Circuit containing Capacitor and Resistor in Series (C-R Series Circuit )

Mathematical Analysis of C-R Series Circuit :

Let us consider, a circuit containing capacitor $C$ resistor $R$ and these are connected in series. If an alternating voltage source is applied across it then the resultant voltage of the C-R circuit

$V=\sqrt{ V_{C} ^{2} + V^{2}_{R}} \qquad(1)$

We know that:

$V_{R} = iR$
$V_{C} = iX_{C}$

So from equation $(1)$

$V=\sqrt{\left( iX_{C} \right)^{2} + \left(iR\right)^{2}} $

$V=i\sqrt{\left( X_{C} \right)^{2} + R^{2}} $

$\frac{V}{i}=\sqrt{\left( X_{C} \right)^{2} + R^{2}} $

$Z=\sqrt{\left( X_{C} \right)^{2} + R^{2}} \qquad(2)$

Where
$Z \rightarrow$ Impedance of C-R circuit.
$X_{C} \rightarrow$ Capacitive Reactance which has value $\frac{1}{\omega C}$

So from equation $(2)$, we get

$Z=\sqrt{\left( \frac{1}{\omega C} \right)^{2} + R^{2}} \qquad(3)$

The phase of resultant voltage:

If the phase of resultant voltage from from current is $\phi$ then

$tan \phi = \frac{X_{C} }{R} \qquad(4)$

$tan \phi = \frac{\frac{1}{\omega C}}{R} $

$tan \phi = \frac{1}{\omega C R} $

$\phi = tan^{1} \left(\frac{1}{\omega C R}\right) $

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Circuit containing Inductor and Resistor in Series (L-R Series Circuit )

Mathematical Analysis of L-R Series Circuit :

Let us consider, a circuit containing inductor $L$ resistor $R$ and these are connected in series. If an alternating voltage source is applied across it then the resultant voltage of the L-R circuit

$V=\sqrt{ V_{L} ^{2} + V^{2}_{R}} \qquad(1)$

We know that:

$V_{R} = iR$
$V_{L} = iX_{L}$

So from equation $(1)$

$V=\sqrt{\left( iX_{L} \right)^{2} + \left(iR\right)^{2}} $

$V=i\sqrt{\left( X_{L} \right)^{2} + R^{2}} $

$\frac{V}{i}=\sqrt{\left( X_{L} \right)^{2} + R^{2}} $

$Z=\sqrt{\left( X_{L} \right)^{2} + R^{2}} \qquad(2)$

Where
$Z \rightarrow$ Impedance of L-R circuit.
$X_{L} \rightarrow$ Inductive Reactance which has value $\omega L$

So from equation $(2)$, we get

$Z=\sqrt{\left( \omega L \right)^{2} + R^{2}} \qquad(3)$

The phase of resultant voltage:

If the phase of resultant voltage from from current is $\phi$ then

$tan \phi = \frac{X_{L} }{R} \qquad(4)$

$tan \phi = \frac{\omega L }{R} $

$\phi = tan^{1} \left(\frac{\omega L }{R}\right) $

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Circuit containing Inductor, Capacitor, and Resistor in Series (L-C-R Series Circuit )

Mathematical Analysis of L-C-R Series Circuit :

Let us consider, a circuit containing inductor $L$, capacitor $C$, and resistor $R$ and these are connected in series. If an alternating voltage source is applied across it then the resultant voltage of the L-C-R circuit

$V=\sqrt{\left( V_{L} -V_{C} \right)^{2} + V^{2}_{R}} \qquad(1)$

We know that:

$V_{R} = iR$
$V_{L} = iX_{L}$
$V_{C} = iX_{C}$

So from equation $(1)$

$V=\sqrt{\left( iX_{L} - iX_{C} \right)^{2} + (iR)^{2}} $

$V=i\sqrt{\left( X_{L} - X_{C} \right)^{2} + R^{2}} $

$\frac{V}{i}=\sqrt{\left( X_{L} - X_{C} \right)^{2} + R^{2}} $

$Z=\sqrt{\left( X_{L} - X_{C} \right)^{2} + R^{2}} \qquad(2)$

Where
$Z \rightarrow$ Impedance of L-C-R circuit.
$X_{L} \rightarrow$ Inductive Reactance which has value $\omega L$
$X_{C} \rightarrow$ Inductive Reactance which has value $\frac{1}{\omega C}$

