{"id":2681,"date":"2024-08-24T20:14:31","date_gmt":"2024-08-24T20:14:31","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2681"},"modified":"2024-08-24T20:17:26","modified_gmt":"2024-08-24T20:17:26","slug":"addition-and-subtraction-of-alternating-quantities","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/24\/addition-and-subtraction-of-alternating-quantities\/","title":{"rendered":"\u00a0\u00a0ADDITION AND SUBTRACTION OF ALTERNATING QUANTITIES"},"content":{"rendered":"\n

In AC circuits, it is required to add or subtract the alternating quantities. In such cases, it should be proceed as follows:<\/p>\n\n\n\n

6.19.1 Addition of Alternating Quantities<\/h4>\n\n\n\n

The given alternating quantities are represented as phasor, and then, they are added in the same manner as forces are added. Only phasors of the similar quantities are added, that is, either all the currents are added or all the voltages are added. Voltages and currents are never added with each other. For addition, the following methods are accomplished:<\/p>\n\n\n\n

    \n
  1. Parallelogram method<\/li>\n\n\n\n
  2. Method of components<\/li>\n<\/ol>\n\n\n\n
      \n
    1. <\/a>Parallelogram Method:<\/strong> This method is applied for the addition of two phasors at a time. The two quantities are represented in magnitude and direction (phasors) as the adjacent sides of a parallelogram. The diagonal of this parallelogram represents the magnitude of the resultant.Consider an AC parallel circuit having two branches, as shown in Figure 6.25, carrying a current of i<\/em>1<\/sub> and i<\/em>2<\/sub>, respectively. Let the two currents be represented as follows: i<\/em>1<\/sub>= I<\/em>m1<\/sub> sin \u03c9<\/em>t<\/em> and i<\/em>2<\/sub> = I<\/em>m2<\/sub> sin (\u03c9<\/em>t<\/em> + \u03b8<\/em>)\"img\"Fig. 6.25  <\/strong>AC current in parallel circuitsThe maximum values of the two currents I<\/em>m1<\/sub> and I<\/em>m2<\/sub> are represented as the two adjacent sides OA and OB, respectively, of a parallelogram OACB as shown in Figure 6.26. The current I<\/em>m2<\/sub> leads the current I<\/em>m1<\/sub> by an angle \u03b8<\/em> \u00b0. The maximum value of the resultant is say I<\/em>mr<\/sub> represented by the diagonal OC of the parallelogram and leads the phasor I<\/em>m1<\/sub> by an angle \u0278<\/em>.\"img\"Fig. 6.26  <\/strong>Phasor representation (for finding resultant by parallelogram method)\"img\"Phase angle, \"img\"The instantaneous value of the resultant current is given by the relation; i<\/em>r<\/sub><\/small> = I<\/em>mr<\/sub><\/small> sin(\u03c9<\/em> \u00d7 t<\/em> + \u0278<\/em>) as \u0278<\/em> is positive. <\/li>\n\n\n\n
    2. Method of components:<\/strong> In this method, each phasor is resolved into horizontal and vertical components. The horizontal components are added algebraically to obtain the resultant horizontal component I<\/em>XX<\/sub>. Similarly, vertical components are summed up algebraically to obtain the resultant vertical component I<\/em>YY<\/sub>.Consider an AC parallel circuit consisting of three branches each carrying a current of i<\/em>1<\/sub>, i<\/em>2<\/sub>, and i<\/em>3<\/sub>, respectively, as shown in Figure 6.27. Let the three currents be represented as follows: i<\/em>1 <\/sub>= I<\/em>m1<\/sub> sin (\u03c9t<\/em> + \u03b8<\/em>1<\/sub>) i<\/em>2<\/sub> = I<\/em>m2<\/sub> sin \u03c9t<\/em> and i<\/em>3<\/sub> = I<\/em>m3<\/sub> sin (\u03c9t<\/em> \u2212 \u03b8<\/em>2<\/sub>)\"img\"Fig. 6.27  <\/strong>AC current in parallel circuit<\/a>The maximum values of the three currents I<\/em>m1<\/sub>, I<\/em>m2<\/sub>,<\/em> and I<\/em>m3<\/sub> are represented by the phasors as shown in Figure 6.28(a). The components are resolved horizontally and vertically.Algebraic sum of horizontal components I<\/em>XX<\/sub> = I<\/em>m1<\/sub>cos\u03b8<\/em>1<\/sub> + I<\/em>m2<\/sub> + I<\/em>m3<\/sub>cos\u03b8<\/em>2<\/sub> Algebraic sum of vertical components I<\/em>YY<\/sub> = I<\/em>m1<\/sub> sin \u03b8<\/em>1<\/sub> + 0 \u2212 I<\/em>m3<\/sub> sin \u03b8<\/em>2<\/sub> Maximum value of resultant components\"img\"If \u0278<\/em> is the phase difference (leading) between resultant current and horizontal axis as shown in Figure 6.28(b). Then,\"img\"The instantaneous value of the resultant current is given by the relation; i<\/em>r<\/sub> = I<\/em>mr<\/sub> sin (\u03c9t<\/em> + \u0278<\/em>) However, if I<\/em>YY<\/sub> comes out to be negative, the angle of phase difference will be lagging (i.e., \u2212\u0278<\/em>). Then, the instantaneous value of the resultant current will be given by the relation i<\/em>r<\/sub> = I<\/em>mr<\/sub> sin (\u03c9t<\/em> \u2212 \u0278<\/em>)\"img\"Fig. 6.28 (a)  <\/strong>Phasor representation of three current (b) Finding resultant by component method<\/li>\n<\/ol>\n\n\n\n

