Figure 1: Hay’s Bridge.

The schematic of Hay’s bridge is shown in Fig. 1. Hay’s bridge differs from Maxwell’s bridge by having a resistance R₁ in series with a capacitor C_l instead of being parallel.

For large phase angles, R₁ needs to be low; therefore this bridge is more convenient for measuring high Q coils. For Q =10, the error is ± 1% and for Q = 30, the error is ± 0.1%.

When bridge is balanced,

\[{{Z}_{1}}{{Z}_{x}}=\text{ }{{Z}_{2}}{{Z}_{3}}\text{            }……\left( 1 \right)\]

\[{{Z}_{1}}={{R}_{1}}-\frac{j}{\omega {{C}_{1}}}\]

\[{{Z}_{2}}={{R}_{2}}\]

\[{{Z}_{3}}={{R}_{3}}\]

\[{{Z}_{x}}={{R}_{x}}+j\omega {{L}_{x}}\]

From equation (1),

\[\left( {{R}_{1}}-\frac{j}{\omega {{C}_{1}}} \right)\left( {{R}_{x}}+j\omega {{L}_{x}} \right)={{R}_{2}}{{R}_{3}}\]

\[{{R}_{1}}{{R}_{x}}+j\omega {{L}_{x}}{{R}_{x}}-\frac{j{{R}_{x}}}{\omega {{C}_{1}}}+\frac{{{L}_{x}}}{{{C}_{1}}}={{R}_{2}}{{R}_{3}}\]

\[\left( {{R}_{1}}{{R}_{x}}+\frac{{{L}_{x}}}{{{C}_{1}}} \right)+j\left( \omega {{L}_{x}}{{R}_{1}}-\frac{{{R}_{x}}}{\omega {{C}_{1}}} \right)={{R}_{2}}{{R}_{3}}\]

Equation real and imaginary terms

\[{{R}_{1}}{{R}_{x}}+\frac{{{L}_{x}}}{{{C}_{1}}}={{R}_{2}}{{R}_{3}}\text{ }…\left( 2 \right)\]

\[\omega {{L}_{x}}{{R}_{1}}-\frac{{{R}_{x}}}{\omega {{C}_{1}}}=0…\left( 3 \right)\]

Thus,

\[{{L}_{x}}=\frac{{{R}_{x}}}{{{\omega }^{2}}{{R}_{1}}{{C}_{1}}}\]

Substituting Equation (3) in equation (2)

\[{{R}_{1}}{{R}_{x}}+\frac{{{R}_{x}}}{{{\omega }^{2}}{{R}_{1}}C_{1}^{2}}={{R}_{2}}{{R}_{3}}\]

\[{{R}_{x}}\left[ {{R}_{1}}+\frac{1}{{{\omega }^{2}}{{R}_{1}}C_{1}^{2}} \right]={{R}_{2}}{{R}_{3}}\]

\[{{R}_{x}}\left[ \frac{{{\omega }^{2}}R_{1}^{2}C_{1}^{2}+1}{{{\omega }^{2}}{{R}_{1}}C_{1}^{2}} \right]={{R}_{2}}{{R}_{3}}\]

\[{{R}_{x}}=\frac{{{\omega }^{2}}C_{1}^{2}{{R}_{1}}{{R}_{2}}{{R}_{3}}}{1+{{\omega }^{2}}R_{1}^{2}C_{1}^{2}}…\left( 4 \right)\]

Substitution Equation (4) in equation (3)

\[{{L}_{x}}=\frac{{{\omega }^{2}}C_{1}^{2}{{R}_{1}}{{R}_{2}}{{R}_{3}}\text{ }}{\left( 1+{{\omega }^{2}}C_{1}^{2}R_{1}^{2} \right)\left( {{\omega }^{2}}{{C}_{1}}{{R}_{1}} \right)}\]

\[{{L}_{x}}=\frac{{{C}_{1}}{{R}_{2}}{{R}_{3}}\text{ }}{\left( 1+{{\omega }^{2}}C_{1}^{2}R_{1}^{2} \right)}…\left( 5 \right)\]

From Equation (5) unknown inductance can be calculated.

As ω appears in the expression for L_x, this bridge is frequency sensitive.

Quality factor (Q) for capacitor is

\[\omega {{L}_{x}}{{R}_{1}}-\frac{{{R}_{x}}}{\omega {{C}_{1}}}=0\text{          }\]

∴\[\text{  }\omega {{C}_{1}}{{R}_{1}}=\frac{1}{Q}\text{   }….\left( 6 \right)\]

Substituting Equation (6) in Equation (5)

\[{{L}_{x}}=\frac{{{C}_{1}}{{R}_{1}}{{R}_{3}}}{1+\frac{1}{{{Q}^{2}}}}\]

For a value of Q greater than 10, the term 1/Q² will be smaller than 1/100 and therefore can be neglected.

