{"id":2855,"date":"2024-08-25T19:41:30","date_gmt":"2024-08-25T19:41:30","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2855"},"modified":"2024-08-25T19:41:30","modified_gmt":"2024-08-25T19:41:30","slug":"transformer-on-load","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/25\/transformer-on-load\/","title":{"rendered":"\u00a0\u00a0TRANSFORMER ON LOAD"},"content":{"rendered":"\n

When a certain load is connected across the secondary, a current\u00a0I<\/em>2\u00a0<\/sub>flows through it as shown in\u00a0Figure 10.16. The magnitude of current\u00a0I<\/em>2\u00a0<\/sub>depends upon terminal voltage\u00a0V<\/em>2\u00a0<\/sub>and impedance of the load. The phase angle of secondary current\u00a0I<\/em>2\u00a0<\/sub>with respect to\u00a0V<\/em>2\u00a0<\/sub>depends upon the nature of load, that is, whether the load is resistive, inductive, or capacitive.<\/p>\n\n\n\n

(Neglecting winding resistance and leakage flux)<\/p>\n\n\n\n

When a certain load is connected across the secondary, a current\u00a0I<\/em>2\u00a0<\/sub>flows through it as shown in\u00a0Figure 10.16. The magnitude of current\u00a0I<\/em>2\u00a0<\/sub>depends upon terminal voltage\u00a0V<\/em>2\u00a0<\/sub>and impedance of\u00a0the load. The phase angle of secondary current\u00a0I<\/em>2\u00a0<\/sub>with respect to\u00a0V<\/em>2\u00a0<\/sub>depends upon the nature of load, that is, whether the load is resistive, inductive, or capacitive.<\/p>\n\n\n\n

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

Fig. 10.16<\/strong>  Transformer on load<\/p>\n\n\n\n

The operation of the transformer on load is explained below with the help of number of diagrams:<\/p>\n\n\n\n

    \n
  1. When the transformer is on no-load as shown in\u00a0Figure 10.17(a), it draws no-load current\u00a0I<\/em>0\u00a0<\/sub>from the supply mains. The no-load current\u00a0I<\/em>0\u00a0<\/sub>produces an mmf.\u00a0N<\/em>1<\/sub>I<\/em>0\u00a0<\/sub>which sets up flux in the core.<\/li>\n\n\n\n
  2. When the transformer is loaded, current\u00a0I<\/em>2\u00a0<\/sub>flows in the secondary winding. This secondary current\u00a0I<\/em>2<\/sub>\u00a0produces an mmf\u00a0N<\/em>2<\/sub>I<\/em>2\u00a0<\/sub>which sets up flux\u00a0\u0278<\/em>2<\/sub>\u00a0in the core. This flux opposes the flux which is set up by the current\u00a0I<\/em>0\u00a0<\/sub>as shown in\u00a0Figure 10.17(b), according to Lenz\u2019s law.<\/li>\n\n\n\n
  3. Since\u00a0\u0278<\/em>2<\/sub>\u00a0opposes the flux, and therefore, the resultant flux tends to decrease and causes the reduction of self-induced emf\u00a0E<\/em>1\u00a0<\/sub>momentarily. Thus,\u00a0V<\/em>1\u00a0<\/sub>predominates over\u00a0E<\/em>1\u00a0<\/sub>causing additional primary current\u00a0\"img\"\u00a0drawn from the supply mains. The amount of this additional current is such that the original conditions, that is, flux in the core must be restored to original value\u00a0\u0278<\/em>, so that\u00a0V<\/em>1\u00a0<\/sub>=\u00a0E<\/em>1<\/sub>. The current\u00a0I<\/em>1\u00a0<\/sub>is in phase opposition with\u00a0I<\/em>2<\/sub>\u00a0and is called primary counterbalancing current. This additional current\u00a0\"img\"\u00a0produces an mmf\u00a0N<\/em>1<\/sub>\"img\"\u00a0which sets up flux,\u00a0\u0278<\/em>, in the same direction as that of\u00a0\u0278<\/em>\u00a0as shown in\u00a0Figure 10.17(c)\u00a0and cancels the flux\u00a0\u0278<\/em>2<\/sub>\u00a0set up by mmf\u00a0N<\/em>2<\/sub>I<\/em>2<\/sub>.Now,\u00a0N<\/em>1<\/sub>\"img\"\u00a0=\u00a0N<\/em>2<\/sub>I<\/em>2<\/sub>\u00a0(ampere-turns balance)\u2234\"img\"<\/li>\n\n\n\n
  4. Thus, the flux is restored to its original value as shown in\u00a0Figure 10.17(d). The total primary current\u00a0I<\/em>1<\/sub>\u00a0is the vector sum of current\u00a0I<\/em>0\u00a0<\/sub>and\u00a0I<\/em>\u2032, that is,\u00a0I<\/em>1<\/sub>\u00a0=\u00a0I<\/em>0<\/sub>+\u00a0\"img\".<\/li>\n<\/ol>\n\n\n\n

