Figure 1.

For pure metals, the resistance increases linearly with increase in temperature as shown in Fig. 1.8.2, and for a certain range of temperature (typically 0 to 100ºC) the rate of increase of resistance (slope of the line) remains constant. The resistance of metals reduces with reduction in temperature and it reduces to 0 Ω at a temperature of -234.5ºC as shown in Fig. 1.

Resistance Temperature Coefficient (R.T.C.):

Figure 2.

Consider a conductor having an increasing resistance linearly with temperature as shown in Fig. 2.

Let,

R₀ – Resistance at 0ºC

R₁ – Resistance at t₁ºC

R₂ – Resistance at t₂ºC

Definition :

The resistance temperature coefficient (RTC) at tºC is defined as the ratio of change in resistance of the material per degree Celsius to its resistance at tºC.

It is denoted by α_t and its units are per degree Celsius (/ºC).

\[\text{RTC at t}{}^\circ \text{C = }{{\alpha }_{t}}=\frac{\text{ }\!\!\Delta\!\!\text{ R per  }\!\!{}^\circ\!\!\text{ C}}{{{\text{R}}_{\text{t}}}}\]

Where,

ΔR – Change in resistance

R_t – Resistance at tºC

R.T.C. at 0ºC :

The R.T.C. at 0ºC is denoted by α₀ and it is defined as follows :

\[{{\alpha }_{0}}=\frac{\text{Change in resistence per  }\!\!{}^\circ\!\!\text{ C}}{\text{Resistence at 0 }\!\!{}^\circ\!\!\text{ C}}\]

The change in resistance per ºC is equal to the slope of the characteristics shown in Fig. 1.9.1.

α₀ = Slope of the characteristic/R₀

\[{{\alpha }_{0}}=\frac{({{R}_{2}}-{{R}_{1}})/({{t}_{2}}-{{t}_{1}})}{{{R}_{0}}}\]

R.T.C. at t₁ºC :

The R.T.C. at t₁ºC is denoted by α₁ and it is defined as follows :

\[{{\alpha }_{1}}=\frac{\text{Change in resistance per }{}^\circ \text{C}}{\text{Resistance at }{{\text{t}}_{\text{1}}}{}^\circ \text{C}}\]

\[=\frac{\text{Slope}}{{{\text{R}}_{\text{1}}}}\]

\[{{\alpha }_{1}}=\frac{({{R}_{2}}-{{R}_{1}})/({{t}_{2}}-{{t}_{1}})}{{{R}_{1}}}\]

So in general we can write that the R.T.C. at temperature t_n is given by,

\[{{\alpha }_{n}}=\frac{\text{Slope of the characteristics}}{{{\text{R}}_{\text{n}}}}\]

Where,

R_n = Resistance of the conductor at temperature T_n.

Unit of R.T.C.

We know that,

\[{{\alpha }_{n}}=\frac{\text{Slope of the charecteristics}}{\text{Resistance at }{{\text{t}}_{\text{1}}}{}^\circ \text{C}}\]

\[=\frac{({{R}_{2}}-{{R}_{1}})/({{t}_{2}}-{{t}_{1}})}{{{R}_{n}}}\]

\[=\frac{\Omega /{}^\circ \text{C}}{\Omega }=/{}^\circ \text{C}\]

Thus the R.T.C. is measured in per degree Celsius (/ºC).

Expression for Resistance at tºC

Figure 3.

Consider the R versus temperature characteristics of metal. (Fig. 3).

Let R_t be the value of resistance at tºC.

\[\text{RTC at t}{}^\circ \text{C = }{{\alpha }_{t}}\]

\[=\frac{\text{Slope of the charecteristics}}{{{\text{R}}_{\text{t}}}}\]

\[\text{But slope = }\frac{{{R}_{t}}-{{R}_{0}}}{t-0}=\frac{{{R}_{t}}-{{R}_{0}}}{t}\]

And,

\[{{\alpha }_{0}}=\frac{\left( {{R}_{t}}-{{R}_{0}} \right)/t}{{{R}_{0}}}\]

\[{{\text{ }\!\!\alpha\!\!\text{ }}_{\text{0}}}{{\text{R}}_{\text{0}}}\text{t = }{{\text{R}}_{\text{t}}}-{{\text{R}}_{\text{0}}}\]

\[{{\text{R}}_{\text{t}}}\text{= }{{\text{R}}_{\text{0}}}\text{( 1 + }{{\text{ }\!\!\alpha\!\!\text{ }}_{\text{0}}}\text{t )}\]

We can generalize this expression by replacing t by t₂ and 0 by t₁ to get,

\[{{\text{R}}_{\text{t2}}}\text{= }{{\text{R}}_{\text{t1}}}\text{( 1 + }{{\text{ }\!\!\alpha\!\!\text{ }}_{\text{t1}}}\Delta \text{t )}\]

Where,

Δt = Change in temperature = (t₂ – t₁)

α_t1 = R.T.C. at t₁ºC and R_t2 = Resistance at t₂ºC.