L_x = C₁R₂R₃ (for Q >10)

Thus for Q > 10, the equation obtained for L_x in Hay’s bridge is same as that in Maxwell’s bridge.

For inductors with Q less than 10, the 1/Q² term can not be neglected. Hence this bridge is not suited for measurement of inductors having Q less than 10.

For measuring inductance using Hay’s bridge, unknown inductance is connected in one of the arms of Hay’s bridge and resistors R₁ and R₃ are adjusted to obtain balance of the bridge. Then using the Equation (5), unknown inductance can be calculated.

Advantages of Hay’s Bridge

1. It can measure inductances with high Q i.e. Q > 10.
2. A commercial bridge measures inductance from 1 μH to 100 H with ± 2% error.
3. It can be used for measuring incremental inductance.

Disadvantages of Hay’s Bridge

1. The unknown value of inductance depends on loss of inductor (Q) and also on operating frequency.
2. It can not measure inductance with low Q i.e. Q < 10.

The post What is Hay’s Bridge? Circuit Diagram, Derivation & Advantages appeared first on Electrical and Electronics Blog.

Figure 1: Maxwell’s Bridge.

When bridge is balanced,

\[{{Z}_{1}{Z}_{x}}={{Z}_{2}{Z}_{3}}\]

 \[{{Z}_{x}}=\frac{{{Z}_{2}}{{Z}_{3}}}{{{Z}_{1}}}={{Z}_{2}}{{Z}_{3}}{{Y}_{1}}\text{            }……\left( 1 \right)\]

\[{{Z}_{1}}={{R}_{1}}\parallel \frac{1}{j\omega {{C}_{1}}}\]

\[{{Y}_{1}}=\frac{1}{{{R}_{1}}}+j\omega {{C}_{1}}\]

\[{{Z}_{2}}={{R}_{2}}\]

\[{{Z}_{3}}={{R}_{3}}\]

\[{{Z}_{x}}={{Z}_{x}}+j\omega {{L}_{x}}\]

From Equation (1) we have,

\[{{R}_{x}}+j\omega {{L}_{x}}={{R}_{2}}{{R}_{3}}\left( \frac{1}{{{R}_{1}}}+j\omega {{C}_{1}} \right)\]

\[{{R}_{x}}+j\omega {{L}_{x}}=\frac{{{R}_{2}}{{R}_{3}}}{{{R}_{1}}}+j\omega {{R}_{2}}{{R}_{3}}{{C}_{1}}\]

Equation real and imaginary terms,

\[{{R}_{x}}=\frac{{{R}_{2}}{{R}_{3}}}{{{R}_{1}}}\]

And  \[{{L}_{x}}={{R}_{2}}{{R}_{3}}{{C}_{1}}\text{            }……\left( 2 \right)\]

Thus the measurement of unknown inductance is independent of the excitation frequency. The scale of the resistance can be calibrated to read inductance directly.

Quality factor (Q) for inductance is,

\[Q=\frac{\omega {{L}_{x}}}{{{R}_{x}}}=\frac{\omega {{C}_{1}}{{R}_{2}}{{R}_{3}}\times {{R}_{1}}}{{{R}_{2}}{{R}_{3}}}=\omega {{C}_{1}}{{R}_{1}}\]

For high values of Q, R₁ becomes excessively large and it is impractical to obtain a satisfactory variable standard resistance in the range. Therefore Maxwell’s bridge is suitable for measurement of inductance with low Q values.

For measurement of inductance using Maxwell’s bridge, unknown inductor is connected in one of the arms and resistor R₁ and R₃ is adjusted to balance the bridge. Then using the derived Equation (2), unknown inductance can be calculated.

Advantages of Maxwell’s Bridge

1. This bridge is particularly suited for inductance measurement as comparison with capacitor is more ideal than with another inductor.
2. It can measure inductors with high Q.
3. It can measure inductance from 1 H to 1000 H with ± 2% error.
4. The measurement is independent of excitation frequency

Disadvantages of Maxwell’s Bridge

1. It cannot be used for measuring inductance with high Q values.
2. Due to fixed capacitor, there is an interaction between the resistance and reactance balances. This can be avoided by varying the capacitor instead of R₁ and R₃.

The post What is Maxwell’s Bridge? Circuit Diagram, Derivation & Advantages appeared first on Electrical and Electronics Blog.

Equipments for Electroplating

Figure 1: Electroplating.

Fig. 1 shows a small electroplating plant with the following equipments.

