Alternating Current Circuit containing Resistance only (R-Circuit)

Alternating Current Circuit containing Resistance (R-Circuit): Let us consider, An alternating current circuit containing resistance $R$ only. This resistance $R$ is connected with an alternating EMF i.e electromotive force source i.e.

$E=E_{\circ}sin\omega t\qquad(1)$

The potential difference across the circuit

$E=iR$

Then from equation $(1)$

$iR=E_{\circ}sin\omega t $

$i=\frac{E_{\circ}}{R}sin\omega t $

$i=i_{\circ}sin\omega t \qquad(2) $

Where $i_{\circ}$ is the peak value or amplitude of the current in the circuit which has value $i_{\circ}=\frac{E_{\circ}}{R}$.

Now compare the equation $(1)$ and equation $(2)$ which shows that if a circuit is containing a resistor only then the current is always in phase with the applied EMF i.e electromotive force. The phase diagram between EMF and the current of resistance is shown below-

The phasor diagram between the EMF and current of resistance is also shown in the given figure below-

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Derivation of Addition of Velocity in Special Relativity

Addition of Velocities:

Let us consider two frames $S$ and $S'$, frame $S'$ is moving with constant velocity $v$ relative to frame $S$ along the positive direction of the X-axis.

Let us express the velocity of the body in these frames. Suppose that body moves a distance $dx$ in time $dt$ in frame $S$ and through a distance $dx'$ in the time $dt'$ in the system $S'$ from point $P$ to $Q$. Then

Addition of velocity

$\frac{dx}{dt} = u\qquad \frac{dx'}{dt'}= u'\qquad (1)$

From Lorentz's inverse transformation

$x=\alpha\left ( x'+vt' \right )\qquad(2)$

$t=\alpha '\left ( t'+\frac{v\cdot x'}{c^{2}} \right )\qquad(3)$

Differentiate the equation $(2)$ and equation $(3)$ with respect to $t'$

$\frac{dx}{dt'}= \alpha \left ( \frac{dx'}{dt'}+v \right )\qquad(4)$

$\frac{dt}{dt'}= \alpha ' \left ({1}+\frac{v}{c^{2}} \cdot \frac{dx'}{dt'} \right )\quad (5)$

Now divide the equation $(4)$ by equation $(4)$. So

$\frac{\frac{dx}{dt'}}{\frac{dt}{dt'}} = \left ( \frac{\frac{dx'}{dt}+v}{1+\frac{v}{c^{2}}\cdot \frac{dx'}{dt'} } \right ) \qquad $

From Lorentz Transformation $\alpha =\alpha '$. So above equation can be written as:

$\frac{dx}{dt} = \frac{u' +v}{1+ \frac{u'v}{c^{2}}}$

From equation $(1)$. The above equation can be written as:

$u = \frac{u' +v}{1+ \frac{u'v}{c^{2}}} $

This equation represents the relativistic law of addition of velocities with respect to an observer at frame-$S$ whereas in classical mechanics it is simply $u = u'+v$. There is the following point observed from the addition of the velocities equation.

1. When $u'$ and $v$ are the smaller as compare to $c$, then $\frac{v\cdot u'}{c^{2}}$ can be negligible so
   
   
   $u = u'+v$
   
   
   which is classical formula.


2. When $u' = c$Then $u =\frac{c+v}{1+\frac{v}{c}}$ So,
   
   
   $u=c$
   
   
   Therefore a object moves with velocity of light $c$ with respect to other, then their relative velocity is always $c$.


3. When $v=u'=c$Then $u =\frac{c+c}{1+\frac{c^{2}}{c^{2}}}=c$ So,
   
   
   $u=v$
   
   
   This shows that the addition of the velocity of light simply reproduces the velocity of light. It means that the velocity of light in a vacuum is the maximum achievable velocity in nature and no signal and any object can travel faster than the velocity of light in a vacuum.