Note :

Hence we conclude that the R.T.C. changes with the change in temperature. Higher the value of R.T.C. and R.T.C is maximum at 0ºC.

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Series Connection of Batteries

Connection diagram :

The series connection of batteries is shown in Fig. 1(a). N number of identical batteries with terminal voltage of V volts and current capacity of I ampere each are connected in series. The load is connected directly across the series combination of N batteries as shown in Fig. 1(a). The load voltage is given by,

\[{{V}_{L}}=(V+V+……+V)\text{    }…..\text{N terms}\]

\[{{V}_{L}}=NV\text{ Volts}\]

However the series connection does not improve the current sourcing capacity. The current sourcing capacity of the series string is same as that of a single battery connected in the string, i.e. I amperes.

Figure 2. Series connection of batteries with different terminal.

It is not always necessary to connect all the batteries of same terminal voltages in series with each other. The batteries of different terminal voltages can be connected in series as shown in Fig. 2.

\[{{V}_{L}}={{V}_{1}}+{{V}_{2}}+{{V}_{3}}+{{V}_{4}}\]

Parallel Connection of Batteries

Connection diagram :

Figure 3.

The parallel connection of batteries is shown in Fig. 3. Batteries are connected in parallel in order to increase the current supplying capacity. If the load current is higher than the current rating of individual batteries, then the parallel connection of batteries is used. The terminal voltage of all the batteries connected in parallel must be the same. The load current is equal to the sum of currents drawn from the individual batteries.

\[{{I}_{L}}={{I}_{1}}+{{I}_{2}}+{{I}_{3}}+{{I}_{4}}\]

If all the batteries are of same current rating then they supply equal amount of current. But, if they are of different current ratings, then they share current in proportion with their current ratings.

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Figure 1: Setup to understand the concept of self inductance.

Definition of Self inductance

As per the Lenz’s law, the self-induced emf opposes any current change taking place. This property of the coil to oppose any change in current flowing through it is known as the self-inductance or inductance. As per the Faraday’s law of electromagnetic induction, the magnitude of the self induced emf in a coil (refer Fig. 1) due to change in current flowing through it is given by,

\[e=-N\frac{d\phi }{dt}\]

where the negative sign indicates that the self induced voltage opposes the change in current through the coil.

Meaning of Self inductance

If the current through the coil (I) produces a flux (ϕ) Webers then the self inductance is given by,

\[\text{Inductance (L) = }\frac{N\times \phi }{I}….(1)\]

Thus, self inductance (L) is defined as the ratio of the flux linkage Wb meter (N × ϕ) to the current I. The units of inductance is Wb Turn / Ampere or Henry. Substituting \(\phi =\frac{L\times I}{N}\) into equation (1) we get,

\[e=-L\left[ \frac{dI}{dt} \right]\text{   Volts}\]

Formula for Self Inductance

We have defined the co-efficient of inductance (L) as,

\[\text{L=}\frac{N\times \phi }{I}\text{ Henry}\]

\[\phi =\frac{m.m.f}{{Re}luc\tan ce}=\frac{N\times I}{S}\]

\[ L=\frac{N}{I}\left[ \frac{N\times I}{S} \right]=\frac{{{N}^{2}}}{S}\text{  Henry}\]

But reluctance,

\[S=\frac{l}{{{\mu }_{0}}{{\mu }_{r}}a}\]

\[L=\frac{{{N}^{2}}\times {{\mu }_{0}}{{\mu }_{r}}a}{l}\text{ Henry}…(1)\]

Where

l = Length of the magnetic circuit ,

µ_r = Relative permeability ,

a = Cross sectional area of magnetic circuit.

N = Number of turns.

Factors Affecting the Self Inductance (L)

From Equation (1) the factors influencing the self inductance are :

1. The self-inductance L is proportional to the square of the number of turns of the coil i.e. N².
2. L is directly proportional to the cross sectional area “a” of the magnetic circuit.
3. L is inversely proportional to the length (l) of the magnetic circuit.
4. L is directly proportional to the relative Permeability µ_r of the material of the magnetic circuit (core).
5. The value of relative permeability µ_r is dependent on the magnetic flux density (B) because µ = B/H. Hence, the self-inductance (L) also is dependent on the value of flux density B.

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Figure 1: Admittance, conductance, susceptance.

