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.

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Fig, 1: Internal Diagram of Megger.

Working Principle of Megger

The principle of working of megger is based on electromagnetic induction. When a conductor carrying current is kept in a magnetic field. then the conductor exerts a force which is proportional to the strength and capacity of the current and magnetic field. The direction of the force is along the direction of current and magnetic field.

Construction of Megger

Megger consists of current coil or PMMC instrument and pressure coils (see Fig. 1). There are two pressure or voltage coils. All these coils are placed around an annular iron ring on a common shaft which rotates freely. Ligaments are the flexible leads which connect the coils. The C-shaped iron core has a deflecting coil on to which pointer is attached. The pointer indicates the deflection over the graduated scale. A hand driven generator is provided in instrument in order to generate the operation of megger. If the scale is calibrated in reverse, the pointer indicates ‘∞’ and if the scale is fully deflected, then the pointer indicates ‘0’ resistance.

Working of Megger

Current is passed to the coils from the hand driven generator. The pressure coils are set in such a way that it stands perpendicular to the magnetic field. If the test terminals are kept open corresponding to ‘∞’ Ω then the current does not flow through the coil (deflection). The pressure coil controls the movement of deflecting coil and makes it to come in opposite direction, when the pointer is at ‘∞’ position, small torque is exerted by the coil. If the test terminals are kept shorted corresponding to ‘0’ Ω, then the large current flows through the deflecting coil. The torque exerted by the coil is increased by making it to move in strong magnetic field. When the pointer is at ‘0’ position, maximum torque is exerted under the pole piece.

The effect of this instrument decreases the “low resistance” and sets up ‘high resistance” portion of the scale. This effect is beneficial to use as insulation test because the insulation resistances are large.

Applications of Megger

1. Megger checks the continuity between any two points in a circuit.
2. It determines the resistance between the equipment and the earth.
3. It performs the various tests in industries such as.

(i) Open circuit (O.C.) tests

(ii) Short – circuit (S.C.) tests

(iii) Ground tests.

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Statement of Thevenin’s Theorem

Thevenin's Theorem states that for the purpose of determining the current through a resistor RL (called load resistance) connected across any two terminals A and B of a linear bilateral network with constant resistances, and constant voltage and current sources (Fig. 1 a), the network excluding RL (enclosed by dotted line in Fig. 1 a), can be replaced by a single source of e.m.f. and a series resistor (enclosed by dotted line in Fig. 1 b) where

(i) this e.m.f is equal to the open-circuit voltage, say Vo, between the terminals under consideration (i.e. the voltage across the two terminals A and B with RL removed) and,

(ii) the resistance of the series resistor, say Ro, is equal to the resistance of the network as viewed from the terminals A and B with the load resistance RL removed amid all sources replaced by their internal resistances.

Hence, the current through R_L,

\[\text{I }=\text{ }\frac{{{\text{V}}_{\text{o}}}}{{{\text{R}}_{\text{o}}}+{{\text{R}}_{\text{L}}}}\]

(a)

(b)

Fig. 1: General two-terminal network with Thevenin’s equivalent circuit.

Thevenin’s Theorem Formula

Current through the load resistance connected across any two terminals A and B of a linear bilateral network with constant resistances, constant voltage and current sources,

\[\text{I }=\text{ }\frac{{{\text{V}}_{\text{o}}}}{{{\text{R}}_{\text{o}}}+{{\text{R}}_{\text{L}}}}\]

where

R_L = Load resistance connected across the terminals A and B.

V₀ = Thevenin’s equivalent voltage i.e. the open-circuit voltage between the terminals A and B with the load resistance R_L removed.

R₀ = Thevenin’s equivalent resistance i.e. the resistance of the network as viewed from the terminals A and B with the load resistance R_L removed and all sources replaced by their internal resistances.

