })^{ 2 } } (\hat { i } )\)\(\)
= 2.25 x 10⁹\(\hat { j } \) + 2.25 x 10⁹\(\hat { i } \) = 2.25 x 10⁹ (\((\hat { i } +\hat { j } )\)
The magnitude of electric field
\(|\vec { { E }_{ A } } |=\sqrt { (2.25\times { 10 }^{ 9 })^{ 2 }+(2.25\times 10^{ 9 })^{ 2 } } \)
= 2.25 x \(\sqrt { 2 } \) x 10⁹ NC^-1
The direction of \(\vec { { E }_{ A } } \) is given by \(\frac { \vec { { E }_{ A } } }{ |\vec { { E }_{ A } } | } =\frac { 2.25\times 10^{ 9 }(\hat { i } +\hat { j } ) }{ 2.25\times \sqrt { 2 } \times { 10 }^{ 9 } } =\frac { (\hat { i } +\hat { j } ) }{ \sqrt { 2 } } \), which is the unit vector along OA as shown in the figure.

The acceleration experienced by an electron placed at point A is
\(\vec { a_{ A } } =\frac { \vec { F } }{ m } =\frac { q\vec { { E }_{ A } } }{ m } \)
=\(\frac { (-1.6\times 10^{ -19 })\times (2.25\times 10^{ 9 })(\hat { i } +\hat { j } ) }{ 9.1\times { 10 }^{ -31 } } \)
= -3.95 x 10²⁰ \((\hat { i } +\hat { j } )\)Nkg^-1
The electron is accelerated in a direction exactly opposite to \(\vec { { E }_{ A } } \).

The relation between drift velocity of electrons and current in a wire of cross- sectional area A is
\({ v }_{ d }=\frac { I }{ neA } =\frac { 0.2 }{ 8.4\times { 10 }^{ 28 }\times 1.6\times { 10 }^{ -19 }\times 0.5\times { 10 }^{ -6 } }\)
v_d = 0.03 x 10^-3 m s^-1

(i) The effect of diffraction has an adverse effect in the sharpness of the image tormed.
(ii) There is always a spread of central maximum in the image for every point of the object, for every point of the object acts as a point source.
(iii) The condition for central maximum (or first minimum) produced by rectangular slit is given by the equation,
\(a \sin \theta=\lambda\)   ....(1)
(iv) But, a circular slit (aperture) produces diffraction pattern of concentric circles as shown in Figure.
(v) These are known as Airy's discs. Most of the optical instruments form images of objects only through the circular slits.

(vi) The condition for central maximum (or) first minimum for circular slit is,
\(\text { a } \sin \theta=1.22 \lambda\)  .....(2)
(vii) Here, the numerical value 1.22 appears in the expression for central maximum (or) first minimum formed by circular slits.
For small angles,\(sin\theta = \theta\), the above equation becomes,
\(a\theta = 1.22\lambda\)
Rewriting further,
\(\theta=\frac{1.22 \lambda}{a}\)   ......(3)
Form thegeometry, \(\theta=\frac{r_0}{f}\)
Substituting for in equation (3) and rearranging gives
\(r_0=\frac{1.22 \lambda f}{a}\)  ....(4)
(viii) For example, let two point-sources of light close to cach other form image on a screen. The diffraction pattern of one point-source may overlap with another and produce a blurred image (or) un-resolved image as shown in Figure (a). To obtain a quality image (or) well resolved image, the two point-sources must be kept apart in such a way that their diffraction patterns do not overlap as shown in Figure (c).
(ix) According to Rayleigh's criterion, the two points on an image are said to be just resolved when the central maximum of one diffraction pattern coincides with the first minimum of the other and vice-versa as shown in Figure (b).

(i) Let us consider a thin lens made up of a medium of refractive index n₂ is placed in a medium of refractive index n₁. Let R₁ and R₂ be the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole.