The concept of admittance simplifies the analysis of parallel a.c. circuits to a large extent. Now, consider the case of a simple circuit containing a resistance R and an inductive reactance X (Fig. 1). Let the current drawn by a circuit be I amperes at the phase angle of when connected across the supply of V volts. Then,

\[I=\frac{V}{Z}=V.Y\]

Further, the active component of the current in the circuit is given by,

\[I\cos \phi =\frac{V}{Z}\times \frac{R}{Z}=V.\frac{R}{{{Z}^{2}}}=V.G\]

Where, the quantity G which is the ratio of resistance to the square of the impedance is known as the conductance of the circuit (in the special case when X = 0, then G = 1/R).

Also, the reactive component of the current in the circuit is given by,

\[I\sin \phi =\frac{V}{Z}\times \frac{X}{Z}=V.\frac{X}{{{Z}^{2}}}=V.B\]

The quantity B which is the ratio of reactance to the square of impedance is known as susceptance of the circuit. The unit of the conductance as well as susceptance is again the siemens. Fig. 1 (a) shows the phasor diagram taking the voltage as a reference quantity.

By dividing each side of the current triangle by V, a similar triangle as shown in Fig. 1 (b) can be obtained. As sides of this triangle represent the conductance, susceptance and admittance of the circuit, it is known as admittance triangle. If the circuit has resistance and capacitive reactance in series, then the current leads the voltage and in that case, the admittance triangle is as shown in Fig. 1 (c).

From the admittance triangle, we have the following relationships :

\[G=Y\cos \phi ,\text{   }B=Y\sin \phi ,\text{     }Y=\sqrt{{{G}^{2}}+{{B}^{2}}}\]

And,

\[\tan \phi =\frac{B}{G}\]

It should be remembered that the capacitive susceptance is always considered as positive and inductive susceptance as negative. This will obviously be clear from the corresponding admittance triangles.

APPLICATION OF ADMITTANCE METHOD TO PARALLEL CIRCUITS

Consider the general parallel circuit consisting of three branches shown in Fig. 2 (a).

Figure 2: (a) Circuit containing impedances in parallel, (b) Phasor diagram.

Let V be the applied voltage, I the total circuit current and let I₁, I₂ and I₃ be the branch currents. These branch currents will differ in phase with respect to the applied voltage. Let the phase difference between V and the branch currents I₁, I₂ and I₃ be ϕ₁, ϕ₂ and ϕ₃ respectively, the values of which are obtained from reactance and resistance of each branch. e.g.

\[\tan {{\phi }_{1}}=\frac{\left( \omega {{L}_{1}}-\frac{1}{\omega {{C}_{1}}} \right)}{{{R}_{1}}}\]

Fig. 2 (b) shows the phasor diagram for the given circuit in which V is considered as reference quantity. The total current is the phasor addition of all the branch currents. Hence, from the phasor diagram,

\[{{I}^{2}}={{({{I}_{1}}\cos {{\phi }_{1}}+{{I}_{2}}\cos {{\phi }_{2}}+{{I}_{3}}\cos {{\phi }_{3}})}^{2}}\]

\[+{{({{I}_{1}}\sin {{\phi }_{1}}+{{I}_{2}}\sin {{\phi }_{2}}+{{I}_{3}}\sin {{\phi }_{3}})}^{2}}\]

\[={{(V{{G}_{1}}+V{{G}_{2}}+V{{G}_{3}})}^{2}}\]

\[+{{(V{{B}_{1}}+V{{B}_{2}}+V{{B}_{3}})}^{2}}\]

Where,

\[{{G}_{1}}=\frac{{{R}_{1}}}{Z_{1}^{2}},\text{      }{{G}_{2}}=\frac{{{R}_{2}}}{Z_{2}^{2}}\text{,      }{{G}_{3}}=\frac{{{R}_{3}}}{Z_{3}^{2}}\]

\[{{B}_{1}}=\frac{{{X}_{1}}}{Z_{1}^{2}},\text{      }{{B}_{2}}=\frac{{{X}_{2}}}{Z_{2}^{2}}\text{,      }{{B}_{3}}=\frac{{{X}_{3}}}{Z_{3}^{2}}\]

Here, the branch reactances X₁, X₂ and X₃ are obviously,

\[{{X}_{1}}=\omega {{L}_{1}}-\frac{1}{\omega {{C}_{1}}},\text{      }{{X}_{2}}=\omega {{L}_{2}}-\frac{1}{\omega {{C}_{2}}}\]

\[\text{      and       }{{X}_{3}}=\omega {{L}_{3}}-\frac{1}{\omega {{C}_{3}}}\]