Explanation of Thevenin’s Theorem

To understand the application of the above theorem more clearly, let us consider the network shown in Fig. 2 (a). It consists of resistors R₁, R₂, and R_L, and a battery having an e.m.f. E and an internal resistance r. If it is required to find the current through a load resistance R_L using Thevenin’s theorem, the procedural steps are as follows:

(i) Remove the load resistance R_L from the terminals A and B and find the open-circuit voltage V₀ between these terminals. Referring to Fig. 2 (b), it is obvious that

\[\text{Current through }{{\text{R}}_{\text{1}}}=\text{ }\frac{\text{E}}{{{\text{R}}_{\text{1}}}+\text{r}}\]

and

\[\text{p}\text{.d across }{{\text{R}}_{\text{1}}}=\text{ }\frac{\text{E }{{\text{R}}_{\text{1}}}}{{{\text{R}}_{\text{1}}}+\text{r}}\]

Since, no current is flowing through R₂, open-circuit voltage across the terminals A and B is

\[{{\text{V}}_{\text{o}}}=\text{ }\frac{\text{E }{{\text{R}}_{\text{1}}}}{{{\text{R}}_{\text{1}}}+\text{r}}\]

(a)

(b)

(c)

(d)

Fig. 2: Circuits to illustrate Thevenin’s theorem

(ii) With the load resistance R_L removed and the battery replaced by its internal resistance r as shown in Fig. 2 (c), find the Thevenin’s equivalent resistance R₀ of the network as viewed from the terminals A and B. Obviously,

\[{{\text{R}}_{\text{o}}}=\text{ }{{\text{R}}_{\text{2}}}+\text{ }\frac{\text{r }{{\text{R}}_{\text{1}}}}{{{\text{R}}_{\text{1}}}+\text{r}}\]

(iii) Replace the entire network enclosed by the dotted line in Fig. 2 (a) by the Thevenin’s equivalent circuit consisting of single voltage source with e.m.f. equal to V₀ and internal resistance equal to R₀ as illustrated in Fig. 2 (d).

(iv) Finally, from the simple circuit of Fig. 2 (d), calculate the current I flowing through the load resistance R_L using the expression

\[\text{I }=\text{ }\frac{{{\text{V}}_{\text{o}}}}{{{\text{R}}_{\text{1}}}+{{\text{R}}_{\text{L}}}}\]

Examples of Thevenin’s Theorem

Example 1 : Using Thevenin’s theorem, calculate the current in the 15 Ω resistor in the network shown in Fig. 3.

Fig. 4.

Solution 1: For convenience, the network under consideration is redrawn in Fig. 3. Open circuiting the branch containing 15 Ω resistor gives the network shown in Fig. 4 (a). To calculate the open-circuit voltage V₀ across the terminals A and B, assume a current I₁ in the direction shown. Applying the Kirchhoff’s voltage law to the outer loop, we get

120 – 25I₁ – 5I₁ – 90 = 0

∴                                           I₁ = 1 A

(a)

(b)

Fig. 4: (a) Network of Fig. 4 with the 15  resistor removed, (b) Network of Fig. 4 (a) with voltage sources suppressed.

From the voltage law, the open-circuit voltage across the terminals A and B is then

V_o = 120 – 25 × 1 = 95 V

To determine the Thevenin’s equivalent resistance R_o, the voltages of the sources are reduced to zero and the resistance between the terminals A and B is calculated. Fig. 4 (b) shows the network after the suppression of the voltage sources. The 25 Ω and 5 Ω resistances being in parallel between the terminals A and B,

Fig. 5 : Thevenin’s equivalent of the circuit of Fig. 3.

The Thevenin’s equivalent circuit for the network of Fig. 4 (a) between the terminals A and B with 15 Ω resistance added is then as shown in Fig. 5. Hence, current through 15 Ω resistance is

\[\text{I }=\text{ }\frac{{{\text{V}}_{\text{o}}}}{{{\text{R}}_{\text{1}}}+{{\text{R}}_{\text{L}}}}\text{ }=\text{ }\frac{\text{95}}{\text{401667}+\text{15}}\]

\[=\text{ 4}\text{.9565 A}\]

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