The post Double Subscript Notation in Single Phase System appeared first on Electrical Academia.

]]>The voltage phasor with double subscripts represents the voltage across the two nodes identified by the two subscripts. The voltage is positive during the half-cycle in which the node named by the first subscript is at a higher potential than the node named by the second subscript.

The current phasor with the double subscripts represents the current flowing between two nodes of a circuit. The current is considered positive when it flows in the direction from the node identified by the first subscript toward the node identified by the second subscript.

In this article, we shall find extremely useful the double subscript notation. In this case of phasors, the notation is V_{ab} for the voltage of point a with respect to point b. we shall also use a double subscript notation for current, taking, for example, I_{ab} as the current flowing in the direct path from point a to point b. these quantities are illustrated in figure (1), where the direct path from a to b is distinguished from the alternative path from a to b through c.

Fig.1: Illustration of Double Subscript Notation

Because of the simpler expression for average power that results, we shall use RMS values of voltage and current throughout this tutorial. That is, if

\[\begin{matrix} \text{V=}\left| \text{V} \right|\angle {{\text{0}}^{\text{o}}}\text{ V rms} & {} & {} \\ {} & {} & (1) \\ \text{I=}\left| \text{I} \right|\angle \text{- }\!\!\theta\text{ A rms} & {} & {} \\\end{matrix}\]

Are the phasors associated with an element having impedance,

$\begin{matrix} \text{Z=}\left| \text{Z} \right|\angle \text{ }\text{ }\!\!\theta\!\!\text{ }\text{ }\text{ }\Omega \text{ } & \cdots & (2) \\\end{matrix}$

The average power delivered to the element is

$\begin{matrix} P=\left| \text{V} \right|\text{.}\left| \text{I} \right|\text{cos }\theta\!\!\text{ } & {} & {} \\ {} & {} & (3) \\ ={{\left| I \right|}^{2}}\operatorname{Re}(Z)\text{ W} & {} & {} \\\end{matrix}$

In the time domain the voltage and current are

$\begin{matrix} v=\sqrt{\text{2}}\left| \text{V} \right|\text{cos }\!\!\omega\!\!\text{ t V} \\ i=\sqrt{\text{2}}\left| \text{I} \right|\text{cos( }\!\!\omega\!\!\text{ t- }\!\!\theta\!\!\text{ ) A} \\\end{matrix}$

The use of double subscript makes it easier to handle phasors both analytically and geometrically. For example, in figure 2 (a), the voltage Vab is

${{\text{V}}_{\text{ab}}}\text{=}{{\text{V}}_{\text{an}}}\text{+}{{\text{V}}_{\text{nb}}}$

This is evident without referring to a circuit since by KVL the voltage between two points a and b is the same regardless of the path, which in this case is the path a,n,b. also, since V_{nb}=-V_{bn}, we have

${{\text{V}}_{\text{ab}}}\text{=}{{\text{V}}_{\text{an}}}\text{-}{{\text{V}}_{\text{bn}}}=100-100\angle -{{120}^{o}}$

Which, after simplification, is

${{\text{V}}_{\text{ab}}}=100\sqrt{3}\angle {{30}^{o}}\text{ V rms}$

These steps are shown graphically in figure 2(b).

Fig.2: (a) Phasor Circuit

Fig.2: (b) Corresponding Phasor Diagram

A single-phase, three-wire source, as shown in figure 3, is one having three output terminals a, b, and a neutral terminal n, for which the terminal voltages are equal. That is,

\[\begin{matrix} {{\text{V}}_{\text{an}}}\text{=}{{\text{V}}_{\text{nb}}}\text{=}{{\text{V}}_{\text{1}}} & \cdots & (4) \\\end{matrix}\]

This is a common arrangement in a normal house supplied with both 115 V and 230 V rms, since if |V_{an}|=|V_{1}|=115V, then |V_{ab}|=|2V_{1}|=230V.

Fig.3: Single-Phase, Three-Wire Source

Let us now consider the source of figure 3 loaded with two identical loads, both having an impedance Z_{1}, as shown in figure 4.