Electroplating Tank : For the solutions employed in electroplating, chemically resistant containers are generally necessary. Welded steel tanks suitably lined are used for the majority of electroplating processes. Wooden, R.C.C., fibre glass, stainless steel and enamelled iron tanks are also in use. The lining materials include lead, rubber, P.V.C. etc. The tank is filled with suitable electrolyte and has arrangement for hanging anodes and articles to be plated which normally form cathode. Apart from the electrolyte, some additional reagents like glue, gum, sugar, etc. are also added to obtain good results.

Power Supply Unit : For electroplating, direct current ranging from 50 to 1000 amperes or even more at 3 to 24 volts is required. Static rectifiers are normally employed for this purpose.

General Procedure for Electroplating

Before electroplating, the object must be perfectly clean, free from dirt, paint, scales, rust, grease, oil, etc.

• Cleaning is carried out using solvents like paraffin, hot alkaline cleaners (e.g. caustic potash) and acids.
• For smooth surface, mechanical operations like grinding, polishing are also carried out if necessary.

The object is then dipped in the bath for electroplating. The actual process of deposition of metal on the object forming the cathode by electrolysis has already been discussed previously. The metal to be deposited is supplied either by the anode of that metal or the electrolyte itself. The cathode (the article to be plated) is surrounded by the set of anodes or it is rotated slowly to ensure an even deposit all over. As a general rule, slower the deposition, the harder and more adherent it is. Hence, small currents are employed. The concentration and temperature of the electrolyte are other vital factors which affect the quality of the plating. Plating time depends on the mass of metal to be deposited. After plating, the article is washed and dried. In the cases of nickel, silver and chromium plating, the plated article is finally buffed with polishing mop rotated at high speed, for a bright metallic finish.

Applications of Electroplating

Gold or silver plating is many times done for decorative or ornamental purposes. The cast iron and steel parts are normally coated with zinc, nickel or chromium to prevent atmospheric corrosion. Nickel or chromium plating also gives shining white surface. Copper plating is mostly used on iron articles to prevent rusting or as preliminary coating for nickel or silver plating.

The post What is Electroplating? Process, Parts & Diagram appeared first on Electrical and Electronics Blog.

Electrolysis Process

Figure 1: Electrolysis.

To understand the process of electrolysis, consider the solution of copper sulphate (CuSO₄) in water. As the moment, copper sulphate (CuSO₄) is dissolved in water, its molecules dissociate into positively charged copper (Cu) ions and negatively charged sulphate (SO₄) ions. The ions thus formed start wandering freely in the solution.

Now, if two electrodes (P and K) are immersed in the solution and a battery (B) is connected as shown in Fig. 1, let us see the result. As soon as the potential difference is applied across the electrodes with the help of the battery, the haphazard movement of the ions becomes a directed movement. The positive copper ions are attracted to the negative electrode (K) called cathode, and the negative sulphate ions to the positive electrode (P) called anode.

On reaching the respective electrodes, they give up their charges and become ordinary molecules. Thus the copper ions become ordinary molecules of copper, which are deposited on the cathode. The sulphate ions also become ordinary sulphate molecules at the anode.

As the uncharged sulphate ion is incapable of separate existence, it immediately reacts with water (H₂O) forming sulphuric acid (H2SO₄) and liberating oxygen gas (O₂) at the anode. The chemical reaction can be stated as follows :

\[2S{{O}_{4}}=2{{H}_{2}}O=2{{H}_{2}}S{{O}_{4}}+{{O}_{2}}\uparrow \]

If the anode is a suitable copper plate, the sulphate ion acts on it to form copper sulphate as given below :

\[S{{O}_{4}}=Cu=CuS{{O}_{4}}\]

This newly formed copper sulphate goes into solution and thus the density of the electrolyte is maintained constant. The net effect is that the copper from the anode is transferred to the cathode. The above process involving chemical decomposition of the electrolyte due to passage of current is known as electrolysis. It should be noted that the conduction of current in electrolyte is due to oppositely directed movements of positive and negative ions and it is always accompanied by a chemical change.

Faraday’s Laws of Electrolysis

In 1834, Faraday formulated following two basic laws based on his experimental work on electrolysis.

First Law

It states that the weight of a substance liberated from an electrolyte in a given time during the process of electrolysis is directly proportional to the quantity of electricity which has assed through the electrode.

Thus, if

W = Weight of the substance liberated in kg

Q =  Quantity of electricity in coulombs

= Current (I) in amperes × Time (t) in seconds. Then, according to this law,

\[W\propto Q\propto I.t\]

Or,

\[W=Z.I.t\]

Where, Z is a constant called the electrochemical equivalent.

It is defined as the mass of the particular substance liberated in unit time by unit current and its unit is the kilogram per coulomb.

Second Law

It states that if the same current flows through several electrolytes, the weights of the substances liberated from each are proportional to the chemical equivalents of these substances.