Hence, the total admittance, of the whole circuit is given by

\[Y=\frac{I}{V}\]

\[=\sqrt{{{({{G}_{1}}+{{G}_{2}}+{{G}_{3}})}^{2}}+{{({{B}_{1}}+{{B}_{2}}+{{B}_{3}})}^{2}}}\]

\[=\sqrt{{{G}^{2}}+{{B}^{2}}}….(1)\]

And,

\[\tan \phi =\frac{{{({{B}_{1}}+{{B}_{2}}+{{B}_{3}})}^{2}}}{{{({{G}_{1}}+{{G}_{2}}+{{G}_{3}})}^{2}}}=\frac{B}{G}\]

Where,

\[G={{G}_{1}}+{{G}_{2}}+{{G}_{3}}\text{     and      }B={{B}_{1}}+{{B}_{2}}+{{B}_{3}}\]

Thus, total conductance of a parallel circuit can be found by merely adding the branch conductances. Similarly, its total susceptance can be found by algebraically adding the individual susceptance of different branches. If Y₁, Y₂ and Y₃ are admittances of the different branches, then, Equation (1) clearly shows that the total admittance of the parallel circuit is given by their phasor addition i.e.

\[\overline{Y}={{\overline{Y}}_{1}}+{{\overline{Y}}_{2}}+{{\overline{Y}}_{3}}\text{ }\]

It should be remembered that similar to impedance, admittance is always treated as a phasor.

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Construction and Symbol of Inductor

Figure 1: Inductor.

Fig. 1(a) shows the construction of an inductor and Fig. 1(b) shows its symbol. It is a fixed value inductor. An inductor consists of N turns of a laminated copper wire are wound around an iron core.

Unit of Inductor

Inductance is measured in Henry or millihenry or microhenry and it is denoted by L. Henry is a very large unit. Therefore millihenry and microhenry are the another small units used for inductors.

\[\text{1 mH = 1 }\times \text{ 1}{{\text{0}}^{-\text{3}}}\text{ H}\]

\[\text{1  }\!\!\mu\!\!\text{ H = 1 }\times \text{ 1}{{\text{0}}^{-6}}\text{ H}\]

The inductance of a coil is given by,

\[\text{L = }\frac{N\times \phi }{I}\]

Where,

N = Number of turns,

ϕ = Flux

I = Current through the coil.

So the factors affecting the inductance are number of turns, flux linkage and current.

Types of Inductor

Inductors are basically categories,

1. Fixed inductors.
2. Variable inductors.

1. Types of Fixed Inductor :

The fixed inductors are classified as follows:

1. Air-core inductor.
2. Iron-core inductor.
3. Ferrite-core inductors.

1. Air-core inductor :

(a) Symbol

(b) Construction

Figure 2: Air-core Inductor.

In this inductor, the coil is wound on a plastic or cardboard core. Therefore, effectively the air acts as core. The symbol of air core inductor is shown in Fig. 2.

Construction :

The construction of an air-core inductor is shown in Fig. 2. In the construction of air core inductors, a core is made up of ceramics, plastic or cardboard type insulating material. The conductive wire is wound on this core hence there is air inside the coil.

Applications :

1. They are used for intermediate or radio frequency (I.F. or R.F.) applications in tuning coils.
2. For inter-stage coupling.
3. IF. coils.
4. Iron-core inductor :

2. Iron-core inductor :

(a) Symbol

(b) Construction

Figure 3: Iron core Inductor.

An iron core inductor is a coil in which solid or laminated iron or other magnetic material forms a part or all of the magnetic circuit linking its winding. It is also known as iron-core choke. Iron core inductors have a high inductance value but they cannot operate at high frequency due to hysteresis and eddy current losses. Iron core increases the magnetic induction of a coil of wire. Because iron has high permeability, it allows more magnetic lines of flux to concentrate the core thereby increasing the electromagnetic induction.

Construction :

Iron core inductor consists of coil wound over a solid or laminated iron core. The construction of iron core inductor is shown in Fig. 3. The material used for the iron core inductor is Silicon steel which is composed of iron with some percent of silicon. The iron core is laminated to avoid eddy current losses. The laminated iron-core consists of thin iron laminations pressed together but insulated from each other. Low frequency iron cored chokes are used as filter chokes to smooth out ripple in the rectified ac supply amplifier stages and in other d.c. applications. The core materials most commonly used for smoothing chokes are, silicon iron laminations and grain oriented silicon iron.

Applications :

The iron core inductors are used in the dc power supply filter circuits and other low frequency applications.

3. Ferrite core inductor :

Figure 4: Ferrite core Inductor.

Ferrite is an artificially prepared non-metallic material using sintered iron oxide with other metal ions to control magnetic properties. If the coil of wire is wound on a solid core made of highly ferromagnetic substance called ferrite. Fig. 4 shows the symbol of ferrite core inductor. Ferrite is a ferrous magnetic material. In this type of inductor, wire is wound on a ferrite core.