(ii) Consider a point object 'O' on the principal axis. A paraxial ray from 'O' which falls very close to P, after refraction at the surface (1) forms image at 1'.
(iii) Before it does so, it is again refracted by the surface (2). Therefore the final image is formed at I.
(iv) The general equation for the refraction at a single spherical surface is given from Equation,
 \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ v} =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ { R } } \)
For the refracting surface (1), the light goes from n₁ to n₂
\(\cfrac { { n }_{ 2 } }{ v' } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ { R }_{ 1 } } \)   .....(1)
For the refracting surface (2), the light goes from n₂ to n₁
\(\frac{n_{1}}{v}+\frac{n_{2}}{v^{\prime}}=\frac{\left(n_{1}-n_{2}\right)}{R_{2}}\)    ......(2)
For surface (2) I' acts as virtual object.
Adding the above two equations (1) and (2)
\(\cfrac { { n }_{ 1 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
on further simplifying and rearranging,
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 }-{ n }_{ 1 } }{ { n }_{ 1 } } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 } }{ n_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ....(3)
If the object is at infinity, the image is formed at the focus of the lens. Thus, for u = \(\infty\), v = f. Then the equation becomes.
\(\cfrac { 1 }{ f } -\cfrac { 1 }{ \infty } =\left( \cfrac { { n }_{ 2 } }{ { { n }_{ 1 } } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ f } =\left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ......(4)
If the lens is kept in air, then we can take n₂ = n and n₁ = 1. So the equation (4) becomes,
\(\\ \cfrac { 1 }{ f } =\left( n-1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)   ..(5)
The above equation is called the lens maker's formula.
Significance:
It tells the lens manufacturers what curvature is needed to make a lens of desired focal length with a material of particular refractive index to make a lens of desired focal length. This formula holds good also for a concave lens.

When electron jumps from higher energy stationary orbit (m) to lower energy stationary orbit (n), emit radiation.
Wave number (or) Wave length of the emitted radiation is given by,
\(\frac{1}{\lambda}=R\left[\frac{1}{n^{2}}-\frac{1}{m^{2}}\right]=\bar{v}\)
V - Ware number; R- Rydberg constant  = 1.09737 x 10⁷ m^-1
m > n (where m, n are integers) 
(a) Lyman series:
When electron jumps from any outer orbit to first orbit, (n = 1 and m = 2,3,4.......) the wave number or wavelength of spectral lines lies in ultra-violet region is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1^{2}}-\frac{1}{\mathrm{~m}^{2}}\right]\)
(b) Balmer series:
Wher electron jumps from any outer orbit to second orbit, (n = 2 and m = 3, 4, 5.......) the wave number or wavelength of spectral lines lies in visible region is,
\(\bar{v}=\frac{1}{\lambda}=R\left[\frac{1}{2^{2}}-\frac{1}{m^{2}}\right]\)
(c) Paschen series:
When electron jumps from any outer orbit to third orbit, (n = 3 and m = 4, 5, 6.......) the wave number or wavelength of spectral lines lies in infra red (Near IR) region is
\(\bar{v}=\frac{1}{\lambda}=R\left[\frac{1}{3^{2}}-\frac{1}{m^{2}}\right]\)
(d) Bracket series:
When electron jumps from any outer orbit to fourth orbit, (n = 4 and m = 5, 6, 7.......) the wave number or wavelength of spectral lines lies in infra red (Middle IR) region is
\(\bar{v}=\frac{1}{\lambda}=R\left[\frac{1}{4^{2}}-\frac{1}{m^{2}}\right]\)
(e) Pfund series:
When electron jumps from any outer orbit to fifth orbit, (n - 5 and m = 6, 7, 8.......) the wave number or wavelength of spectral lines lies in infra red (far IR) region is
\(\bar{v}=\frac{1}{\lambda}=R\left[\frac{1}{5^{2}}-\frac{1}{m^{2}}\right]\)