The chemical equivalent of a substance is the weight of the substance which can displace or combine with unit weight of the substance in a chemical reaction.

With the above definition, the chemical equivalent of hydrogen itself will be obviously one.

Extraction and Electro refining of Metals

The metals like aluminium, copper, magnesium, sodium, zinc, etc. can be extracted from their ores and refined by electrolytic processes.

In electrolytic extraction process, the ore is either melted or treated with strong acid. The pure metal is then obtained by the electrolysis of this solution. The metals obtained by electrolytic extraction methods being 98 to 99 percent pure, these methods are always preferable in comparison with the other metallurgical processes available for the most of these metals. Sometimes, however, this electrolytic extraction cannot be carried out economically. Therefore, in such cases, only the refining is done electrolytically.

Copper is extracted by metallurgical as well as electrolytic processes. However, it is always refined electrolytic ally to its purest form as required by the electrical industries. For refining of metals, ingots of impure metal obtained from metallurgical process are used as anodes. They are placed in an electrolytic cell containing a suitable electrolyte. The current is then passed through the bath. Due to electrolysis, the pure metal gets deposited on the cathode made of pure metal.

Power Supply: Large amount of d.c. power is required for extraction and refining of metals. Voltage per electrolytic cell is only about 10 V. However, a number of such cells can be connected in series, so that the voltage required is 500 to 800 volts, with currents upto several thousand amperes. Special heavy current motor-generator sets, rotary convertors or mercury arc rectifiers are used for conversion of a.c. to d.c.

The post What is Electrolysis? Definition, Process, Diagram & Law appeared first on Electrical and Electronics Blog.

The selection of a particular bus-bar arrangement is done depending upon the factors such as voltage level, simplicity, reliability, safety, cost of installation and maintenance, etc.

Bus-bar Arrangements

Different types of bus-bar arrangements available are,

1. Single bus-bar system
2. Single bus-bar with sectionalizer
3. Main and transfer bus-bar system.

Single Bus-bar System

Single bus-bar system is the simplest and cheapest arrangement of bus-bars. It consists of a single bus-bar to which all the electrical equipments viz., generators, transformers, isolators, etc., are connected. A circuit breaker is provided for every individual equipment for protection against fault currents. Similarly, isolators helps in the isolation of generators, transformers etc., during fault clearance and maintenance.

Single bus-bar system is used for voltages below 33 kV. Usually, it is employed for 11 kV indoor substations. Single line diagram of a single bus-bar system is shown in the following figure.  Further, it is to be noted that all the incoming and outgoing lines are connected to the single bus-bar only. The incoming lines at a voltage of 11 kv are connected to the bus-bar through isolators and circuit breakers. In the outgoing line, the voltage is stepped down to 400 V using 11 kV/400 V transformer, thereby making the voltage at outgoing lines to 400 V

Advantages of Single Bus-bar System

1. Due to the simplicity and low initial cost, single bus-bar systems are used.
2. It is easy to operate since, the connections of single bus-bar system are simple.
3. Single bus-bar system can be conveniently used where there is no future expansion of the substation is expected.

Disadvantages of Single Bus-bar System

1. In case of fault on the bus-bars, the supply to the whole system, including healthy feeders gets interrupted.
2. It is not possible to carry out repairs without interrupting the supply.

Single Bus-bar with Sectionalizer

Sectionalized single busbar means single busbar with 2 to 3 sections. Sections of busbar are separated by isolator with circuit breaker combination as shown in figure.  Sectionalization of busbar with only isolator is not preferable. We know that isolator is also called as no-load switch i.e., which works on no-load only. Hence, a circuit breaker is necessary to remove the load on bus-bar. It is clear that sectionalization of busbar prefers isolator with circuit breaker.

Sectionalized single bus-bar has following advantages (over single bus-bar arrangement),

Advantages of Single bus-bar with sectionalizer

1. Flexible operation can be achieved using single bus-bar scheme with sectionalization.
2. More reliable than simple bus-bar scheme.
3. The faulted section can be isolated without affecting the other section’s supply.
4. The maintenance or repair of one section is possible as it is shut downed, without affecting the other section supply.

Disadvantages of Single bus-bar with sectionalizer

1. If the circuit breakers are not used as the sectionalizing switch, then the coupling of the bus-bar may occur during the load transfer and the use of circuit breakers make the system expensive.
2. In order to maintain the continuity of supply to the system, during the maintenance of the circuit breaker, the circuit breaker must be provided with isolators on both sides, which further increases the cost of the system.
3. Single bus-bar system with sectionalization is uneconomical for small substations.

Main and transfer bus-bar system

Bus-bars are the copper rods, that are used to collect electrical energy at one place. The generators and feeders that are operating at same voltage (or) constant voltage are connected directly to these busbars. In order to avoid the interruption of power flow the study of bus-bar arrangement is very important.