      6.19.2  Subtraction of Alternating Quantities<\/h4>\n\n\n\n

      The methods explained above (i.e., parallelogram method and method of components) are also applied for the subtraction of an alternating quantity. The only difference is that in this case, the phasor of the alternating quantity which is to be subtracted is reversed or represented 180\u00b0 out of phase. Then it is added with the other alternating quantity (or quantities) as usual.<\/p>\n\n\n\n

      <\/a>Example 6.15<\/strong><\/p>\n\n\n\n

      Calculate (i) the maximum value and (ii) the root-mean-square value of the following quantities:<\/p>\n\n\n\n

      (i) 40 sin \u03c9t<\/em> (ii) B sin (\u03c9t<\/em> \u2212 \u03c0<\/em>\/2) (iii) 10 sin \u03c9t<\/em> \u2212 17.3 cos\u03c9t<\/em><\/p>\n\n\n\n

      Draw the vectors showing the phase difference with respect to A sin (\u03c9t<\/em> \u2212 \u03c0<\/em> \/6).<\/p>\n\n\n\n

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

      The instantaneous value of an alternating quantity is given by the relation<\/p>\n\n\n\n

       <\/p>\n\n\n\n

      i<\/em> = I<\/em>m<\/sub> sin \u03c9t<\/em><\/p>\n\n\n\n

        \n
      1. The given alternating quantity is 40 sin \u03c9t<\/em>\u2234 Maximum value = 40 (Ans.)RMS value = Max. value\/\"img\" = 40\/\"img\" = 28.284 (Ans.)The vector lies on the horizontal axis as shown in Figure 6.29(a).\"img\"Fig. 6.29  <\/strong>(a) Phasor diagram as per data<\/li>\n\n\n\n
      2. The given alternating quantity is B sin (\u03c9t<\/em> \u2212 \u03c0<\/em>\/2)\u2234 Maximum value = B (Ans.)RMS value = B\/\"img\"(Ans.)The vector lags behind the horizontal axis by 90\u00b0 as shown in Figure 6.29(b).\"img\"Fig. 6.29  <\/strong>(b) Resultant phasor diagram<\/li>\n\n\n\n
      3. The given alternating quantity is 10 sin \u03c9t<\/em> \u2212 17.3 cos\u03c9t<\/em>. In fact, this quantity has two components displaced from each other by 90\u00b0 as shown in Figure 6.29(a).Resultant maximum value \"img\"<\/li>\n<\/ol>\n\n\n\n

        Let the phase angle of the resultant with the horizontal be \u03b8<\/em> \u00b0.<\/p>\n\n\n\n

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

         <\/p>\n\n\n\n

        tan\u03b8<\/em> = 17.3\/10 = 1.73<\/p>\n\n\n\n

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

         <\/p>\n\n\n\n

        \u03b8<\/em> = tan\u2212<\/sup>1<\/sup> 1.73 = 60\u00b0<\/p>\n\n\n\n

        Hence, this vector lags behind the horizontal axis by 60\u00b0 as shown in Figure 6.29(b). This vector is also represented in Figure 6.29(a).<\/p>\n\n\n\n

        The quantity A sin (\u03c9t<\/em> \u2212 \u03c0<\/em>\/6) makes as angle of lag, \u03b8<\/em>1<\/sub> = \u03c0<\/em>\/6 = 30\u00b0 with the horizontal as shown in Figure 6.29(a).<\/p>\n\n\n\n

        Considering phasor diagram shown in Figure 6.29(a);<\/p>\n\n\n\n

        The phase difference between first quantity (i.e., 40) and A = 30\u00b0 (Ans.)<\/p>\n\n\n\n

        The phase difference between second quantity (i.e., B) and A = 60\u00b0 (Ans.)<\/p>\n\n\n\n

        The phase difference between third quantity (i.e., 20) and A = 60\u00b0 \u2212 30\u00b0 = 30\u00b0 (Ans.)<\/p>\n\n\n\n