So from equation $(2)$, we get

$Z=\sqrt{\left( \omega L - \frac{1}{\omega C} \right)^{2} + R^{2}} \qquad(3)$

The phase of resultant voltage:

If the phase of resultant voltage from from current is $\phi$ then

$tan \phi = \frac{X_{L} - X_{C}}{R} \qquad(4)$

$tan \phi = \frac{\omega L - \frac{1}{\omega C}}{R} $

$ \phi = tan^{-1} \left( \frac{\omega L - \frac{1}{\omega C}}{R} \right) $

The Impedance and Phase at Resonance Condition:($X_{L} = X_{C}$):

At resonance $X_{L} = X_{C} \qquad(5)$

$\omega L = \frac{1}{\omega C}$

$\omega^{2} = \frac{1}{L C}$

$\omega = \sqrt{\frac{1}{L C}}$

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

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

Where $f \rightarrow$ Natural frequency of the circuit

1.) The Impedance of the circuit at resonance condition:

Substitute the resonance condition i.e. $X_{L} = X_{C}$ in equation $(2)$ then the impedance of the L-C-R Circuit

$Z=R$

The impedance of the L-C-R circuit at resonance condition is equal to the resistance of the resistor applied in a circuit.

2.) The Phase of resultant voltage at resonance condition:

$tan\phi =0$

$tan\phi = tan 0^{\circ}$

$\phi=0^{\circ}$

The phase of resultant voltage at resonance condition is zero. i.e. the direction of resultant voltage in the direction of current in the circuit.

Note: There are following cases arise in the L-C-R circuit at resonance condition

Case -1:

If $X_{L} \gt X_{C}$, the $tan \phi$ is positive, i.e. $\phi$ is positive. In this case, the voltage leads to the current. Therefore, the circuit is more inductive rather than capacitive or resistive.

Case -2:

If $X_{L} \lt X_{C}$, the $tan \phi$ is negative, i.e. $\phi$ is negative. In this case, the voltage lags behind the current. Therefore, the circuit has a more capacitance-dominated circuit.

Case -3:

If $X_{L} = X_{C}$, the $tan \phi$ is zero, i.e. $\phi$ is negative. In this case, the voltage and the current are in phase. Therefore, the circuit is purely resistive.

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Merits and Demerits of Alternating Current in Comparison to Direct Current

Merits and Demerits of AC in Comparison to DC

(a) Merits

(i) Alternating current can be increased or decreased by using a transformer.

This is the reason that Alternating current can be transmitted from one place to other place at relatively lower expenditure and minimum loss of energy. In Direct current, it is not possible.

(ii) Alternating current can be controlled by choke coil or capacitor at very small loss of energy. To control Direct current resistance is required in which energy loss is very high.

(ii) Alternating current can easily be converted into Direct current by using a rectifier but converting Direct current into Alternating current is not easy.

(iv) Alternating current is cheaper than DC. (life of a cell or battery is very limited).

(b) Demerits

(i) Alternating current is more dangerous as compared to Direct current.

(ii) Alternating current cannot be used in electrolysis.

(iii) Most of the Alternating current of high frequency flows on the surface of the wire, therefore, a thick wire by joining number of thin insulated wires in parallel are to be used.

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Power in Alternating Current Circuit

Definition of Power in Alternating Current Circuit:

The rate of power consumption in an alternating current circuit is known as power in the circuit.

Mathematical Analysis:

Let us consider an alternating current circuit in which the voltage and current at any instant are given by

$V=V_{\circ} sin \omega t \qquad(1)$

$i=i_{\circ} sin \left( \omega t - \phi \right) \qquad(2)$

Where $\phi$ $\rightarrow$ Phase difference between voltage and current

So instantaneous Power

$P=Vi$

$P=\left\{V_{\circ} sin \omega t \right\} \left\{ i_{\circ} sin \left( \omega t - \phi \right) \right\}$

$P=V_{\circ} i_{\circ} \: sin \omega t \: sin \left( \omega t - \phi \right)$

$P=\frac{V_{\circ} i_{\circ}}{2} \left[2\: sin \: \omega t \: sin \left( \omega t - \phi \right)\right]$

$P=\frac{V_{\circ} i_{\circ}}{2} \left[ cos \left( \omega t -\omega t + \phi \right) \\ \qquad - cos \left( \omega t + \omega t - \phi \right) \right]$