The main and transfer busbar arrangement uses two buses, one as main bus and other as transfer (or) auxiliary (or) spare bus. The generators and feeders are connected to both main bus and as well as transfer (or) auxiliary bus.

Under normal conditions the generator and feeder, are connected to main bus. Suppose, assume that the fault has occurred in any breaker or main bus then the power flow gets interrupted. In order to avoid this, the entire equipment that are connected to main bus is shifted (or) transferred to a transfer bus (or) auxiliary bus without any interruption of power flow by using a bus- coupler, which uses double isolating switches. Thus the generator and feeders are transferred from main bus to auxiliary bus without any interruption of power.

Advantages of Main and Transfer Bus-bar System

1. There is no interruption of power supply i.e., the power supply is continuous even under fault condition.
2. The maintenance and repair of bus or circuit breaker is easy.
3. During fault conditions the circuit is transferred easily to an auxiliary bus.

Disadvantages of Main and Transfer Bus-bar System

1. The main and transfer bus-bar arrangement is very expensive than other bus-bar arrangements.
2. The service may be interrupted during switch over from one bus to another bus.

The post What is Electrical Bus Bar? Types, Advantages & Disadvantages appeared first on Electrical and Electronics Blog.

Operation of Transformer On load

Figure (1) shows a transformer with a load connected across the secondary winding. The load current I₂ flowing through the secondary turns sets up its own m.m.f N₂I₂ which produces the flux ϕ₂.

According to Lenz’s law this flux is in such a direction that it opposes the flux ϕ, produced by the m.m.f N₁I₀ which is set up by the no-load current I₀ flowing throw the primary turns. Consequently the flux is momentarily reduced due to opposing flux ϕ. This in turn causes reduction in induced e.m.f (E₁) in primary according to Faraday’s law E₁ reduces, the difference between applied voltage (V₁) and E₁ increases.

Consequently, the primary will draw more current. Consider \({{{I}’}_{1}}\) to be this additional primary current. It is also known as counter balancing current as it balances between applied voltage and primary e.m.f or it is known as load component of primary current and it is antiphase with secondary current I₂. Now this current \({{{I}’}_{1}}\) sets up its own m.m.f N₁\({{{I}’}_{1}}\) which produces the flux and it is equal in magnitude in such a direction that it opposes the flux ϕ₂. Hence ϕ’₁ and ϕ₂ cancel each other and only flux ϕ flows in the core. Therefore the total flux produced during loaded condition is approximately equal to the flux at no-load.

\[{{\phi }_{2}}=-{{{\phi }’}_{1}}\]

As secondary ampere turns of I₂ are neutralized by primary ampere turns of \({{{I}’}_{1}}\).

\[{{N}_{2}}{{I}_{2}}={{N}_{1}}{{{I}’}_{1}}\]

\[{{{I}’}_{1}}=\frac{{{N}_{2}}}{{{N}_{1}}}{{I}_{2}}\]

The net primary current is the vector sum of primary counter balancing current \({{{I}’}_{1}}\) and the no-load current I₀.

\[{{I}_{1}}={{{I}’}_{1}}+{{I}_{0}}\]

Since the no-load current I₀ is very small compared to the counter balancing current \({{{I}’}_{1}}\), therefore the net primary current is approximately equal to the current \({{{I}’}_{1}}\).

\[{{I}_{1}}={{{I}’}_{1}}\]

\[=\frac{{{N}_{2}}}{{{N}_{1}}}{{I}_{2}}=K{{I}_{2}}\]

‘K’ represents transformation ratio.

\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{{{I}’}}_{1}}}{{{I}_{2}}}=\frac{{{N}_{2}}}{{{N}_{1}}}=K\]

\[{{I}_{1}}=K{{I}_{2}}={{{I}’}_{1}}\]

Therefore, the primary and secondary currents are inversely proportional to their turns ratio. The total primary current is in anti-phase with I₂ and K times the current I₂.

Phasor Diagram of Transformer with Resistive Load

The phasor diagram for resistive load is drawn as shown in the following figure (2). For purely resistive load, the secondary load current I₂ is in phase with the secondary’ terminal voltage V₂. The counter balancing current \({{{I}’}_{1}}\) is in opposition and equal in magnitude with the secondary load current I₂. The primary current I₁ is the vector sum of \({{{I}’}_{1}}\) and no-load current I₀ respectively. I₀ lags behind V₁ by no-load power factor angle ϕ₀ and I₁ lags behind the voltage V₁ by primary power factor angle ϕ₁.

The post Transformer on Load – Circuit Diagram & Phasor Diagram appeared first on Electrical and Electronics Blog.

