{"id":2769,"date":"2024-08-25T11:37:11","date_gmt":"2024-08-25T11:37:11","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2769"},"modified":"2024-08-25T11:37:11","modified_gmt":"2024-08-25T11:37:11","slug":"notation-of-phasor-on-rectangular-co-ordinate-axes","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/25\/notation-of-phasor-on-rectangular-co-ordinate-axes\/","title":{"rendered":"NOTATION OF PHASOR ON RECTANGULAR CO-ORDINATE AXES"},"content":{"rendered":"\n

Consider a phasor\u00a0\"image\"\u00a0lying along OX-axis as shown in\u00a0Figure 7.61. The phasor is reversed when it is multiplied by \u22121, that is, the phasor is rotated through 180\u00b0 in counter clockwise (CCW) direction and attains the position along OX\u2032-axis. Let us consider\u00a0j<\/em>\u00a0as a factor which when multiplied by the phasor\u00a0\"image\", the phasor is rotated through 90\u00b0 in CCW direction. This means that multiplying the phasor by\u00a0j<\/em>2<\/sup>\u00a0is the same as multiplying by \u22121. Therefore, it follows that<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Fig. 7.61<\/strong>  Vector representation by j-notation<\/p>\n\n\n\n

j2<\/sup><\/em> = \u22121 or 1<\/a><\/sup>j<\/em> = \"image\"<\/p>\n\n\n\n

Therefore, it is concluded that j<\/em> is just an operator that is when multiplied with a phasor, it shows that the phasor is rotated through 90\u00b0 in CCW direction. Each successive multiplication of j<\/em>, rotates the phasor further by 90\u00b0 as<\/p>\n\n\n\n

 <\/p>\n\n\n\n

j<\/em> = \"image\"………………..90\u00b0 CCW rotation from OX-axis<\/p>\n\n\n\n

j<\/em>2<\/sup> = \u22121………………….180\u00b0 CCW rotation from OX-axis<\/p>\n\n\n\n

j<\/em>3<\/sup> = j<\/em>2<\/sup>j<\/em> = \u2212\"image\"………..270\u00b0 CCW rotation from OX-axis<\/p>\n\n\n\n

j<\/em>4<\/sup> = j<\/em>2<\/sup>j<\/em>2<\/sup> = 1…………….360\u00b0 CCW rotation from OX-axis<\/p>\n\n\n\n

 <\/p>\n\n\n\n

The symbol\u00a0j<\/em>\u00a0is used to represent the vertical (quadrature) components of phasor quantities. For instance, consider a phasor\u00a0V<\/em>\u00a0rotated through\u00a0\u03b8<\/em>\u00b0 counter clockwise from OX-axis as shown in\u00a0Figure 7.62(a). The phasor has two rectangular components (i) the horizontal component ‘a<\/em>‘ along X-axis and (ii) the vertical component ‘b<\/em>‘ rotated through 90\u00b0 in CCW direction from OX-axis and is expressed as ‘jb<\/em>‘. Therefore, in rectangular form, the phasor\u00a0\"image\"\u00a0is represented as:<\/p>\n\n\n\n

\"image\" = a<\/em> + jb<\/em> having magnitude \"image\" and angle \u03b8<\/em> = tan\u2212<\/sup>1<\/sup> (b<\/em>\/a<\/em>) [positive]<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Fig. 7.62<\/strong>  (a) Position of vector V at an instant (b) Position of vector V at an instant<\/p>\n\n\n\n

If the phasor\u00a0\"image\"\u00a0is displaced through an angle\u00a0\u03b8<\/em>\u00b0 in clockwise direction as shown in\u00a0Figure 7.62(b), the vertical component will be expressed as ‘\u2212jb<\/em>‘. Therefore, in rectangular form, the phasor is represented as<\/p>\n\n\n\n

\"image\" = a<\/em> + jb<\/em> having magnitude \"image\" and angle \u03b8<\/em> = tan\u2212<\/sup>1<\/sup> (\u2212b<\/em>\/a<\/em>) [negative]<\/p>\n\n\n\n

7.21.1  Mathematical Representation of Phasors<\/h4>\n\n\n\n

In the mathematical form, a phasor can be represented in (i) rectangular form, (ii) trigonometric form, and (iii) polar form. Consider a voltage phasor\u00a0\"image\"\u00a0displaced\u00a0\u03b8<\/em>\u00b0CCW from the reference axis (i.e., OX\u2212axis) as shown in\u00a0Figure 7.62(a). Let us see how this phasor is represented in different forms.<\/p>\n\n\n\n

Rectangular form<\/em><\/p>\n\n\n\n

This method is also known as symbolic notation. In this method, the phasor is resolved into horizontal and vertical components and expressed in the complex form, i.e.,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\"image\" = a<\/em> + jb<\/em><\/p>\n\n\n\n

Magnitude of phasor,<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n

Its angle with OX-axis, \u03b8<\/em> = tan\u2212<\/sup>1<\/sup> (b<\/em>\/a<\/em>)<\/p>\n\n\n\n

If angle\u00a0q<\/em>\u00a0would have been negative as shown in\u00a0Figure 7.62(b), the vertical component would be negative. Then, the phasor\u00a0\"image\"\u00a0would have been represented as<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\"image\" = a<\/em> \u2212 jb<\/em><\/p>\n\n\n\n