$P=\frac{V_{\circ} i_{\circ}}{2} \left[ cos \left(\phi \right) - cos \left( 2\omega t - \phi \right) \right]$

The average power for one cycle is:

$P_{avg}$ = The average value of $\left\{ \frac{V_{\circ} i_{\circ}}{2} \left[ cos \left(\phi \right) - cos \left( 2\omega t - \phi \right) \right] \right\}$

The term $cos \left( 2\omega t - \phi \right)$ is time-dependent and the average value of term $cos \left( 2\omega t - \phi \right)$ for complete one cycle is zero. So the average power dissipation in one cycle is

$P_{avg}= \frac{V_{\circ} i_{\circ}}{2} cos \phi $

$P_{avg}= \frac{V_{\circ}}{\sqrt{2}} \frac{i_{\circ}}{\sqrt{2}} cos \phi $

$P_{avg}= V_{rms} \: i_{rms} cos \phi $

Case -1 The circuit containing Pure Resistor only:

If the circuit contains a pure resistance only then voltage $V$ and current $i$ are always in the same phase. i.e. $\phi=0^{\circ}$ and $cos \phi = +1$, then average power

$P_{avg}= V_{rms} \: i_{rms} $

Case -2 The circuit containing Pure Inductor only:

If the circuit containing pure inductor only then voltage $V$ leads over current $i$ by $90^{\circ}$ i.e. $\phi=90^{\circ}$ and $cos 90^{\circ} = 0$, then average power

$P_{avg}= 0 $

Case -3 The circuit containing Pure Capacitor only:

If the circuit containing pure capacitor only then voltage $V$ lags behind current $i$ by $90^{\circ}$ i.e. $\phi=90^{\circ}$ and $cos 90^{\circ} = 0$, then average power

$P_{avg}= 0 $

Case -4 The circuit containing inductor and resistor only (L-R Circuit):

We know that the phase in the L-R circuit is

$tan \phi = \frac{X_{L}}{R}$

From the above equation, Draw the triangle as shown in the figure below:

From the above figure, the value of $cos\phi$ is

$cos \phi = \frac{R}{\sqrt{R^{2} + X^{2}_{L}}}$

So average power

$P_{avg}= V_{rms} \: i_{rms} \frac{R}{\sqrt{R^{2} + X^{2}_{L}}} $

$\left(P_{avg}\right)_{L-R} \lt \left(P_{avg}\right)_{R}$

Thus, the average power consumption in an L-R circuit is less than the power consumed in a purely resistive circuit

Case -5 The circuit containing capacitor and resistor only (C-R Circuit):

We know that the phase in the C-R circuit is

$tan \phi = \frac{X_{C}}{R}$

From the above equation, Draw the triangle as shown in the figure below:

From the above figure, the value of $cos\phi$ is

$cos \phi = \frac{R}{\sqrt{R^{2} + X^{2}_{C}}}$

So average power

$P_{avg}= V_{rms} \: i_{rms} \frac{R}{\sqrt{R^{2} + X^{2}_{C}}} $

$\left(P_{avg}\right)_{C-R} \lt \left(P_{avg}\right)_{R}$

Thus, the average power consumption in a C-R circuit is less than the power consumed in a purely resistive circuit

Case -6 The circuit containing capacitor and inductor only (L-C Circuit):

We know that the phase in the L-R circuit is

$\phi = 90^{\circ}$ and $cos 90^{\circ} = 0$

So average power

$P_{avg}= 0 $

Case -7 The circuit containing inductor, capacitor, and resistor only (L-C-R Circuit):

We know that the phase in the L-C-R circuit is

$tan \phi = \frac{X_{L} - X_{C}}{R}$

From the above equation, Draw the triangle as shown in the figure below:

From the above figure, the value of $cos\phi$ is

$cos \phi = \frac{R^{2}}{\sqrt{R^{2} + \left( X^{2}_{L} - X^{2}_{C} \right) }}$

So average power

$P_{avg}= V_{rms} \: i_{rms} \frac{R^{2}}{\sqrt{R^{2} + \left( X^{2}_{L} - X^{2}_{C} \right) }} $

$\left(P_{avg}\right)_{L-C-R} \lt \left(P_{avg}\right)_{R}$

Thus, the average power consumption in an L-C-R circuit is less than the power consumed in a purely resistive circuit

Wattless Current:

If the circuit contains only resistor or only inductor or only inductor and capacitor then power consumption in the circuit will be zero (i.e. there will be no energy dissipation in the circuit) due to the phase difference $90^{\circ}$ between voltage(or impedance) and current. So the current in the circuit is called the "Wattless current".