Fig.4: Single-Phase, Three-Wire System with two Identical Loads

The currents in the lines aA and bB are

$\begin{align}& {{\text{I}}_{\text{aA}}}\text{=}\frac{{{\text{V}}_{\text{an}}}}{{{\text{Z}}_{\text{1}}}}\text{=}\frac{{{\text{V}}_{\text{1}}}}{{{\text{Z}}_{\text{1}}}} \\& \text{and} \\& {{\text{I}}_{\text{bB}}}\text{=}\frac{{{\text{V}}_{\text{bn}}}}{{{\text{Z}}_{\text{1}}}}\text{=-}\frac{{{\text{V}}_{\text{1}}}}{{{\text{Z}}_{\text{1}}}}\text{=-}{{\text{I}}_{\text{aA}}} \\\end{align}$

Therefore the current in the neutral wire, nN, by KCL is

${{\text{I}}_{\text{nN}}}\text{=-(}{{\text{I}}_{\text{aA}}}\text{+}{{\text{I}}_{\text{bB}}}\text{)=0}$

Thus the neutral could be removed without changing any current or voltage in the system.

If the lines aA and bB are not perfect conductors but have equal impedances Z_{2}, then InN is still zero because we simply add the series impedances Z_{1} and Z_{2} and have essentially the same situation as in figure 4. Indeed, in the more general case shown in figure 5, the neutral current I_{nN} is still zero. This may be seen by writing the two mesh equations.

$\begin{align}& ({{Z}_{1}}+{{Z}_{2}}+{{Z}_{3}}){{I}_{aA}}+{{Z}_{3}}{{I}_{bB}}-{{Z}_{1}}{{I}_{3}}={{V}_{1}} \\& {{Z}_{3}}{{I}_{aA}}+({{Z}_{1}}+{{Z}_{2}}+{{Z}_{3}}){{I}_{bB}}+{{Z}_{1}}{{I}_{3}}=-{{V}_{1}} \\\end{align}$

And adding the result, which yield

$({{Z}_{1}}+{{Z}_{2}}+{{Z}_{3}})({{I}_{aA}}+{{I}_{bB}})+{{Z}_{3}}({{I}_{aA}}+{{I}_{bB}})=0$

Or

$\begin{matrix}{{I}_{aA}}+{{I}_{bB}}=0 & \cdots & (5) \\\end{matrix}$

Since by KCL the left member of the last equation is –I_{nN}, the neutral current is zero. This is, of course, a consequence of the symmetry of figure 5.

Fig.5: Symmetrical Single-Phase, Three-Wire System

If the symmetry of figure 5 is destroyed by having unequal loads at terminals A-N and N-B or unequal line impedances in lines aA and bB, then there will be a neutral current.

For example, let us consider the situation in figure 6, which has two loads operating at approximately 230 V. the mesh equations are

Fig.6: Single Phase, Three Wire System

$\begin{align}& 43{{I}_{1}}-2{{I}_{2}}-40{{I}_{3}}=115 \\& -2{{I}_{1}}+63{{I}_{2}}-60{{I}_{3}}=115 \\& -40{{I}_{1}}-60{{I}_{2}}+(110+j10){{I}_{3}}=0 \\\end{align}$

Solving for the currents, we have

$\begin{align}& {{I}_{1}}=16.32\angle -{{33.7}^{o}}\text{ A rms} \\& {{I}_{2}}=15.73\angle -{{35.4}^{o}}\text{ A rms} \\& {{I}_{3}}=14.46\angle -{{39.9}^{o}}\text{ A rms} \\\end{align}$

Therefore the neutral current is

${{I}_{nN}}={{I}_{2}}-{{I}_{1}}=0.76\angle {{184.3}^{o}}\text{ }A\text{ }rms$

And, of course, is not zero.

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]]>The post Three Phase Star Connection (Y): Three Phase Power,Voltage,Current appeared first on Electrical Academia.

]]>In star connections, fundamentally we connect the same phase sides to a mutual (common) point known as neutral point and provide supply to its free ends which stay thereafter as shown in figure 1. As far as line and phase voltages are concerned, they are related to each other as:

${{\text{V}}_{\text{line}}}\text{=}\sqrt{\text{3}}{{\text{V}}_{\text{phase}}}$

Which means that whatever supply voltage we have, we need to insulate the windings for ${}^{1}/{}_{\sqrt{3}}$ times the line voltage only. Whereas, **in a ****Star**** connection, line and phase currents remain the same** as:

${{\text{I}}_{\text{line}}}\text{=}{{\text{I}}_{\text{phase}}}$

Let us consider the three-phase source of figure 1, which has line terminals a, b, and c and a neutral terminal n. in this case, the source is said to be Y-connected.

Fig.1: Y-Connected Source

The voltages V_{an}, V_{bn}, and V_{cn} between the line terminals and the neutral terminal are called phase voltages and in most cases, we shall consider are given by

In both cases, each phase voltage has the same RMS magnitude V_{p}, and the phases are displaced 120^{o}, with V_{an} arbitrarily selected as the reference phasor. Such a set of voltages is called a balanced set and is characterized by

$\begin{matrix} {{V}_{an}}+{{V}_{bn}}+{{V}_{cn}}=0 & \cdots & (3) \\\end{matrix}$

As may be seen from (1) or (2).