(i) Consider a circuit containing a resistor of resistance R, a inductor of inductance L and a capacitor of capacitance C connected across an alternating voltage source (Figure ). The instantaneous value of the alternating voltage is given by the equation
ሀ = V_m sin ωt   ......(1)

(ii) Let i be the resulting circuit current in the circuit at that instant. As a result, the voltage is developed across R, Land C.
(iii) We know that voltage across R (V_R) is in phase with i, voltage across L (V_L) leads i by \(\frac { \pi }{ 2 } \) and voltage across C (V_c) lags behind i  by \(\frac { \pi }{ 2 } \)
(iv) The phasor diagram is drawn with current as the reference phasor. The current is represented by the phasor \(\vec { OI } \), V_R by \(\vec { OA } \); V_L by \(\vec { OB } \); V_c by \(\vec { OC } \) as shown in Figure.
(v) The length of these phasors are OI = I_m, OA = I_mR, OB = I_mX_L; OC = I_mX_C 
The circuit is either effectively inductive or capacitive or resistive that depends on the  value of V_L or V_C. Let us assume that V_L>V_C so that nef voltage drop across L-C combination is V_L - V_C which is represented by a phasor \(\vec { OD } \)
(vi) By parallelogram law, the diagonal \(\vec { OE } \) gives the resultant voltage ሀ of V_R and (V_L - V_C) and its length OE is equal to V_m Therefore,
V²_m = V²_R + (V_L - V_C)² = \(\sqrt { { ({ { I }_{ m }R) } }^{ 2 }+{ ({ I }_{ m }{ X }_{ L }-{ I }_{ m }{ X }_{ C }) }^{ 2 } }=I_m \sqrt {R^2+({X_L-X_C)}^2}\) 
or  \({ I }_{ m }=\frac { { V }_{ m } }{ \sqrt { R^{ 2 }+({ { X }_{ L }-{ X }_{ C }) }^{ 2 } } } \)   ......(2)
\((or) { I }_{ m }=\frac { { V }_{ m } }{ Z } \) where z = \(\sqrt { { R }^{ 2 }+({ { X }_{ L }-{ X }_{ C }) }^{ 2 } } \)   ......(3)
(vii) Z is called impedance of the circuit which refers to the effective opposition to the circuit current by the series RLC circuit. The voltage triangle and impedance triangle are given in the Figure.


(viii) From phasor diagram, the phase angle between v and i is found out from the following relation
\(tan\phi =\frac { V_{ L }-{ V }_{ C } }{ { V }_{ R } } =\frac { X_{ L }-{ V }_{ C } }{ R } \)
Special cases:
(i) If X_L > X_C (X_L - X_C) is positive and phase angle \(\phi \) is also positive. It means that the applied voltage leads the current by \(\phi \) (or current lags behind voltage by \(\phi\)). The circuit is inductive.
∴v = V_m sin ωt; i = I_m sin(ωt - \(\phi \))
(ii) If X_L < X_C (X_L - X_C) is negative and\(\phi \) is also negative. Therefore current leads voltage by \(\phi \) (or voltage lags behind current by\(\phi \)) and the circuit is capacitive.
∴ = V_m sin ωt; i = I_m sin(ωt + \(\phi \))
(ii) If X_L = X_C \(\phi \) is zero. Therefore current and voltage are in the same phase and the circuit is resistive
∴v = V_m sin ωt, i = I_m sinωt 

Distribution of charges in a conductor:
(i) Consider two conducting spheres A and B of radii r₁ and r₂ respectively connected to each other by a thin conducting wire as shown in the Figure. The distance between the spheres is much greater than the radii of either spheres.