Construction :

The construction of a ferrite core inductor is as shown in Fig. 4. Ferrites are ceramic materials composed of oxides of iron and other magnetic material. It is used at a high and medium frequency because it has high permeability with low loss, so it is more effective than iron core inductor. These inductors usually employ pot cores i.e. cores consisting of an outer cylinder with closed end. The winding is placed in annular space. The air- gap is introduced in the central core. We can choose a suitable length of this air gap, in order to change the properties of the pot to suit a wide range of design requirements.

Applications :

1. These are used at high and medium frequencies.
2. Ferrite rod antenna.

Specifications of inductor

1. Inductance value.
2. Q factor value.
3. Operating frequency range.
4. Power dissipation.
5. Core type.
6. Size and mounting requirements.
7. Stary capacitance.

The post What is an Inductor? Definition, Meaning, Unit, Symbol, Types, Uses & Applications appeared first on Electrical and Electronics Blog.

Figure 1. Force acting on a Current Carrying Conductor placed in a Magnetic Field.

It is observed that whenever a current carrying straight conductor is placed in a magnetic field, it experiences a mechanical force. This is a very important magnetic effect of an electric current as the operation of the electric motors and other electrical appliances is entirely based on this effect. The magnitude of this force is dependent on the following factors:

1. Flux density (B) of the magnetic field in which the conductor is placed (in tesla).
2. Magnitude of the current (I) in the conductor (in amperes).
3. Active length (l) of the conductor (in metres). Active length is the part of the total length of the conductor which actually lies in the magnetic field.

If the conductor is at right angles to the magnetic field, then this force (F) is given by the following expression :

\[\text{F=B I }l\text{ newton}….(1)\]

It will be interesting to see how this force is actually produced. Fig. 1 (a) shows a straight conductor carrying current in the direction towards the observer and placed in a uniform magnetic field at right angles to it. The original field and that due to the conductor are also shown in Fig. 1 (a). These two fields combine to form a single resultant field. It will be seen that in the region below the conductor, both the fields act in the same direction and therefore give a resultant field at any point equal to the sum of the individual fields at that point. In the region above the conductor, one field acts in the direction opposite to that of other and so the resultant is the difference of the two fields. In general, the original field of the magnet is strengthened at the bottom of conductor and weakened at the top. The resultant field pattern is shown in Fig. 1 (b) effect, it seems as if some of the lines of force from the top region are transferred to bottom region.

Due to crowding, the lines of force in the lower region get stretched like rubber bands. Therefore, they try to contract and thereby push the conductor upwards. Thus, conductor experiences a mechanical force in the upward direction. If the direction of the current in the conductor or the direction of the main field reversed, the direction of force exerted on the conductor is also reversed. The direct of the force can be easily found by Fleming’s left hand rule.

Fleming’s Left Hand Rule :

Arrange the first finger, the second finger and the thumb of your left hand mutually at right angles to one another (Fig. 1 c). Point the first finger in the direction of the field and the second finger in the direction of the current, then the thumb will point in the direction of the force on the conductor.

On applying the above rule to the conductor which we have considered, it will be readily found that it experiences a mechanical force in the upward direction. In general, if the current carrying conductor under consideration in the above case is assumed to be making an angle θ with the direction of the field, then the Expression (1) for the force experienced by it will obviously get modified as follows :

\[\text{F = B I }l\text{ sin}\theta \text{ newton}\]

Where,

l sinθ = Component of the active length of the conductor at right- angles to the flux.

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Circuit Diagram & Derivation for Voltage Divider Rule

Figure 1: Series Circuit. 

In order to understand the concept of voltage division rule. Consider a series electrical circuit as shown in figure 1. In this circuit, the current ‘I ‘ remains constant and is flowing through each resistor whereas voltage ‘ V ‘ is dropping at each resistor.

Hence, a series circuit acts as a voltage divider. The respective voltage drop in each resistor is proportional to the value of that resistor.

Voltage division rule is used to determine the various voltages across the circuit from a single source. From Ohm’s law, we know that,

Current,

\[I=\frac{V}{R}\]

In the circuit considered, the value of total resistance is equal to sum of all the resistances.

i.e.,

\[{{R}_{eq}}={{R}_{a}}+{{R}_{b}}+{{R}_{c}}+{{R}_{d}}+{{R}_{e}}\]

Current,

\[I=\frac{V}{{{R}_{a}}+{{R}_{b}}+{{R}_{c}}+{{R}_{d}}+{{R}_{e}}}\]

The voltage across each resistor is given by the product of the voltage across that resistor and the current passing through it.