    This shows that flux in the core of a transformer remains the same from no-load to full load; this is the reason why iron losses in a transformer remain the same from no-load to full load<\/p>\n\n\n\n

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

    Fig. 10.17<\/strong>  (a), (b), (c) and (d) : Effect on the magnetic flux set-up in the core when transformer is loaded<\/p>\n\n\n\n

    <\/a>10.10  PHASOR DIAGRAM OF A LOADED TRANSFORMER<\/h5>\n\n\n\n

    (Neglecting voltage drops in the winding; ampere-turns balance)<\/p>\n\n\n\n

    Since the voltage drops in both the windings of the transformer are neglected,<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    V<\/em>1<\/sub> = E<\/em>1<\/sub> and E<\/em>2<\/sub> = V<\/em>2<\/sub><\/p>\n\n\n\n

    While drawing the phasor diagram, the following important points are to be considered.<\/p>\n\n\n\n

      \n
    1. For simplicity, let the transformation ratio K <\/em>= l be considered, and therefore, E<\/em>2<\/sub> <\/sup>= E<\/em>1<\/sub>.<\/li>\n\n\n\n
    2. The secondary current I<\/em>2<\/sub> <\/sup>is in phase, lags behind, and leads the secondary terminal voltage V<\/em>2 <\/sub>by an angle \u0278<\/em>2<\/sub> for resistive, inductive, and capacitive load, respectively.<\/li>\n\n\n\n
    3. The counterbalancing current \"img\"(i.e., \"img\" = KI<\/em>2<\/sub> here K<\/em> = 1 \u2234 \"img\" = I<\/em>2<\/sub>) and is 180\u00b0 out of phase with I<\/em>2<\/sub>.<\/li>\n\n\n\n
    4. The total primary current I<\/em>1<\/sub> is the vector sum of no-load primary current I<\/em>0 <\/sub>and counter balancing current \"img\"That is,\"img\"Where \u03b8<\/em> is the phase angle between I<\/em>0<\/sub> and \"img\"<\/li>\n\n\n\n
    5. The p.f. on the primary side is cos \u0278<\/em>1<\/sub> which is less than the load p.f. cos \u0278<\/em>2<\/sub> on the secondary side. Its value is determined by the relation. \"img\"<\/li>\n<\/ol>\n\n\n\n
      \"img\"\/<\/figure>\n\n\n\n

      Fig. 10.18<\/strong>  Phasor diagram (a) for resistive load (b) for inductive load (c) for capacitive load<\/p>\n\n\n\n

      The phasor diagrams of the transformer for resistance, inductive, and capacitive loads are shown in\u00a0Figure 10.18(a),\u00a0(b)<\/a>, and\u00a0(c)<\/a>, respectively.<\/p>\n\n\n\n

      <\/a>Alternately,<\/em><\/p>\n\n\n\n

      The primary current I<\/em>1<\/sub> <\/sup>can also be determined by resolving the vectors, that is,<\/p>\n\n\n\n

       <\/p>\n\n\n\n

      I<\/em>v<\/sub> = I<\/em>0<\/sub> cos \u0278<\/em>0<\/sub> + \"img\" cos\u0278<\/em>2<\/sub> [where sin \u0278<\/em>0<\/sub> = sincos\u22121<\/sup>(cos \u0278<\/em>0<\/sub>)<\/p>\n\n\n\n