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Self Induction Phenomenon and its Coefficient

Self Induction:

When a changing current flows in a coil then due to the change in magnetic flux in the coil produces an electro-motive force $\left(emf \right)$ in that coil. This phenomenon is called the principle of Self Induction.

The direction of electro-motive force can be found by applying "Lenz's Law".

Mathematical Analysis of Coefficient of Self Induction:

Let us consider that a coil having the number of turns is $N$. If the change in current is $i$, then linkage flux in a coil will be

$N \phi \propto i$

$N\phi = L i \qquad(1)$

Where $L$ $\rightarrow$ Coefficient of Self Induction.

According to Faraday's law of electromagnetic induction. The electro-motive force $\left(emf \right)$ in a coil is

$e=-N\left( \frac{d \phi}{dt} \right)$

$e=-\frac{d \left(N \phi \right)}{dt} \qquad(2)$

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

$e=-\frac{d \left(L i\right)}{dt} $

$e=-L \left(\frac{d i}{dt} \right) $

$L = \frac{e}{\left(\frac{d i}{dt} \right)}$

If $\left(\frac{d i}{dt} \right) = 1$

Then

$L = e$

The above equation shows that If the rate of flow of current in a coil is unit then the coefficient of self-induction in that coil will be equal to the induced electro-motive force $\left( emf \right)$.

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Mutual Induction Phenomenon and its Coefficient

Mutual Induction:

When two coils are placed near each other then the change in current in one coil ( Primary Coil) produces electro-motive force $\left( emf \right)$ in the adjacent coil ( i.e. secondary coil). This phenomenon is called the principle of Mutual Induction.

The direction of electro-motive force $\left( emf \right)$ depends or can be found by "Lenz's Law"

Mathematical Analysis of Coefficient of Mutual Induction:

Let us consider that two coils having the number of turns are $N_{1}$ and $N_{2}$. If these coils are placed near to each other and the change in current of the primary coil is $i_{1}$, then linkage flux in the secondary coil will be

$N_{2}\phi_{2} \propto i_{1}$

$N_{2}\phi_{2} = M i_{1} \qquad(1)$

Where $M$ $\rightarrow$ Coefficient of Mutual Induction.

According to Faraday's law of electromagnetic induction. The electro-motive force $\left( emf \right)$ in the secondary coil is

$e_{2}=-N_{2}\left( \frac{d \phi_{2}}{dt} \right)$

$e_{2}=-\frac{d \left(N_{2} \phi_{2} \right)}{dt} \qquad(2)$

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

$e_{2}=-\frac{d \left(M i_{1} \right)}{dt} $

$e_{2}=-M \left(\frac{d i_{1}}{dt} \right) $

$M = \frac{e_{2}}{\left(\frac{d i_{1}}{dt} \right)}$

If $\left(\frac{d i_{1}}{dt} \right) = 1$

Then

$M = e_{2}$

The above equation shows that If the rate of flow of current in the primary coil is unit then the coefficient of mutual induction will be equal to the induced electro-motive force $\left( emf \right)$ in the secondary coil.

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Faraday's laws of electromagnetic induction

Faraday's Laws of Electromagnetic Induction:

The Faraday's experiment shows the two laws which are known as Farday's laws of electromagnetic induction

First Law (Neumann's Law): The rate of change of magnetic flux through a circuit is equal to the emf produced in the circuit. This is also known as "Neumann Law"

$e=-\frac{\Delta \phi}{ \Delta t}$

Here negative sign shows the direction of emf.

If $\Delta t \rightarrow 0$

$e=-\frac{d \phi}{ d t}$

This equation represents an independent experimental law that cannot be derived from other experimental laws.

If the circuit is a tightly wound coil of $N$ turns, then the induced emf

$e=-N\frac{d \phi}{ d t}$

$e=-\frac{d \left(N \phi\right)}{ dt}$

Here $N \phi$ is called the 'Linkage magnetic flux'.

Note: The change in flux induces emf, not the current.

Second Law (Lenz's Law): The direction of induced EMF produced in a closed circuit is such that it opposes the original cause that produces it. It is also called "Lenz's law". The direction of induced EMF is described by Fleming's right-hand rule.