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Work done by a rotating electric dipole in uniform electric field

Derivation of Work done by a rotating electric dipole in uniform electric field :

Let us consider an electric dipole AB, made up of two charges $+q$ and $-q$, placed at a very small distance $l$ in a uniform electric field $\overrightarrow{E}$. If dipole $AB$ rotates at an angle $θ$ from its equilibrium position. If $A'$ and $B'$ are the new positions of a dipole in the electric field. Then the force on a $+q$ charge particle due to the electric field→

$ \overrightarrow{F_{+q}}=q\overrightarrow{E}\qquad (1)$

Then force on $-q$ charge particle due to electric field→

$ \overrightarrow{F_{-q}}=q\overrightarrow{E}\qquad(2)$

So work done by a force on a $+q$ charge particle to bring it from position $A$ to position $A'$→

$ \overrightarrow{W_{+q}}=\overrightarrow{F_{+q}}·\overrightarrow{AC}$

$ \overrightarrow{W_{+q}}=q\overrightarrow{E}· \overrightarrow{AC}\qquad(3)$

Similarly, work done by the force on $-q$ charge particle to bring from position $B$ to position $B'$→

$ \overrightarrow{W_{-q}}=\overrightarrow{F_{-q}}·\overrightarrow{BD}$

$ \overrightarrow{W_{-q}}=q\overrightarrow{E}· \overrightarrow{BD}\qquad (4)$

So the total work is done by the force on the dipole→

$ \overrightarrow{W}=\overrightarrow{W_{+q}}\:+\:\overrightarrow{W_{-q}}$

$\overrightarrow{W}=q\overrightarrow{E}·(\overrightarrow{AC}+\overrightarrow{BD})\qquad (5)$

From the figure, there is symmetry, so

$ \overrightarrow{AC}=\overrightarrow{BD}$

So from equation $(5)$

$ \overrightarrow{W}=q\overrightarrow{E}·(2\overrightarrow{AC})$

$ \overrightarrow{W}=2q\overrightarrow{E}(\overrightarrow{AO}-\overrightarrow{CO})\qquad (6)$

From figure→

$ \left | \overrightarrow{AO} \right |=\frac{l}{2}$

$ \left | \overrightarrow{CO} \right |=\frac{l}{2}\:cos \theta$

Now substitute the values in equation (6). So equation (6) can be written as in magnitude form →

$ W=2qE(\frac{l}{2}-\frac{l}{2}\:cos\theta )$

$ W=2qE\left(\frac{l}{2}\right)(1-cos\theta )$

$ W=pE(1-cos\theta )$

The above expression shows that work is done on a rotating electric dipole in a uniform electric field.

Case (i):

If $\theta=0^{\circ}$, Then work done will be minimum

$W_{min}=0$

Case (ii):

If $\theta=90^{\circ}$, Then work done

$W=pE$

Case (iii):

If $\theta=180^{\circ}$, Then work done will be maximum

$W_{max}=2pE$

Alternative Method

Let us consider an electric dipole $AB$ rotates from angle $\theta_{1}$ to $\theta_{2}$ by applying torque on an elecric dipole $AB$ in uniform electric field $\overrightarrow{E}$, if change in angle is $d\theta$, then Work done

$dW= \tau d\theta$

$dW= pE sin \theta d\theta \quad \left\{ \because \tau= p E sin\theta \right\}$

Now integrate the above equation from angle $\theta_{1}$ to $\theta_{2}$

$\int dW= \int_{\theta_{1}}^{\theta_{2}} pE sin \theta d\theta$

$\int dW= pE \int_{\theta_{1}}^{\theta_{2}} sin \theta d\theta$

$ W= pE \left[ - cos \theta \right]_{\theta_{1}}^{\theta_{2}}$

$ W= pE \left[ -cos {\theta_{2}} + cos{\theta_{1}}\right]$

$ W= pE \left[ cos {\theta_{1}} - cos{\theta_{2}}\right]$

This is the equation of work done by rotating an electric dipole from angle $\theta_{1}$ to $\theta_{2}$ in a uniform electric field

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Electric field Intensity due a uniformly charged Spherical Shell

Electric field intensity at a different point in the field due to the uniformly charged spherical shell:

Let us consider, a spherical shell of radius $R$ in which $+q$ charge is distributed uniformly on the surface of the sphere. Now find the electric field intensity at different points due to the spherical shell. These different points are:

1.) Electric field intensity at an external point of the spherical shell
2.) Electric field intensity on the surface of the spherical shell
3.) Electric field intensity at an internal point of the spherical shell

1. Electric field intensity at an external point of the spherical shell:

If point $O$ is the center of the spherical shell, The electric field at the outside of the spherical shell can be determined by the following steps →

1.) First, take the point $P$ outside the sphere
2.) Draw a spherical surface of radius r which passes through point $P$.This hypothetical surface is known as the Gaussian surface.
3) Now take a small area $\overrightarrow {dA} $ around point $P$ on the Gaussian surface to find the electric flux passing through it.
4.) Now find the direction between the electric field vector and small area vector.

Due to uniform charge distribution, the electric field intensity will be the same at every point on the Gaussian surface. So from the figure,

Electric field intensity due to the uniformly charged

The direction of electric field intensity on the Gaussian surface is radially outward which is in the direction of the area vector of the Gaussian surface. i.e. ($\theta=0^{\circ}$).Here $\overrightarrow {dA}$ is small area around point $P$ so the small electric flux $d\phi_{E}$ will pass through this small area $\overrightarrow {dA}$. so this flux can be found by applying the Gauss's law in question given below:

$ d\phi_{E}= \overrightarrow {E}\cdot \overrightarrow{dA}$

$ d\phi_{E}= E\:dA\: cos 0^{\circ} \qquad \left \{\because \theta=0^{\circ} \right \}$

$ d\phi_{E}= E\:dA \qquad (1) \quad \left\{\because cos0=1 \right \}$

The electric flux passing through the entire Gaussian surface, So integrate the equation $(1)$ →

$ \phi_{E}=\oint E\:dA\qquad(2)$

According to Gauss's law:

$ \phi_{E}=\frac{q}{\epsilon_{0}}\qquad (3)$

From equation (1) and equation (2), we can write as

$ \frac{q}{\epsilon_{0}}=\oint E\:dA$

$ \frac{q}{\epsilon_{0}}= E\oint dA$

The area of entire Gaussian spherical is $\oint {dA}=4\pi r^{2}$. Now substitute this value in above equation. So above equation can be written as:

$ \frac{q}{\epsilon_{0}}= E(4\pi r^{2})$

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{r^{2}}$

From the above equation, we can conclude that the behavior of the electric field at the external point due to the uniformly charged spherical shell is the same as, like the entire charge is placed at the center, point charge

If the surface charge density is $\sigma$, Then total charge $q$ on the surface of a spherical shell is →

$ q=4\pi R^{2}\: \sigma$

Substitute this value of charge $q$ in above equation, so we can write the equation as:

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{4\pi R^{2}\: \sigma}{r^{2}}$

$ E=\frac{\sigma}{\epsilon_{0}}\frac{R^{2}}{r^{2}}$

This equation describes the electric field intensity at the external point of the spherical shell.

2. Electric field intensity on the surface of the spherical shell:

If point $P$ is placed on the surface of the spherical shell i.e ($r=R$). so electric field intensity on the surface of the spherical shell can be found by putting $r=R$ in the formula of the electric field intensity at the external point of the spherical shell:

$ E=\frac{1}{4\pi \epsilon_{0}}\frac{q}{R^{2}}$

$ E=\frac{\sigma}{\epsilon_{0}}$

3. Electric field intensity at an internal point of the spherical shell:

If point $P$ is placed inside the spherical shell then the electric field intensity will be zero at that point because the charge is distributed uniformly on the surface of the solid sphere so there will not be any charge on the Gaussian surface and electric flux will be zero inside the solid sphere. i.e.

$ \phi_{E}=\oint E\:dA$

$ 0=E\oint dA \qquad \left \{ \because \phi_{E}=0 \right \}$

$ E=0$

Electric field intensity distribution with distance for Spherical Shell:

Electric field intensity distribution with distance shows that the electric field is maximum on the surface of the sphere and zero inside the sphere. Electric field intensity distribution outside the sphere reduces with the distance according to $E=\frac{1}{r^{2}}$.