When the transformer is on no-load, the secondary winding is opened as in figure (a) and current I₂ is zero. In this condition, the primary winding draws a no-load current ‘I₀‘ which has two components i.e.,

1. Magnetizing component (I_µ), and
2. Working component (I_w).

1. Magnetizing Component (I_µ) :

It lags behind the applied voltage on primary winding ‘V₁‘ by 90º. It is also called as reactive or wattless component of no-load current and is responsible to develop an e.m.f to maintain the flux ‘ϕ’ in the core. It is expressed as \({{I}_{\mu }}=\text{ }{{I}_{0}}\sin {{\phi }_{0}}\).

2. Working Component (I_w) :

It is in phase with the primary applied voltage ‘V₁‘. The component is also called as active component or iron loss component, and is used for describing the core losses such as hysteresis loss and eddy current loss. It is expressed as \({{I}_{w }}=\text{ }{{I}_{0}}\cos {{\phi }_{0}}\).

From the phasor diagram of figure (b),

\[\sin {{\phi }_{0}}=\frac{{{I}_{\mu }}}{{{I}_{0}}}\]

Thus, \({{I}_{\mu }}={{I}_{0}}\sin {{\phi }_{0}}\) is the reactive component of no-load current I₀ and

\[\cos {{\phi }_{0}}=\frac{{{I}_{w }}}{{{I}_{0}}}\]

Thus, \({{I}_{w }}={{I}_{0}}\cos {{\phi }_{0}}\) is the active component of no load current I₀.

Hence,

\[{{I}_{0}}=\sqrt{I_{w}^{2}+I_{\mu }^{2}}\]

cosϕ₀ is the no-load power factor and ϕ₀ is the hysteresis angle of advance.

The post Transformer on No Load – Circuit Diagram & Phasor Diagram appeared first on Electrical and Electronics Blog.

Consider the two winding single-phase transformer shown in figure (1).

\({{{I}}_{1}}\) = Current in the primary

\({{{E}}_{1}}\) = Induced e.m.f in the primary

\({{{V}}_{1}}\) = Voltage applied to the primary

\({{{I}}_{2}}\) = Current in the secondary

\({{{E}}_{2}}\) = Induced e.m.f in the secondary

\({{{V}}_{2}}\) = Terminal voltage of secondary

Here, the primary current \({{{I}}_{1}}\) has two components, one is no-load primary current, \({{{I}}_{0}}\) and the other one is load component of primary current \({{{I}’}_{2}}\). The function of current \({{{I}’}_{2}}\) is to counter balance the secondary current \({{{I}}_{2}}\). The no-load primary current \({{{I}}_{0}}\) leads to the production of losses in the core while magnetizing the core of the transformer. The no- load primary current \({{{I}}_{0}}\) can be resolved into two components i.e., active (or) working component I_w and reactive (or) magnetizing component ‘I_µ‘. The working component ‘I_w’ of no-load current \({{{I}}_{0}}\) leads to the core loss, hence it can be represented by a resistance R₀. The magnetizing current ‘I_µ‘ produces flux which induces e.m.f E₁.

The reactance due to flux is represented by X₀. To account for the core loss and the magnetizing current, an equivalent circuit can be represented by a shunt branch in the primary side as shown in the figure (2).

\[\text{Core loss = }I_{w}^{2}{{R}_{0}}=\frac{E_{1}^{2}}{{{R}_{0}}}\]

To make transformer calculations simpler, transfer voltage, current and impedance either to the primary or secondary

Equivalent Circuit of Transformer as Referred to Primary Side

Secondary parameters transferred to primary side are given as follows,

\[{{{R}’}_{2}}=\frac{{{R}_{2}}}{{{K}^{2}}}\]

\[{{{X}’}_{2}}=\frac{{{X}_{2}}}{{{K}^{2}}}\]

\[{{{Z}’}_{2}}=\frac{{{Z}_{2}}}{{{K}^{2}}}\]

\[{{{I}’}_{2}}=K{{I}_{2}}\]

\[{{{E}’}_{2}}=\frac{{{E}_{2}}}{K}\]

\[{{{V}’}_{2}}=\frac{{{V}_{2}}}{K}\]

Where,

\[K=\frac{{{N}_{2}}}{{{N}_{1}}}\]

We know that,

\[{{R}_{01}}={{R}_{1}}+{{{R}’}_{2}}={{R}_{1}}+\frac{{{R}_{2}}}{{{K}^{2}}}\]

\[{{X}_{01}}={{X}_{1}}+{{{X}’}_{2}}={{X}_{1}}+\frac{{{X}_{2}}}{{{K}^{2}}}\]

\[{{Z}_{01}}=\sqrt{R_{01}^{2}+X_{01}^{2}}=\frac{{{Z}_{02}}}{{{K}^{2}}}\]

The equivalent circuits referred to primary side are as shown in figures (3) and (4).