The sequence of voltages in (1) is called the positive sequence, or ABC sequence, while that of (2) is called the negative, or ACB sequence. Phasor diagrams of two sequences are shown in figure 2, where we may see by inspection that (3) holds. Evidently, the only difference between the positive and the negative sequence is the arbitrary choice of terminal labels a,b, and c. this without loss of generality i shall consider only the positive sequence.

Fig.2: (a) Positive and (b) Negative Sequence

By (1), the voltages in ABC sequence may each be related to V_{an}. The relationships are

$\begin{matrix} \begin{align} & {{V}_{bn}}={{V}_{an}}\angle -{{120}^{o}} \\ & {{V}_{cn}}={{V}_{an}}\angle {{120}^{o}} \\\end{align} & \cdots & (4) \\\end{matrix}$

The line-to-line voltages, or simply line voltages, in figure (1) are V_{ab}, V_{bc}, and V_{ca}, which may be found from the phase voltages. For example

$\begin{align} & {{V}_{ab}}={{V}_{an}}+{{V}_{nb}} \\ & ={{V}_{p}}\angle {{0}^{o}}-{{V}_{p}}\angle -{{120}^{o}} \\ & =\sqrt{3}{{V}_{p}}\angle {{30}^{o}} \\\end{align}$

In line manner,

${{V}_{bc}}=\sqrt{3}{{V}_{p}}\angle -{{90}^{o}}$

${{V}_{ca}}=\sqrt{3}{{V}_{p}}\angle -{{210}^{o}}$

If we denote the magnitude of the line voltages by V_{L,} then we have

$\begin{matrix} {{V}_{L}}=\sqrt{3}{{V}_{p}} & \cdots & (5) \\\end{matrix}$

And thus

These results may also be obtained graphically from the phasor diagram shown in figure (3).

Fig.3: Phasor Diagram showing phase and line voltages

Let us consider the system of the figure (4), which is balanced Y-Y, three-phase, four wire system if the source voltages are given by (1). The term Y-Y applies since both the source and the load are Y-connected. The system is said to be balanced since the source voltages constitute a balanced set and the load is balanced (each phase impedance Z_{p} is equal). The fourth wire is the neutral line n-N, which may be omitted to form a Three-phase, three-wire system.

Fig. 4: Balanced Y-Y System

The line currents of figure 4 are evidently

$\begin{matrix} \begin{align} & {{I}_{aA}}=\frac{{{V}_{an}}}{{{Z}_{p}}} \\ & {{I}_{bB}}=\frac{{{V}_{bn}}}{{{Z}_{p}}}=\frac{{{V}_{an}}\angle -{{120}^{o}}}{{{Z}_{p}}}={{I}_{aA}}\angle -{{120}^{o}} \\ & {{I}_{cC}}=\frac{{{V}_{cn}}}{{{Z}_{p}}}=\frac{{{V}_{an}}\angle {{120}^{o}}}{{{Z}_{p}}}={{I}_{aA}}\angle {{120}^{o}} \\\end{align} & \cdots & (7) \\\end{matrix}$

The last two results are a consequence of (4) and show that the line currents also form a balanced set. Therefore their sum is

$-{{I}_{nN}}={{I}_{aA}}+{{I}_{bB}}+{{I}_{cC}}=0$

Thus the neutral carries no current in a balanced Y-Y four-wire system.

In the case of Y-connect loads, the currents in the lines aA, bB, and cC are also the phase currents ( the currents carried by the phase impedances). If the magnitudes of the phase and line currents are I_{p} and I_{L}, respectively, then I_{L}=I_{p}, and (7) becomes

Where θ is the angle of Z_{p}.

The average power P_{p} delivered to each phase of figure 4 is

$\begin{matrix} \begin{align} & {{P}_{p}}={{V}_{p}}{{I}_{p}}\cos \theta \\ & =I_{p}^{2}\operatorname{Re}({{Z}_{p}}) \\\end{align} & \cdots & (9) \\\end{matrix}$

And the total power delivered to the load is

${{P}_{p}}=3{{P}_{p}}$

The angle θ of the phase impedance is thus the power factor angle of the three-phase load as well as that of a single phase.

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]]>The post Three Phase Delta Connection: Three Phase Power,Voltage,Current appeared first on Electrical Academia.