<\/a>Trigonometric form<\/em><\/p>\n\n\n\n

In this case, the horizontal and vertical components of the phasor are expressed in the trigonometric form. For example, in Figure 7.62(a)<\/a>, we get horizontal component, a<\/em> = V<\/em> cos \u03b8<\/em> and vertical component, b<\/em> = V<\/em> sin \u03b8<\/em><\/p>\n\n\n\n

 <\/p>\n\n\n\n

\"image\" = V<\/em> cos \u03b8<\/em> + j V<\/em> sin \u03b8<\/em> or \"image\" = V<\/em> (cos \u03b8<\/em> + j<\/em> sin \u03b8<\/em> )<\/p>\n\n\n\n

\u2234<\/p>\n\n\n\n

If angle\u00a0\u03b8<\/em>\u00a0is negative, as shown in\u00a0Figure 7.62(b), then<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\"image\" = V<\/em> (cos \u03b8<\/em> \u2212 j<\/em> sin \u03b8<\/em> )<\/p>\n\n\n\n

Polar form<\/em><\/p>\n\n\n\n

The short form of trigonometric representation of a phasor is called polar form.<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\"image\" = v \u2220 \u03b8\u00b0<\/em><\/p>\n\n\n\n

where V<\/em> = the magnitude of the phasor and \u03b8<\/em> = phase angle measured in CCW direction from the reference axis, that is, OX-axis. There is no mathematical explanation for this form.<\/p>\n\n\n\n

If angle\u00a0\u03b8<\/em>\u00a0is negative, as shown in\u00a0Figure 7.62(b), then<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\"image\" = V<\/em> \u2220 \u2212\u03b8<\/em> \u00b0<\/p>\n\n\n\n

In fact, all the above mentioned three mathematical forms of representing a phasor convey the same information, that is, magnitude of the phasor and its direction with the horizontal axis. Therefore, one form is converted into the other form rapidly as per the requirement to speed up the calculations.<\/p>\n\n\n\n

7.22  ADDITION AND SUBTRACTION OF PHASOR QUANTITIES<\/h5>\n\n\n\n

The rectangular form is the best suited for addition and subtraction of phasor quantities. Therefore, if the phasor quantities are given in polar form, they are first converted into rectangular form and then added or subtracted.<\/p>\n\n\n\n

Consider two voltage phasors represented as<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\"image\" = a<\/em>1<\/sub> + jb<\/em>1<\/sub> and \"image\" = a<\/em>2<\/sub> \u2212 jb<\/em>2<\/sub><\/p>\n\n\n\n

7.22.1  Addition<\/h4>\n\n\n\n

In this case, the in-phase components of the quantities are added together, that is, horizontal components are added separately and the vertical components are added separately as<\/p>\n\n\n\n

Resultant voltage, \"image\" = \"image\" + \"image\" = (a<\/em>1<\/sub> + jb<\/em>1<\/sub>) + (a<\/em>2<\/sub> \u2212 jb<\/em>2<\/sub>) = (a<\/em>1<\/sub> + a<\/em>2<\/sub>) + j<\/em> (b<\/em>1<\/sub> \u2212 b<\/em>2<\/sub>)<\/p>\n\n\n\n

Magnitude of resultant, \"image\"<\/p>\n\n\n\n

Its angle with OX-axis, \"image\"<\/p>\n\n\n\n

7.22.2  Subtraction<\/h4>\n\n\n\n

Similar to addition, ordinary rules of phasor algebra are followed while subtracting the phasor quantities. Let phasor \"image\" be subtracted from phasor \"image\"<\/p>\n\n\n\n

<\/a>\u2234 Resultant voltage \"image\" = \"image\" + \"image\" = = (a<\/em>1<\/sub> + jb<\/em>1<\/sub>) \u2212 (a<\/em>2<\/sub> \u2212 jb<\/em>2<\/sub>) = (a<\/em>1<\/sub> \u2212 a<\/em>2<\/sub>) + j<\/em> (b<\/em>1<\/sub> + b<\/em>2<\/sub>)<\/p>\n\n\n\n

Magnitude of resultant, \"image\"<\/p>\n\n\n\n

Its angle with OX-axis, \"image\"<\/p>\n","protected":false},"excerpt":{"rendered":"

Consider a phasor\u00a0\u00a0lying along OX-axis as shown in\u00a0Figure 7.61. The phasor is reversed when it is multiplied by \u22121, that is, the phasor is rotated through 180\u00b0 in counter clockwise (CCW) direction and attains the position along OX\u2032-axis. Let us consider\u00a0j\u00a0as a factor which when multiplied by the phasor\u00a0, the phasor is rotated through 90\u00b0 […]<\/p>\n","protected":false},"author":1,"featured_media":2481,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[410],"tags":[],"class_list":["post-2769","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-single-phase-ac-circuits"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/08\/singlephase-network-energy-meter-connection-260nw-2444369485.jpg","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2769","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/comments?post=2769"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2769\/revisions"}],"predecessor-version":[{"id":2770,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2769\/revisions\/2770"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/2481"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=2769"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2769"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2769"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}