{"id":2859,"date":"2024-08-25T19:43:37","date_gmt":"2024-08-25T19:43:37","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2859"},"modified":"2024-08-25T19:43:38","modified_gmt":"2024-08-25T19:43:38","slug":"equivalent-reactance","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/25\/equivalent-reactance\/","title":{"rendered":"\u00a0\u00a0EQUIVALENT REACTANCE"},"content":{"rendered":"\n

To make the calculations easy, the reactance of the two windings can be transferred to any one side. The reactance from one side to the other is transferred in such a manner that percentage voltage drop remains the same when represented on either side.<\/p>\n\n\n\n

Let the primary reactance\u00a0X<\/em>1\u00a0<\/sub>be transferred to the secondary, and the new value of this reactance is\u00a0\"image\"\u00a0called equivalent reactance of primary referred to secondary, as shown in\u00a0Figure 10.22(a).<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

Fig. 10.22<\/strong>  (a) and (b) Equivalent reactance when referred to secondary side (c) and (d) Equivalent reactance when referred to primary side<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

or<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

\u2234Total equivalent reactance referred to secondary.<\/p>\n\n\n\n

 <\/p>\n\n\n\n

X<\/em>es<\/sub> = X<\/em>2<\/sub> +  = X<\/em>2<\/sub> + K<\/em>2<\/sup> X<\/em>1<\/sub><\/p>\n\n\n\n

Now, let us consider secondary reactance\u00a0X<\/em>2<\/sub>\u00a0when it is transferred to primary side its new value is\u00a0\"image\"\u00a0called equivalent reactance of secondary referred to primary, as shown in\u00a0Figure 10.22(c).<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

\u2234 Total equivalent reactance referred to primary.<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

<\/a>Example 10.14<\/strong><\/p>\n\n\n\n

A 63 kVA, 1100\/220 V single-phase transformer has R<\/em>1 <\/sub>= 0.16 ohm, X<\/em>1 <\/sub>= 0.5 ohm, R<\/em>2 <\/sub>= 0.0064 ohm and X<\/em>2<\/sub> = 0.02 ohm. Find equivalent resistance and reactance as referred to primary winding.<\/p>\n\n\n\n

(P.T.U. May 2009)<\/strong><\/p>\n\n\n\n

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

Here, transformer rating = 63 kVA; V<\/em>1 <\/sub>= 1100 V; V<\/em>2 <\/sub>= 220 V<\/p>\n\n\n\n

 <\/p>\n\n\n\n

R<\/em>1 <\/sub>= 0.16 ohm; X<\/em>1<\/sub> = 0.5 ohm; R<\/em>2 <\/sub>= 0.0064 ohm; X<\/em>2 <\/sub>= 0.02 ohm<\/p>\n\n\n\n

Transformation ratio,<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

Equivalent resistance referred to secondary side,<\/p>\n\n\n\n

R<\/em>es<\/sub> = R<\/em>2<\/sub> + \"img\" = R<\/em>2<\/sub> + R<\/em>1<\/sub> \u00d7 K<\/em>2<\/sup> = 0.0064 + 0.16 \u00d7 (0.2)2<\/sup> = 0.0128 ohm<\/p>\n\n\n\n

Equivalent reactance referred to secondary side,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

X<\/em>es<\/sub> = X<\/em>2<\/sub> +\"image\" = X<\/em>2<\/sub> + X<\/em>1<\/sub> \u00d7 K<\/em>2<\/sup> = 0.02 + 0.5 \u00d7 (0.2)2<\/sup> = 0.04 ohm<\/p>\n\n\n\n

Example 10.15<\/strong><\/p>\n\n\n\n

A 33 kVA, 2200\/220 V, 50 Hz single-phase transformer has the following parameters. Primary winding resistance r<\/em>1<\/sub> = 2.4 \u03a9, Leakage reactance x<\/em>1<\/sub> = 6 \u03a9 Secondary winding resistance r<\/em>2 <\/sub>= 0.03 \u03a9 Leakage reactance x<\/em>2 <\/sub>= 0.07 \u03a9. Then, find primary, secondary, and equivalent resistance and reactance.<\/em><\/p>\n\n\n\n