Equivalent Circuit of Transformer Referred to Secondary Side

Primary parameters transferred to secondary side are given as follows,

\[{{{R}’}_{1}}={{K}^{2}}{{R}_{1}}\]

\[{{{X}’}_{2}}={{K}^{2}}{{X}_{1}}\]

\[{{{Z}’}_{1}}={{K}^{2}}{{Z}_{1}}\]

\[{{{E}’}_{1}}=K{{E}_{1}}\]

\[{{{V}’}_{1}}=K{{V}_{1}}\]

\[{{{I}’}_{1}}=\frac{{{I}_{1}}}{K}\]

\[{{{I}’}_{0}}=\frac{{{I}_{0}}}{K}\]

\[{{{R}’}_{0}}=\frac{{{R}_{0}}}{{{K}^{2}}}\]

\[{{{X}’}_{0}}=\frac{{{X}_{0}}}{{{K}^{2}}}\]

We know that,

\[{{R}_{02}}={{R}_{2}}+{{{R}’}_{2}}={{R}_{2}}+{{K}^{2}}{{R}_{1}}\]

\[{{X}_{02}}={{X}_{2}}+{{{X}’}_{1}}={{X}_{2}}+{{K}^{2}}{{X}_{1}}\]

\[{{Z}_{02}}=\sqrt{R_{02}^{2}+X_{02}^{2}}={{K}^{2}}{{Z}_{01}}\]

The equivalent circuit diagrams referred to secondary side are shown in figure (5) and figure (6).

The post Equivalent Circuit of Transformer – Circuit Diagram & Derivation appeared first on Electrical and Electronics Blog.

Objectives of Tariff :

Like other commodities, electrical energy is also sold at such a rate that it not only returns the cost but also earns reasonable profit. Therefore, a tariff should include the following terms:

1. Recovery of cost of producing electrical energy at the power station.
2. Recovery of cost on the capital investment in transmission and distribution systems.
3. Recovery of cost of operation and maintenance of supply of electrical energy e.g., metering equipment, billing etc.
4. A suitable profit on the capital investment.

Types of Tariff

The different types of tariff are as follows,

1. Simple tariff
2. Flat rate tariff
3. Block rate tariff
4. Three-part tariff
5. Two-part tariff
6. Maximum demand tariff
7. Power factor tariff.

1. Simple Tariff

In this type of tariff, the rate per unit of energy consumed is fixed and every consumer will be charged the same. The simple tariff is also called as uniform tariff and is the simplest form of tariff. The simple tariff remains constant and does not vary with the increase or decrease in number of units consumed by a consumer.

2. Flat Rate Tariff

In this type of tariff different types of consumers are charged at different rates. The rate for each type of consumer is arrived at by taking its load factor, diversity factors into account. The bill will be total units consumed × rate/unit.

Advantage :

1. It is easy to understand by different types of consumers and calculations at suppliers end are simple.

Disadvantages :

1. If the consumer has got two types of loads i.e., (i) fan and lighting load (ii) power load, then two meters are installed at his premises i.e., for different types of power supply, separate meters
   are required.
2. It is very difficult to derive at the load factor and diversity factor to be used in deciding the traffic.

3. Block Rate Tariff

Here, cost of units consumed will change as the consumer crosses particular number of units its. If the number of units generated increases, then the cost of generation per unit, automatically decreases. For example, for first 50 unit there may be one ratio, 51 to 200 there may be other rate. From 201 to 400 there may be some another rate. Such type of tariff is commonly used now a days for domestic and small industrial consumers.

Advantage :

If the consumer has large demand of  number of units, then he has to pay less amount only. Due to that reason, the consumers show more interest for consuming more electrical energy.

Disadvantage :

It lacks a measure of the consumers demand.

The overall annual cost of electrical energy generated by a power station can be expressed in two forms, three-part tariff forms and two-part tariff form.

4. Three-part Tariff

In this method, the overall annual cost of electrical energy generated is divided into three parts i.e., Fixed cost, Semi-fixed cost and Running cost.

Total annual cost of energy =Fixed cost + Semi-fixed cost + Running cost

= Constant + Proportional to maximum demand+ Proportional to kWh generated

= (a+ b kW + c kWh)

Where,

a – Annual fixed cost independent of maximum demand and energy.

b – Constant which when multiplied by maximum kW demand on the station gives the annual semi- fixed cost.

c – A constant which when multiplied by kwh output per annum gives the annual cost

It may be seen that by adding fixed charge or consumer ‘s charge to two-part tariff, becomes three-part tariff.