Voltage across resistor ‘R_a‘

\[{{V}_{{{R}_{a}}}}=I{{R}_{a}}\]

\[=\frac{V}{R}.{{R}_{a}}\text{   }\left[ I=\frac{V}{R} \right]\]

\[=\frac{V.{{R}_{a}}}{{{R}_{a}}+{{R}_{b}}+{{R}_{c}}+{{R}_{d}}+{{R}_{e}}}\text{   }\left[ R={{R}_{a}}+{{R}_{b}}+{{R}_{c}}+{{R}_{d}}+{{R}_{e}} \right]\]

Similarly,

Voltage across resistor ‘R_b‘

\[{{V}_{{{R}_{b}}}}=\frac{V.{{R}_{b}}}{{{R}_{a}}+{{R}_{b}}+{{R}_{c}}+{{R}_{d}}+{{R}_{e}}}\]

Voltage across resistor ‘R_c‘

\[{{V}_{{{R}_{C}}}}=\frac{V.{{R}_{b}}}{{{R}_{a}}+{{R}_{b}}+{{R}_{c}}+{{R}_{d}}+{{R}_{e}}}\]

Voltage across resistor ‘R_d‘

\[{{V}_{{{R}_{d}}}}=\frac{V.{{R}_{d}}}{{{R}_{a}}+{{R}_{b}}+{{R}_{c}}+{{R}_{d}}+{{R}_{e}}}\]

Voltage across resistor ‘R_e‘

\[{{V}_{{{R}_{e}}}}=\frac{V.{{R}_{e}}}{{{R}_{a}}+{{R}_{b}}+{{R}_{c}}+{{R}_{d}}+{{R}_{e}}}\]

The post What is Voltage Divider Rule? Derivation, Circuit Diagram & Formula appeared first on Electrical and Electronics Blog.

Suppose that the application of voltage ‘ V ‘ across PP’ produces current is through the branch QQ’ having impedance Z₃. Then, the ratio of the response to excitation is given by,

\[\frac{R}{E}=\frac{{{I}_{3}}}{V}…(1)\]

Now, the positions of source and response are interchanged. The voltage source is connected across the terminals QQ’ and it produces current response I₁ through the branch PP’ having impedance Z₁ as shown in the figure (b). Now, the ratio of the response to the excitation is given by,

\[\frac{R}{E}=\frac{{{I}_{1}}}{V}…(2)\]

According to reciprocity theorem, the ratio of response to the excitation is constant in both cases.

Hence,

Equation (1) = Equation (2)

\[\frac{{{I}_{3}}}{V}=\frac{{{I}_{1}}}{V}=\frac{I}{V}\left( Constant \right)\]   \[As \left[ i.e,{{I}_{3}}={{I}_{1}} \right]\]

Consider the network shown in figure (1). Suppose application of voltage V across AA’ produces current I through BB’. Now, if the positions of source and responses are interchanged by connecting the voltage source across BB’, the resultant current I will flow through terminals AA’. According to reciprocity theorem, the ratio of response to excitation is same in both the cases. The reciprocity theorem is not valid for a network with two sources.

i.e,\[\frac{I}{V}=\text{ Constant }\left( \text{In both the cases} \right)\]

Reciprocity Theorem Example

Consider the network shown in figure (2). Let 10 V voltage source be the excitation (or input) and current I be the response (or output). By applying KCL at node ‘a’, we get,

\[\frac{{{V}_{a}}-10}{2}+\frac{{{V}_{a}}}{4}+\frac{{{V}_{a}}-20}{30}=0\]

\[{{V}_{a}}\left( \frac{1}{2}+\frac{1}{3}+\frac{1}{4} \right)-\frac{10}{2}-\frac{20}{3}=0\]

\[{{V}_{a}}\times \frac{13}{12}=\frac{35}{3}\]

\[{{V}_{a}}=\frac{35}{3}\times \frac{12}{13}=\frac{140}{13}\text{V}\]

\[I=\frac{{{V}_{a}}}{4}=\frac{140}{13\times 4}=\frac{35}{13}\text{A}\]

The ratio of response to the excitation is,

\[\frac{I}{V}=\frac{35}{13}\times \frac{1}{10}=\frac{7}{26}…(3)\]

When the input and output terminals are interchanged, the circuit until be modified as shown in figure (3).