(P.T.U. Dec. 2009)<\/strong><\/p>\n\n\n\n

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

Here, rating of transformer = 33 kVA; V<\/em>1 <\/sub>= 2200 V; V<\/em>2 <\/sub>= 220 V<\/p>\n\n\n\n

 <\/p>\n\n\n\n

f<\/em> = 50 Hz; R<\/em>1<\/sub> = 2.4\u03a9; X<\/em>1<\/sub> = 6\u03a9; R<\/em>2<\/sub> = 0.03\u03a9; X<\/em>2<\/sub> = 0.07\u03a9<\/p>\n\n\n\n

Transformation ratio,<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

Transformer resistance referred to primary side<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

Transformer reactance referred to primary side<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

Transformer resistance referred to secondary side<\/p>\n\n\n\n

R<\/em>es<\/sub> = R<\/em>2<\/sub> + \"image\" = R<\/em>2<\/sub> + R<\/em>1<\/sub> \u00d7 K<\/em>2<\/sup> = 0.03 + 2.4 \u00d7 (0.1)2<\/sup> = 0.054\u03a9<\/p>\n\n\n\n

Transformer reactance referred to secondary side<\/p>\n\n\n\n

X<\/em>es<\/sub> = X<\/em>2<\/sub> + \"image\" = X<\/em>2<\/sub> + X<\/em>1<\/sub> \u00d7 K<\/em>2<\/sup> = 0.07 + 6 \u00d7 (0.1)2<\/sup> = 0.054\u03a9<\/p>\n\n\n\n

<\/a>Example 10.16<\/strong><\/p>\n\n\n\n

A single-phase transformer having voltage ratio 2500\/250 V (primary to secondary) has a primary resistance and reactance 1.8 ohm and 4.2 ohm, respectively. The corresponding secondary values are 0.02 and 0.045 ohm. Determine the total resistance and reactance referred to secondary side. Also, calculate the impedance of transformer referred to secondary side.<\/em><\/p>\n\n\n\n

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

Here,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

R<\/em>1<\/sub> = 1.8 \u03a9; X<\/em>1<\/sub> = 4.2 \u03a9; R<\/em>2<\/sub> = 0.02 \u03a9; X<\/em>2<\/sub> = 0.045 \u03a9<\/p>\n\n\n\n

Transformation ratio,<\/p>\n\n\n\n

\"img\"\/<\/figure>\n\n\n\n

Total resistance referred to secondary side,<\/p>\n\n\n\n

R<\/em>es<\/sub> = R<\/em>2<\/sub> +  = R<\/em>2<\/sub> + R<\/em>1<\/sub> \u00d7 K<\/em>2<\/sup> = 0.02 + 1.8 \u00d7 (0.1)2<\/sup> = 0.038\u03a9<\/p>\n\n\n\n

Total reactance referred to secondary side,<\/p>\n\n\n\n

X<\/em>es<\/sub> = X<\/em>2<\/sub> +  = X2<\/sub> + X1<\/sub>\u00d7 K<\/em>2<\/sup> = 0.045 + 4.2 \u00d7 (0.1)2<\/sup> = 0.087\u03a9<\/p>\n\n\n\n

Impedance of transformer referred to secondary side,<\/p>\n\n\n\n

\"img\"\/<\/figure>\n","protected":false},"excerpt":{"rendered":"

To make the calculations easy, the reactance of the two windings can be transferred to any one side. The reactance from one side to the other is transferred in such a manner that percentage voltage drop remains the same when represented on either side. Let the primary reactance\u00a0X1\u00a0be transferred to the secondary, and the new […]<\/p>\n","protected":false},"author":1,"featured_media":2841,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[413],"tags":[],"class_list":["post-2859","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-single-phase-transformers"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/08\/power-transformer.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2859","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=2859"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2859\/revisions"}],"predecessor-version":[{"id":2860,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2859\/revisions\/2860"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/2841"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=2859"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2859"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2859"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}