{"id":2765,"date":"2024-08-25T11:34:25","date_gmt":"2024-08-25T11:34:25","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2765"},"modified":"2024-08-25T11:34:27","modified_gmt":"2024-08-25T11:34:27","slug":"admittance-method","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/25\/admittance-method\/","title":{"rendered":"\u00a0\u00a0ADMITTANCE METHOD"},"content":{"rendered":"\n

Before applying this method for the solution of parallel AC circuits, the reader should be familiar with the following important terms:<\/p>\n\n\n\n

7.19.1  Admittance<\/h4>\n\n\n\n

The reciprocal of impedance of an AC circuit is called admittance of the circuit. Since impedance is the total opposition to the flow of AC in an AC circuit, the admittance is the effective ability of the circuit due to which it allows the AC to flow through it. It is represented by letter ‘Y<\/em>‘.<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

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

The unit of admittance is mho (i.e., ohm spelled backward and its symbol is \u01b1).<\/p>\n\n\n\n

7.19.2  Admittance Triangle<\/h4>\n\n\n\n

Admittance can also be represented by a triangle similar to that of impedance.<\/p>\n\n\n\n

Impedance\u00a0Z<\/em>\u00a0of the circuit has two rectangular components resistance\u00a0R<\/em>\u00a0and reactance\u00a0X<\/em>. Similarly, admittance\u00a0Y<\/em>\u00a0also has two rectangular components conductance\u00a0g<\/em>\u00a0and susceptance\u00a0b<\/em>, as shown in\u00a0Figure 7.57.<\/p>\n\n\n\n

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

Fig. 7.57<\/strong>  (a) Admittance triangle for inductive circuit (b) Admittance triangle for capacitive circuit<\/p>\n\n\n\n

7.19.3  Conductance<\/h4>\n\n\n\n

The base of an admittance triangle is representing conductanceas shown in\u00a0Figure 7.57.<\/p>\n\n\n\n

Conductance,<\/p>\n\n\n\n

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

Conductance is always positive irrespective of the circuit parameters. The unit of conductance is mho<\/em>.<\/p>\n\n\n\n

<\/a>7.19.4  Susceptance<\/h4>\n\n\n\n

The perpendicular or an admittance triangle is representing susceptance, as shown in\u00a0Figure 7.57.<\/p>\n\n\n\n

Susceptance,<\/p>\n\n\n\n

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

Susceptance is positive for capacitive reactance (see Fig. 7.57(b)<\/a>) and negative for inductive reactance (see Fig. 7.57(a)<\/a>). The unit of susceptance is mho.<\/p>\n\n\n\n

7.19.5  Solution of Parallel AC Circuits by Admittance Method<\/h4>\n\n\n\n

Consider a parallel AC circuit shown in\u00a0Figure 7.58. For its solution, we shall proceed as follows:<\/p>\n\n\n\n

Step I:\u00a0<\/strong>Draw the circuit as per the given problem as shown in\u00a0Figure 7.58.<\/p>\n\n\n\n

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

Fig. 7.58<\/strong>  AC circuits connected in parallel<\/p>\n\n\n\n

Step II: <\/strong>Find impedance and phase angle of each branch.<\/p>\n\n\n\n

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

Step III: <\/strong>Find conductance, susceptance, and admittance of each branch.<\/p>\n\n\n\n

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

Step IV: <\/strong>Find the algebraic sum of conductance and susceptance.<\/p>\n\n\n\n

 <\/p>\n\n\n\n

G<\/em> = g<\/em>1<\/sub> + g<\/em>2<\/sub>; B<\/em> = b<\/em>1<\/sub> \u2212 b<\/em>2<\/sub><\/p>\n\n\n\n

Step V: <\/strong>Find total admittance of the circuit.<\/p>\n\n\n\n

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

Step VI: <\/strong>Find branch currents and total current.<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>1<\/sub> = VY<\/em>1<\/sub>; I<\/em>2<\/sub> = VY<\/em>2<\/sub>; I<\/em> = VY<\/em><\/p>\n\n\n\n

Step VII: <\/strong>Find the phase angle and the pf of the whole circuit.<\/p>\n\n\n\n

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

Example 7.38<\/strong><\/p>\n\n\n\n

The active and lagging reactive components of current taken by an AC circuit from a 250\u2212V supply are 50 A and 25 A, respectively. Calculate the conductance, susceptance, admittance, and power factor of the circuit.<\/p>\n\n\n\n

(U.P.T.U. Tut.)<\/strong><\/p>\n\n\n\n

<\/a>Solution:<\/em><\/p>\n\n\n\n

Active component of current, I<\/em>a<\/sub> = I<\/em> cos \u0278 <\/em>= 50 A<\/p>\n\n\n\n

Leading reactive component of current, I<\/em>r<\/sub>= I<\/em> sin \u0278<\/em> = 25<\/p>\n\n\n\n

Circuit current, \"image\"<\/p>\n\n\n\n

Admittance of the circuit, \"image\"<\/p>\n\n\n\n

Power factor of the circuit, \"image\"<\/p>\n\n\n\n

Conductance of the circuit, G<\/em> = Y cos \u0278<\/em> = 0.223 \u00d7 0.8944 = 0.2 mho<\/p>\n\n\n\n

