In single phase electric power, the **double-subscript notation** eliminates the need for both the polarity markings for voltages and direction arrows for currents. It is even more useful for representing voltages and currents in three-phase circuits, resulting in greater clarity and less confusion.

## Voltage Phasor

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.

**Current Phasor**

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.

**Double Subscript Notation**

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

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

**Single Phase Voltage**

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

**Single Phase Current**

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.

**Single Phase, Three Wire System Example**

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.