5. Two-part Tariff

The total charge under this kind of tariff is divided into two parts i.e., a fixed charge based on the maximum demand and variable charge (running charge) per unit of energy consumed. The expression for the annual cost of energy then becomes,

Total annual cost of energy = (A kW + B kWh)

Where,

A – A constant which when multiplied by maximum kW demand on the station gives the annual cost of the first part.

B – A constant which when multiplied by the annual kWh generated gives the annual running cost.

The fixed charges are dependent upon the maximum demand of the consumer, while the running charges are dependent upon the number of units consumed by the consumer. It is easy to understand and this type of tariff applies to consumers industrial consumers.

6. Maximum Demand Tariff

Maximum demand tariff is almost similar to the two-part tariff. The only difference is that, in a maximum demand tariff, the maximum demand is measured by an indicator known as maximum demand indicator.

The drawbacks of two-part tariff are eliminated in maximum demand tariff. This type of tariff is applicable for bulk supplier and for large industrial consumers, who have control over their maximum demand

7.Power Factor Tariff

when the power factor of the consumer’s load are taken into consideration, then that tariff is known as power factor In A.C. system, the efficiency of a plant and equipment not only depends on kW, but also on the power factor. The power factor plays a major role in A. C systems. So, to increase the utility of the plant to a maximum extent, the plant must be operated at most economical power factor.

The three main classes are,

1. kVA Maximum Demand Tariff: The maximum demand of the consumer is measured in kVA, not in kW. The maximum kVA demand is charged in addition to the charge corresponding to the energy.
2. kWh and kVARh Tariff: In this type of tariff both kWh and kVARh of a consumer are charged separately.
3. Sliding Scale Method: Under this kind of tariff, average power factor say 0.8 lagging is assumed as a reference. If the p.f of a consumer falls by even 1 % then an extra amount will be charged for it. On the other hand, if the p.f of any consumer rises by even 1%, a discount will be given to him.

Requirements of Tariff Method

1. Tariff should be simple.
2. It should give lesser rates for more consumption.
3. It must be same for large population.
4. It should consider both the maximum demand charges and the energy charges.
5. It must encourage the consumers with high load factors.
6. It should motivate consumers for using power during off-peak hours.
7. It should charge more for lighting than power connection.
8. The penalty should be provided for low power factor.

Q1. Why is tariff for power load less than the lighting load?

Ans. Although tariff should include the total cost of producing and supplying electrical energy plus the profit, it cannot be the same for all types of consumers. This is because the cost of producing electrical energy depends to a considerable extent upon the magnitude of electrical energy consumed by the user and his load conditions. Therefore, in all fairness, due consideration has to be given to different types of consumers (e.g., industrial, domestic and commercial) while fixing the tariff. This makes the problem of suitable rate making highly complicated. The power load improves the load factor of a system to greater extent rather than a lighting load. Hence, the reason, why the tariff for power load is less than the lighting load.

The post What is Electricity Tariff? Definition, Meaning, Types & Objectives appeared first on Electrical and Electronics Blog.

Figure 1: Electrical Insulator.

An electrical insulator in an overhead line is to hold the live conductor to prevent leakage of current from the conductor to the pole. These are made of porcelain clay and are thoroughly glazed to avoid the absorption of moisture from the atmosphere.

Properties of an Electrical insulator

1. It should have high electrical resistance in order to prevent the leakage of current.
2. It should have high mechanical strength in order to withstand the weight of conductor and the forces acting due to wind pressure, ice etc.
3. It should have high resistance to temperature variation in order to prevent it from any damage.
4. It should have high permittivity  in order to withstand the electrical stresses due to over voltages, switching surges, lightning
5. It should be non-porous and free from impurities like cavity, void, crack etc.
6. It should have high ratio of puncture strength to flashover.
7. It should not be brittle.
8. It should be non-hygroscopic.

Types of Electrical Insulator

The various types of insulators are as follows,

1. Pin-type insulators
2. Suspension type insulators
3. Strain insulators
4. Shackle insulators and
5. Stay insulators.

Puncture in an Electrical Insulator

Puncture is an electric breakdown in an insulator and it is most severe but least frequent type of electrical breakdown. Whenever, the voltage across the insulator exceeds the puncture voltage, an arc will strike between the conductor and passes through the body of the insulator. Since, the arc produced during puncture passes through the whole body of the insulator, the complete insulator gets damaged and has to be replaced with a new one. In order to avoid puncture of insulators, the thickness of the porcelain disc has to be increased.

Whenever flash over occurs, there are chances that the insulator will continue to work. But, whenever a puncture occurs,  there is no chance, that it will continue to work and the insulator has to be replaced. Hence for safety operation of the insulator, the insulator will be designed in such a way that flash over occurs before puncture. This can be ensured by keeping the value of safety factor as high as possible. The value of safety factor for a pin type insulator is about 10.

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