      I<\/em>H<\/sub> = I<\/em>0<\/sub> sin \u0278<\/em>0<\/sub> + \"img\" sin \u0278<\/em>2 <\/sub>and sin \u0278<\/em>2<\/sub> = sincos\u22121<\/sup>(cos \u0278<\/em>2<\/sub>)]<\/p>\n\n\n\n

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

      Example 10.12<\/strong><\/p>\n\n\n\n

      A single-phase transformer with a ratio of 440\/110 V takes a no-load current of 5 A at 0.2 p.f. lagging. If the secondary supplies a current of 120 A at p.f. of 0.8 lagging, calculate the primary current and p.f.<\/p>\n\n\n\n

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

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

      Let the primary counterbalancing current be \"img\"<\/p>\n\n\n\n

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

      \"img\" = KI<\/em>2<\/sub> = 0.25 \u00d7 120 =30A<\/p>\n\n\n\n

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

       <\/p>\n\n\n\n

      cos \u0278<\/em>0<\/sub> = 0.2; \u0278<\/em>0<\/sub> = cos\u22121<\/sup>0.2 = 78.46\u00b0<\/p>\n\n\n\n

      cos \u0278<\/em>2<\/sub> = 0.8; \u0278<\/em>2<\/sub> = cos\u22121<\/sup>0.8 = 36.87\u00b0<\/p>\n\n\n\n

      \u03b8<\/em> = \u0278<\/em>0<\/sub> \u2212 \u0278<\/em>2<\/sub> = 78.46\u00b0 \u2212 36.87\u00b0 = 41.59\u00b0<\/p>\n\n\n\n

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

      Primary p.f.,<\/p>\n\n\n\n

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

      = 0.7375 lag<\/p>\n\n\n\n

      Example 10.13<\/strong><\/p>\n\n\n\n

      A single-phase transformer with a ratio of 6600\/400 V (primary to secondary voltage) takes to no-load current of 0.7 A at 0.24 power factor lagging. If the secondary winding supplies a current of 120 A at a power factor of 0.8 lagging. Estimate the current drawn by the primary winding.<\/em><\/p>\n\n\n\n

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

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

       <\/p>\n\n\n\n

      I<\/em>0<\/sub> = 0.7 A; cos \u0278<\/em>0<\/sub> = 0.24 lag; I<\/em>2<\/sub> = 120 A; cos \u0278<\/em>2<\/sub> = 0.8 lag<\/p>\n\n\n\n

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

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

      <\/a>Let the primary counterbalance current be I<\/em>1<\/sub>.<\/p>\n\n\n\n

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

      N<\/em>1<\/sub>\"img\" = N<\/em>2<\/sub>I<\/em>2<\/sub><\/p>\n\n\n\n

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

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

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

       <\/p>\n\n\n\n

      cos \u0278<\/em>0<\/sub> = 0.24; \u0278<\/em>0<\/sub> = cos\u22121<\/sup>0.24 = 76.11\u00b0<\/p>\n\n\n\n

      cos \u0278<\/em>2<\/sub> = 0.8; \u0278<\/em>2<\/sub> = cos\u22121<\/sup>0.8 = 36.87\u00b0<\/p>\n\n\n\n

      Angle between vector\u00a0I<\/em>0\u00a0<\/sub>and\u00a0I<\/em>1\u00a0<\/sub>(Figure 10.14(b))<\/p>\n\n\n\n

       <\/p>\n\n\n\n

      \u03b8<\/em> = 76.11\u00b0 \u2212 36.87\u00b0 = 39.24\u00b0<\/p>\n\n\n\n

      Current drawn by the primary,<\/p>\n\n\n\n

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

      When a certain load is connected across the secondary, a current\u00a0I2\u00a0flows through it as shown in\u00a0Figure 10.16. The magnitude of current\u00a0I2\u00a0depends upon terminal voltage\u00a0V2\u00a0and impedance of the load. The phase angle of secondary current\u00a0I2\u00a0with respect to\u00a0V2\u00a0depends upon the nature of load, that is, whether the load is resistive, inductive, or capacitive. (Neglecting winding resistance and […]<\/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-2855","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\/2855","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=2855"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2855\/revisions"}],"predecessor-version":[{"id":2856,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2855\/revisions\/2856"}],"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=2855"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2855"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2855"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}