Electric field intensity distribution with distance

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Energy distribution laws of black body radiation

1.) Wein’s laws of Energy distributions→

A.) Wein's Fifth Power law→ The total amount of the energy emitted by a black body per unit volume at an absolute temperature in the wavelength range $\lambda$ and $\lambda + d\lambda$ is given as

$E\lambda \cdot d\lambda= \frac{A}{\lambda^{5}}f\left ( \lambda T \right ) \cdot d\lambda \qquad (1)$

Where $A$ is a constant and $f(\lambda T)$ is a function of the product $\lambda T$ and is given as

$ f\left ( \lambda T\right )=e^-\frac{hc}{\lambda kT}\qquad (2)$

From equation $(1)$ and $(2)$

$E_\lambda \cdot d\lambda = \frac{A}{\lambda ^{5}}e^\frac{-hc}{\lambda kT} \cdot d\lambda$

$E_\lambda \cdot d \lambda = A \lambda ^{-5} e^\frac{-hc}{\lambda kT} \cdot d \lambda$

Wien’s law energy distribution explains the energy distribution at the short wavelength at higher temperatures and fails for long wavelengths.

B.) Wein's Displacement law→ As the temperature of the body is raised the maximum energy shift toward the shorter wavelength i.e.

$\lambda_{m} \times T = Constant $

Where
$\lambda_m$- Wavelength at which the energy is maximum
$T$-Absolute temperature

Thus, if radiation of a particular wavelength at a certain temperature is adiabatically altered to another wavelength then temperature changes in the inverse ratio.

2.) Rayleigh-Jean’s law→ The total amount of energy emitted by a black body per unit volume at an absolute temperature T in the wavelength range $\lambda $ and $\lambda +d\lambda $ is given as

$E_{\lambda}.d\lambda = \frac{8\pi kt}{\lambda ^{4}}.d\lambda$

Where K– Boltzmann’s Constant which has valve $ 1.381\times 10^{23}\frac{J}{K}$

This law, explains the energy distribution at the longer wavelength at all temperatures and fails totally for the shorter wavelength.

Note→ The energy distribution curves of the black body show a peak while going towards the ultraviolet wavelength (shorter $ \lambda $) and then fall while Rayleigh-Jeans law indicates continuous rise only. This is the failure of classical physics.

3.) Stefan-Boltzmann Law→ The total amount of heat radiated by a perfectly black body per unit area per second is directly proportional to the fourth power of its absolute temperature $(T)$. i.e.

$E \propto T^{4}$

$E = \sigma T^{4}$

Where $\sigma$= Stefan’s Constant which has value $5.67\times 10^{-8} W-\frac{K^{4}}{m^{2}}$

It is a black body at absolute temperature $T$ is surrounded by another black body at absolute temperature $T_{0}$, The net amount of heat $E$ lost by the former per second per $cm^{2}$ is→

$E=\sigma (T^{4}-T_{0}^{^{4}})$

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Relation between group velocity and phase velocity

We know that phase velocity

$V_{p}=\frac{\omega }{k}$

$\omega =V_{p}.k \qquad(1)$

And group velocity

$V_{g}=\frac{d\omega}{dk} \qquad(2)$

Substitute the value of $\omega$ from equation$(1)$ in equation $(2)$

$V_{g}=\frac{d}{dk}(V_{p}.k)$

$V_{g}=V_{p}+k.\frac{dV_{p}}{dk}$

$V_{p}=V_{p}+k.\frac{dV_{p}}{d\lambda}.\frac{d\lambda }{dk} \qquad (3)$

But

$\lambda=\frac{2\pi }{k}$

The above equation can be obtain from following formula i.e. $k=\frac{2\pi}{\lambda }$

Now put the value of $\lambda$ in equation $(3)$

$V_{g}=V_{p}+k\frac{dV_{p}}{d\lambda}\frac{d}{dk}(\frac{2\pi }{k}$

$V_{g}=V_{p}+k\frac{dV_{p}}{d\lambda}(\frac{-2\pi }{k^{2}}$

$V_{g}=V_{p}-\frac{2\pi}{k}\frac{dV_{p}}{d\lambda }$

$V_{g}=V_{p}-\lambda\frac{dV_{p}}{d\lambda }$

Thus, the above equation represents the relation between group velocity and phase velocity.

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