]]>In Delta connection, phase sides are connected in a cyclical arrangement in order to make a closed loop as shown in figure 1. As far as line and phase currents are concerned, they are related to each other as:

${{\text{I}}_{\text{line}}}\text{=}\sqrt{\text{3}}{{\text{I}}_{\text{phase}}}$

Which means that whatever supply current we have, we will have wire cross-section for ${}^{1}/{}_{\sqrt{3}}$ times line current only. Whereas, in Delta connection, line and phase voltage are same:

${{\text{V}}_{\text{phase}}}\text{=}{{\text{V}}_{\text{line}}}$

A balanced delta connected load (with equal phase impedances) is shown in figure 1.

Fig.1: Delta Connected Load

An advantage of a delta connected load over a Y-connected load is that loads may be added or removed readily on a single phase of a delta since the loads are connected directly across the lines. Also, for a given power delivered to the load the phase currents in a delta are smaller than those in a Y. on the other hand, the delta phase voltages are higher than those of the Y connection. Sources are rarely delta connected because if the voltages are not perfectly balanced, there will be a net voltage, and consequently a circulating current, around the delta. This, of course, causes undesirable heating effects in the generating machinery. Also, the phase voltages are lower in the Y-connected generator, and thus less insulation is required. Obviously, systems with delta-connected loads are three wire systems, since there is no neutral connection.

From figure 1, we see that **in the case of delta connected load the line voltages are the same as the phase voltages**. Therefore if the line voltages are given by

$\begin{matrix} \begin{align} & {{V}_{ab}}={{V}_{L}}\angle {{30}^{o}} \\ & {{V}_{bc}}={{V}_{L}}\angle -{{90}^{o}} \\ & {{V}_{ca}}={{V}_{L}}\angle -{{210}^{o}} \\\end{align} & \cdots & (1) \\\end{matrix}$

Then the phase voltages are

Where

$\begin{matrix} {{V}_{L}}={{V}_{p}} & \cdots & (3) \\\end{matrix}$

If Zp=|Zp|∠θ , then the phase currents are

Where

$\begin{matrix} {{I}_{p}}=\frac{{{V}_{L}}}{\left| {{Z}_{p}} \right|} & \cdots & (5) \\\end{matrix}$

The current in line aA by KCL is

${{I}_{aA}}={{I}_{AB}}-{{I}_{CA}}$

Which after simplifications is

${{I}_{aA}}=\sqrt{3}{{I}_{p}}\angle -\theta $

The other line currents, obtained similarly, are

${{I}_{bB}}=\sqrt{3}{{I}_{p}}\angle -{{120}^{o}}-\theta $

${{I}_{cC}}=\sqrt{3}{{I}_{p}}\angle -{{240}^{o}}-\theta $

Evidently the relation between the line and the phase current magnitudes in the delta case is

$\begin{matrix} {{I}_{L}}=\sqrt{3}{{I}_{p}} & \cdots & (6) \\\end{matrix}$

And the line currents are thus

Thus the currents and voltages are balanced sets, as expected. The relationships between line and phase currents for the delta connected load are summed up in the phasor diagram of figure 2.

Now, let us derive a formula for the power delivered to a balanced three phase load with a power factor angle θ. Whether the load is Y-connected or delta connected, we have

$P=3{{P}_{p}}=3{{V}_{p}}{{I}_{p}}\cos \theta $

In the Y-connected case,

$\begin{align} & {{V}_{p}}={}^{{{V}_{L}}}/{}_{\sqrt{3}} \\ & and \\ & {{I}_{p}}={{I}_{L}} \\\end{align}$

And in the delta connected case,

$\begin{align} & {{I}_{p}}={}^{{{I}_{L}}}/{}_{\sqrt{3}} \\ & and \\ & {{V}_{p}}={{V}_{L}} \\\end{align}$

In either case, then,

$P=3\frac{{{V}_{L}}{{I}_{L}}}{\sqrt{3}}\cos \theta $

Or

**Summary**

In **Delta-connected case, we have following relations for current and voltages**:

$\begin{align} & {{I}_{p}}={}^{{{I}_{L}}}/{}_{\sqrt{3}} \\ & and \\ & {{V}_{p}}={{V}_{L}} \\\end{align}$

In **Y-connected case**,

$\begin{align} & {{V}_{p}}={}^{{{V}_{L}}}/{}_{\sqrt{3}} \\ & and \\ & {{I}_{p}}={{I}_{L}} \\\end{align}$

Whereas the** formula to compute the power in both cases remains the same** which is

$P=\sqrt{3}{{V}_{L}}{{I}_{L}}\cos \theta $

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