\[\frac{{{{{V}’}}_{a}}}{2}+\frac{{{{{V}’}}_{a}}-10}{4}+\frac{{{{{V}’}}_{a}}-20}{3}=0\]

\[{{{V}’}_{a}}\left( \frac{1}{2}+\frac{1}{3}+\frac{1}{4} \right)=\frac{10}{2}+\frac{20}{3}\]

\[{{{V}’}_{a}}\times \frac{13}{12}=\frac{55}{6}\]

\[{{{V}’}_{a}}=\frac{55}{6}\times \frac{12}{13}=\frac{100}{13}\text{V}\]

\[{{{I}’}_{a}}=\frac{{{{{V}’}}_{a}}}{2}=\frac{100}{13\times 2}=\frac{55}{13}\text{A}\]

The ratio of response to the excitation is,

\[\frac{{{I}’}}{V}=\frac{55}{13}\times \frac{1}{10}\]

\[=\frac{11}{26}…(4)\]

As equations (3) and (4) are not equal it can be said that the reciprocity theorem is invalid for networks with two or more sources.

The post Reciprocity Theorem – Statement, Derivation, Examples & Formula appeared first on Electrical and Electronics Blog.

\[v={{V}_{m}}\sin \theta ={{V}_{m}}\sin \omega t\]

Due to the voltage applied an alternating current flows through the circuit. Let this instantaneous current be “i”.

Applying KVL to the above circuit, we have,

\[v=iR\] \[i=\frac{v}{R}\] \[i=\frac{{{V}_{m}}\sin \omega t}{R}\]

The value of “i” will be maximum when sinωt = 1

\[{{I}_{m}}=\frac{{{V}_{m}}}{R}\] Instantaneous current, i = I_m sinωt

Phasor Diagram of Purely Resistive Circuit

From the phasor diagram shown in figure (2), it is clear that the A.C voltage and current are in-phase i.e., there is no phase difference between them.

Impedance of Purely Resistive Circuit

The impedance ‘Z’ is given by the quantity \(\sqrt{{{R}^{2}}+{{X}^{2}}}\) which offers opposition to the current flow in the circuit and is measured in ohms (Ω).

\[I=\frac{V}{Z}=\frac{V}{\sqrt{{{R}^{2}}+{{X}^{2}}}}\]

Here the circuit consists of only resistive element.

The impedance of circuit,

\[Z=\sqrt{{{R}^{2}}}=R\Omega \]

Instantaneous Power of Purely Resistive Circuit

Instantaneous power, P = VI

Where,

\[v={{V}_{m}}\sin \omega t\]

\[i={{I}_{m}}\sin \omega t\]

\[P=\left( {{V}_{m}}\sin \omega t \right)\left( {{I}_{m}}\sin \omega t \right)\]

\[={{V}_{m}}{{I}_{m}}\sin \omega t\]

\[={{V}_{m}}{{I}_{m}}\left[ \frac{1-\cos 2\omega t}{2} \right]\text{   }\left[ {{\sin }^{2}}\omega t=\frac{1-\cos 2\omega t}{2} \right]\]

\[=\frac{{{V}_{m}}{{I}_{m}}}{2}-\frac{{{V}_{m}}{{I}_{m}}}{2}\cos 2\omega t\]

Instantaneous power consists of two parts,

1. \(\left( \frac{{{V}_{m}}{{I}_{m}}}{2} \right)\) represents steady power
2. \(\left( \frac{{{V}_{m}}{{I}_{m}}}{2}\cos 2\omega t \right)\) represents fluctuating power.

Average Power of Purely Resistive Circuit

\[{{P}_{avg}}=\frac{1}{2\pi }\int\limits_{0}^{2\pi }{\frac{{{V}_{m}}{{I}_{m}}}{2}d\left( \omega t \right)}+\frac{1}{2\pi }\int\limits_{0}^{2\pi }{\frac{{{V}_{m}}{{I}_{m}}}{2}\cos 2\omega t.d\left( \omega t \right)}\]

\[=\frac{1}{2\pi }\left[ \frac{{{V}_{m}}{{I}_{m}}}{2} \right]\left( \omega t \right)_{0}^{2\pi }+\frac{1}{2\pi }\frac{{{V}_{m}}{{I}_{m}}}{2}\int\limits_{0}^{2\pi }{\cos 2\omega t.d\left( \omega t \right)}\]

\[=\frac{1}{2\pi }\frac{{{V}_{m}}{{I}_{m}}}{2}\left[ 2\pi -0 \right]+\frac{1}{2\pi }\left( \frac{{{V}_{m}}{{I}_{m}}}{2}\left[ \frac{\sin \omega t}{2} \right]_{0}^{2\pi } \right)\]

\[=\frac{{{V}_{m}}{{I}_{m}}}{2}+\frac{1}{2\pi }\left[ \frac{{{V}_{m}}{{I}_{m}}}{4}\left[ 0-0 \right] \right]\]

\[=\frac{{{V}_{m}}{{I}_{m}}}{2}+0\]

Total average power

\[{{P}_{avg}}=\frac{{{V}_{m}}{{I}_{m}}}{2}\]

\[=\frac{{{V}_{m}}}{\sqrt{2}}\times \frac{{{I}_{m}}}{\sqrt{2}}=VI\text{   watts}\]

Power Factor of Purely Resistive Circuit

It is defined as the cosine of the phase angle between current and voltage.