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Paramagnetic Substances and Its properties

Paramagnetic Substances :

Those substances, which are placed in the external magnetic field and they are weakly magnetized in the direction of the external magnetic field, are called paramagnetic substances. The susceptibility $\chi_{m} $ of paramagnetic substances is small and positive. Further, When a paramagnetic substance is placed in the magnetic field, then the flux density of the paramagnetic substance is slightly more than the free space. Thus, the relative permeability of paramagnetic substance $\mu_{r}$, is slightly more than 1.

Properties of Paramagnetic substances:

1. When a rod of a paramagnetic material is suspended freely between external magnetic poles (i.e. Between North and South Poles) then its axis becomes along the direction of the external magnetic field $B$ (Figure). The poles produced on the two sides of the rod are opposite to the poles of the external magnetic field.

2. In a non-uniform magnetic field, a paramagnetic substance tends to move from the weaker magnetic field to the stronger magnetic field. If a paramagnetic liquid is taken in a watch glass placed on two magnetic poles very near to each other, then the liquid rises in the middle as shown in the figure below(Figure) where the field is strongest. Now, if the distance between the poles is increased, the liquid is depressed in the middle, because now the field is strongest near the poles.

3. If the solution of a paramagnetic substance is poured into a U-tube and apply the strong magnetic field into one arm of this U-tube then the level of the solution in that arm rises. As shown in the figure below:

4. When paramagnetic gas molecules are passed between the poles of a magnet then paramagnetic gas molecules are attracted toward the magnetic field.

5. The susceptibility of a paramagnetic substance is inversely dependent on temperature.

$\chi \propto \frac{1}{T_{C}-T}$

Explanation of Paramagnetism on the Basis of Atomic Model:

The property of Paramagnetism is generally found in those substances whose atoms (or ions or molecules) have an 'odd' number of electrons. In these odd numbers of electrons one electron is not able to form a pair because the net magnetic dipole moment of the atoms (or ions or molecules) are not zero. but in the absence of an external magnetic field, these magnetic dipole moments are randomly arranged inside the substance because the net magnetic dipole moment of the material is zero.

When a paramagnetic substance is placed in an external magnetic field $B$ then the magnetic dipole moment of the atoms (or ions or molecules) are weakly aligned in the direction of the external magnetic field. Thus, a small magnetic dipole moment is induced in the substance which is directly proportional to the magnetic field $B$. Hence, the paramagnetic substance is magnetized in the direction of the external magnetic field $B$, and the field lines become less dense inside the paramagnetic substance compared to those outside.

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Work energy theorem Statement and Derivation

Work-energy theorem statement:

The work between the two positions is always equal to the change in kinetic energy between these positions. This is known as the work energy Theorem.

$W=K_{f}-K_{i}$

$W=\Delta K$

Derivation of the Work-energy theorem:

According to the equation of motion:

$v^{2}_{B}=v^{2}_{A}+2as $

$2as=v^{2}_{B}-v^{2}_{A}$

$2mas=m(v^{2}_{B}-v^{2}_{A})$

$mas=\frac{m}{2} (v^{2}_{B}-v^{2}_{A})$

$Fs=\frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv^{2}_{A} \qquad (\because F=ma)$

$W=\frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv^{2}_{A} \qquad (\because W=Fs)$

$W=K_{f}-K_{i}$

Where
$K_{f}$= Final Kinetic Energy at position $B$
$K_{i}$= Initial Kinetic Energy at position $A$

$W=\Delta K$

Alternative Method (Integration Method):

We know that the work done by force on a particle from position $A$ to position $B$ is-

$W=\int F ds$

$W=\int (ma)ds \qquad (\because F=ma)$

$W=m \int \frac{dv}{dt}ds$

$W=m \int dv \frac{ds}{dt}$

$W=m \int v dv \qquad (\because v=\frac{ds}{dt})$

If position $A$ is the initial point where velocity is $v_{A}$ and position $B$ is the final point where velocity is $v_{B}$ then work is done by force under the limit-

$W=m \int_{v_{A}}^{v_{B}} v dv$

$W=m[\frac{v^{2}}{2}]_{v_{A}}^{v_{B}} $

$W=\frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv^{2}_{A}$

$W=K_{f}-K_{i}$

$W=\Delta {K}$

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