Susceptance of the circuit, \"image\"<\/p>\n\n\n\n

 <\/p>\n\n\n\n

= 0.1 mho (inductive)<\/p>\n\n\n\n

Example 7.39<\/strong><\/p>\n\n\n\n

A parallel circuit has two branches. One branch contains a resistance of 8 \u03a9 and inductance of 19.1 mH in series and the other contains a resistance of 6 \u03a9 and capacitor of capacitance 601.55 F in series. This parallel circuit is connected across a supply voltage of 240 V, 50 Hz. Determine (i) current drawn by each branch, (ii) total current drawn from the mains, and (iii) pf of the whole circuit.<\/p>\n\n\n\n

Solution:<\/em><\/p>\n\n\n\n

The circuit is shown in\u00a0Figure 7.59.<\/p>\n\n\n\n

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

Fig. 7.59<\/strong>  Circuit as per data<\/p>\n\n\n\n

Branch I<\/em><\/strong><\/p>\n\n\n\n

 <\/p>\n\n\n\n

X<\/em>L1<\/sub> = 2 \u0278 f L<\/em>1<\/sub> = 2 p <\/em>\u00d7 50 \u00d7 0.0191 = 6 \u03a9<\/p>\n\n\n\n

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

I<\/em>1<\/sub> = VY<\/em>1<\/sub> = 240 \u00d7 0.1 = 24 A<\/p>\n\n\n\n

Branch II<\/em><\/strong><\/p>\n\n\n\n

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

= 5.291 \u03a9<\/p>\n\n\n\n

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

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

I<\/em>2<\/sub> = VY<\/em>2<\/sub> = 240 \u00d7 0.125 = 30 A<\/p>\n\n\n\n

Total conductance, G<\/em> = g<\/em>1<\/sub> + g<\/em>2<\/sub> = 0.08 + 0.09375 = 0.17375 mho<\/p>\n\n\n\n

Total susceptance, B<\/em> = \u2212 b<\/em>1<\/sub> + b<\/em>2<\/sub> = \u22120.06 + 0.08268 = 0.02268 mho (positive)<\/p>\n\n\n\n

Total admittance, \"image\"<\/p>\n\n\n\n

Total current, I<\/em> = VY<\/em> = 240 \u00d7 0.1752 = 42.05 A<\/p>\n\n\n\n

Power factor, cos <\/em>\u0278 = G<\/em>\/Y<\/em> = 0.17375\/0.1752 = 0.9917 leading<\/p>\n\n\n\n

Example 7.40<\/strong><\/p>\n\n\n\n

Calculate the impedance and admittance of the Branch AB and CD of the circuit shown in\u00a0Figure 7.60. Further, find the resistant impedance and admittance of the circuit.<\/p>\n\n\n\n

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

Fig. 7.60<\/strong>  Give circuit<\/p>\n\n\n\n

Solution:<\/em><\/p>\n\n\n\n

Branch AB<\/em><\/strong><\/p>\n\n\n\n

 <\/p>\n\n\n\n

R<\/em>1<\/sub> = 10 \u03a9; L<\/em>1<\/sub> = 0.01 H<\/p>\n\n\n\n

Inductive reactance, X<\/em>L1<\/sub> = 2 \u03c0<\/em> f L<\/em>1<\/sub> = 2\u03c0<\/em> \u00d7 50 \u00d7 0.01 = 3.1416 \u03a9<\/p>\n\n\n\n

Impedance, \"image\"<\/p>\n\n\n\n

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

Conductance, \"image\"<\/p>\n\n\n\n

Susceptance, \"image\" = \u22120.0286 mho (negative, being inductive)<\/p>\n\n\n\n

Admittance, \"image\"<\/p>\n\n\n\n

Branch CD<\/em><\/strong><\/p>\n\n\n\n

 <\/p>\n\n\n\n

R<\/em>2<\/sub> = 20 \u03a9; L<\/em>2<\/sub> = 0.05 H; X<\/em>L2<\/sub> = 2\u03c0<\/em> f<\/em> L<\/em>2<\/sub> = 2\u03c0<\/em> \u00d7 50 \u00d7 0.05 = 15.7 \u03a9<\/p>\n\n\n\n

Impedance, \"image\"<\/p>\n\n\n\n

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

Conductance, \"image\"<\/p>\n\n\n\n

Susceptance, \"image\" \u22120.0243 mho (negative, being inductive)<\/p>\n\n\n\n

<\/a>Admittance, \"image\"<\/p>\n\n\n\n

Total conductance, G<\/em> = g<\/em>1<\/sub> + g<\/em>2<\/sub> = 0.091 + 0.031 = 0.122 mho<\/p>\n\n\n\n

Total susceptance, B<\/em> = b<\/em>1<\/sub> + b<\/em>2<\/sub> = \u22120.086 \u2212 0.0243 = \u22120.0529 mho<\/p>\n\n\n\n

Total admittance, \"image\"<\/p>\n\n\n\n

Impedance of the whole circuit, \"image\"<\/p>\n","protected":false},"excerpt":{"rendered":"

Before applying this method for the solution of parallel AC circuits, the reader should be familiar with the following important terms: 7.19.1  Admittance The reciprocal of impedance of an AC circuit is called admittance of the circuit. Since impedance is the total opposition to the flow of AC in an AC circuit, the admittance is the […]<\/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-2765","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\/2765","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=2765"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2765\/revisions"}],"predecessor-version":[{"id":2766,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2765\/revisions\/2766"}],"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=2765"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2765"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2765"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}