Power factor = cosϕ

The voltage and current pass through their zero values at same instant and attain their positive and negative peaks at the same instant as shown in figure (3).  Therefore, the phase difference between voltage and current is zero i.e ϕ = 0.

The power factor, cosϕ = 1.

Power factor for resistive circuit is unity.

The post What is a Purely Resistive Circuit? Circuit Diagram, Phasor Diagram, Formula & Derivation appeared first on Electrical and Electronics Blog.

\[v={{V}_{m}}\sin \omega t….(1)\]

Due to applied voltage an alternating current flows through the inductor and sets up a self-induced e.m.f ‘e’ of di magnitude \(L\frac{di}{dt}\). At every instant the applied voltage has to dt overcome this self-induced e.m.f.

Thus,

Applied alternating voltage = – (Self-induced e.m.t)

\[v=-e\]

\[v=L\frac{di}{dt}\] \[{{V}_{m}}\sin \omega t=L\frac{di}{dt}\]

\[di=\frac{{{V}_{m}}}{L}\sin \omega t\]

Integrating on both sides, we have,

\[\int{di}=\int{\frac{{{V}_{m}}}{L}\sin \omega t}\]

\[i=\frac{{{V}_{m}}}{L}\left( \frac{-\cos \omega t}{\omega } \right)\]

\[=\left( \frac{{{V}_{m}}}{L} \right)\sin \left( \omega t-90{}^\circ  \right)….(2)\]

The value of ‘i’ will be maximum when sin (ωt – 90º) unity.

\[{{I}_{m}}=\frac{{{V}_{m}}}{\omega L}\] Substituting I_m value in equation (2), we get,

Instantaneous current, i = I_m sin (ωt – 90º)

Phasor Diagram of Purely Inductive Circuit

The phasor diagram of a Purely inductive circuit is shown in figure (2). From figure (2) it is clear that the current in the circuit lag behind the voltage by 90º.

Impedance of Purely Inductive Circuit

Impedance ‘Z’ of any circuit is given by,

\[Z=\sqrt{{{R}^{2}}+{{X}^{2}}}\]

Where,

R = Resistance of the circuit

X = Reactance of the circuit.

Here the circuit consists of only inductive element.

The impedance,

Z = X_L

Where,

X_L = Inductive reactance

X_L = ωL = 2πfL

Inductive reactance is measured in ohms (Ω).

Instantaneous Power of Purely Inductive Circuit

\[P=Vi\] \[=\left( {{V}_{m}}\sin \omega t \right)\left( {{I}_{m}}\sin \left( \omega t-90{}^\circ  \right) \right)\]

\[={{V}_{m}}{{I}_{m}}\sin \omega t\left( -\cos \omega t \right)\] \[=-{{V}_{m}}{{I}_{m}}\sin \omega t\cos \omega t\]

\[=\frac{-{{V}_{m}}{{I}_{m}}}{2}\sin 2\omega t\text{   }\left[ \sin 2\omega t=2\sin \omega t\cos \omega t \right]\]

Average Power of Purely Inductive Circuit

It is the total power over one full cycle.

\[{{P}_{avg}}=\frac{1}{2\pi }\int\limits_{0}^{2\pi }{-}\left( \frac{{{V}_{m}}{{I}_{m}}}{2} \right)\sin \omega td\left( \omega t \right)\]

\[=\frac{1}{2\pi }\left[ -\frac{{{V}_{m}}{{I}_{m}}}{2} \right]\int\limits_{0}^{2\pi }{\sin 2\omega td\omega t}\]

\[=\frac{-{{V}_{m}}{{I}_{m}}}{4\pi }\left[ \frac{-\cos \omega t}{2} \right]_{0}^{2\pi }\] \[=\frac{{{V}_{m}}{{I}_{m}}}{8\pi }\left[ 1-1 \right]=0\]

Therefore, the average power consumed by the circuit is zero.

Power Factor of Purely Inductive Circuit

The phase difference between voltage and current is 90º with the current lagging i.e., the current waveform crosses its zero after the voltage waveform cross 90º as shown in figure (3).

ϕ = 90º lag

Power factor, cosϕ = 0

The post What is a Purely Inductive Circuit? Circuit Diagram, Phasor Diagram, Formula & Derivation appeared first on Electrical and Electronics Blog.