(ii) If a charge Q is introduced into any one of the spheres, this charge Q is redistributed into both the spheres such that the electrostatic potential is same in both the spheres. They are now uniformly charged and attain electrostatic equilibrium. Let q₁ be the charge residing on the surface of sphere A and q₂ is the charge residing on the surface of sphere B such that Q = q₁ + q₂, The charges are distributed only on the surface and there is no net charge inside the conductor.
The electrostatic potential at the surface of the sphere A is given by
\({ V }_{ A }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1} } \quad \quad ...(1)\)
(iii) The electrostatic potential at the surface of the sphere B is given by ,
\({ V }_{ B }=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 2 } }{ { r }_{ 2 } } \)    ...(2)
iv) The surface of the conductor is an equipotential. Since the spheres are connected by the conducting wire, the surfaces of both the spheres together form an equipotential surface.
This implies that
V_A= V_B
or \(\frac { { q }_{ 1 } }{ { r }_{ 1 } } =\frac { { q }_{ 2 } }{ { r }_{ 2 } } \)   ...(3)
v) Let us take the charge density on the surface of sphere A and charge density on the surface of sphere B is σ₂.This implies that q₁ = 4πr₁²σ₁ and 4πr₂²σ₂ substituting these values into equation (3), we get
σ₁r₁ = σ₂r₂    ...(4)
from which we conclude that σr = constant ...(5)
Lightning conductor:
It is used to protect tall buildings from lightning strikes.
Principle:
(i) Action at points (or) corona discharge.
(ii) This device consists of a long thick copper rod passing from top of the building to the ground.
(iii) The upper end of the rod has a sharp spike or a sharp needle as shown in Figure.
(iv) The lower end of the rod is connected to copper plate which is buried deep into the ground.
(v) When a negatively charged cloud is passing above the building, it causes a positive charge on the spike. Since the induced charge density on thin sharp spike is large, it results in a corona discharge. This positive Charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud.
(vi) The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the Earth. The lightning arrester does not stop the lightning; rather it diverts the lightning to the ground safely.

(i) The meter bridge is another form of Wheatstone's bridge. It consists of a uniform manganin wire AB of one meter length.
(ii) This wire is stretched along a meter scale on a wooden board between two copper strips C and D. Between these two copper strips another copper strip E is mounted to enclose two gaps G₁ and G₂.
(iii) An unknown resistance P is connected in G₁ and a standard resistance Q is connected in G_2. A jockey (conducting wire) is connected to the terminal E on the central copper strip through a galvanometer (G) and a high resistance (HR).
(iv) The exact position of jockey on the wire can be read on the scale. A Lechlanche cell and a key (K) are connected across the ends of the bridge wire.

(v) The position of the jockey on the wire is adjusted so that the galvanometer shows zero deflection. Let the position of jockey at the wire be at J.
(vi) The resistances corresponding to AJ and JB of the bridge wire now form the resistance R and S of the Wheatstone's bridge. Then for the bridge balance.
\(\cfrac { P }{ Q } =\cfrac { R }{ S } =\cfrac { { r }.AJ }{ { r }.JB } \)   
where r' is the resistance per unit length of wire
\(\cfrac { P }{ Q } =\cfrac { AJ }{ JB } =\cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \)   
\(P=Q\cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \)
(vii) By interchanging P and Q, another set of readings are taken and the average value of P is value of unknown resistance.   

(i) The ohm's law can be derived from the equation \(J=\sigma E\) Consider a segment of wire of length I and cross-sectional area A as shown in Figure.

(ii) When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current in the wire.
(iii) For simplicity, we assumed that the electric field is uniform in the entire length of the wire, the potential difference (voltage V) can be written as V = EI
(iv) As we know, the magnitude of current density
\(J=\sigma E=\sigma \cfrac { V }{ l } \)   
(v) But \(J=\cfrac { I }{ A } \), so we write the equation as,
\(\cfrac { I }{ A } =\sigma \cfrac { V }{ l } \)       
(vi) By rearranging the above equation we get,
\(V=I\left( \cfrac { I }{ \sigma A } \right) \)  
(vii) The quantity \(\cfrac { l }{ \sigma A } \)is called resistance of the conductor and it is denoted as R. Note that the resistance is directly proportional to the length of the conductor and inversely proportional to area of cross-section.
(viii) Therefore, the macroscopic form of ohm's law can be stated as V = IR.