        Example 6.16<\/strong><\/p>\n\n\n\n

        Two AC voltages are represented as follows:<\/p>\n\n\n\n

         <\/p>\n\n\n\n

        \u03bd<\/em>1 <\/sub>(t<\/em>) = 30 sin (314 t<\/em> + 45\u00b0);<\/p>\n\n\n\n

        \u03bd<\/em>2<\/sub> (t<\/em>) = 60 sin (314 t<\/em> + 60\u00b0)<\/p>\n\n\n\n

        <\/a>Calculate the resultant \u03c5<\/em>(t<\/em>) and express in the form:<\/p>\n\n\n\n

         <\/p>\n\n\n\n

        \u03c5<\/em>(t<\/em>) = V<\/em>m<\/sub> sin (314 t<\/em> \u00b1 \u0278<\/em>)<\/p>\n\n\n\n

        (U.P.T.U. 2003<\/strong>–2004)<\/strong><\/p>\n\n\n\n

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

         <\/p>\n\n\n\n

        \u03b8<\/em> (t<\/em>) = m<\/em> \u00d7 t<\/em> + 0 = 1 \u00d7 t<\/em><\/p>\n\n\n\n

        The phasor representation of the two voltages is shown in Figure 6.30.<\/p>\n\n\n\n

         <\/p>\n\n\n\n

        \u03bd<\/em>1 <\/sub>(t<\/em>) = 30 sin (314 t<\/em> + 45\u00b0); \u03bd<\/em>2 <\/sub>(t<\/em>) = 60 sin (314 t<\/em> + 60\u00b0)<\/p>\n\n\n\n

        Resolving the vectors V<\/em>1m<\/sub> and V<\/em>2m<\/sub> horizontally and vertically, we get,<\/p>\n\n\n\n

         <\/p>\n\n\n\n

        \u03a3 V<\/em>xx<\/sub> = 30 cos 45\u00b0 + 60 cos 60\u00b0<\/p>\n\n\n\n

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

        \u03a3 V<\/em>yy<\/sub> = 30 sin 45\u00b0 + 60 sin 60\u00b0<\/p>\n\n\n\n

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

        Maximum value of resultant voltage,<\/p>\n\n\n\n

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

        Phase angle, \"img\"<\/p>\n\n\n\n

        Resultant voltage, \u03c5<\/em>(t<\/em>) = V<\/em>r(m)<\/sub> sin (\u03c9t<\/em> + \u0278<\/em>r<\/sub>) = 89.31 sin (314 t<\/em> + 55\u00b0) (Ans.)<\/p>\n\n\n\n

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

        Fig. 6.30  <\/strong>Phasor diagram as per data<\/p>\n\n\n\n

        Example 6.17<\/strong><\/p>\n\n\n\n

        Draw a phasor diagram showing the following voltages.<\/p>\n\n\n\n

         <\/p>\n\n\n\n

        \u03bd<\/em>1<\/sub> = 100 sin 500 t<\/em>;<\/p>\n\n\n\n

        <\/sub>\u03bd<\/em>2<\/sub> = 200 sin (500 t<\/em> + \u03c0<\/em> \/3)<\/p>\n\n\n\n

        \u03bd<\/em>3<\/sub> = \u2212 50 cos 500 t<\/em>;<\/p>\n\n\n\n

        \u03bd<\/em>4<\/sub> = 150 sin (500 t<\/em> \u2212 \u03c0<\/em> \/4)<\/p>\n\n\n\n

        Find RMS value of resultant voltage.<\/p>\n\n\n\n

        (U.P.T.U. 2005\u20132006)<\/strong><\/p>\n\n\n\n

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

         <\/p>\n\n\n\n

        \u03bd<\/em>1<\/sub> = 100 sin 500 t;<\/em><\/p>\n\n\n\n

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

        = 50 sin (500 t<\/em> \u2212 \u03c0<\/em>\/2)<\/p>\n\n\n\n

        \u03bd<\/em>4<\/sub> = 150 sin (150 t<\/em> \u2212 \u03c0<\/em>\/4)<\/p>\n\n\n\n

        All the four voltages are shown vectorially in Figure 6.31.<\/p>\n\n\n\n

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

        Fig. 6.31  <\/strong>Phasor diagram as per data<\/p>\n\n\n\n

        <\/a>Resolving the phasors in horizontal axis<\/p>\n\n\n\n

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

        = 100 \u00d7 1 + 200 \u00d7 0.5 + 50 \u00d7 0 + 150 \u00d7 0.707<\/p>\n\n\n\n

        = 306.05 V<\/p>\n\n\n\n

        Resolving the phasors in vertical axis<\/p>\n\n\n\n

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

        = 100 \u00d7 0 + 200 \u00d7 0.866 \u2212 50 \u00d7 1 \u2212 150 \u00d7 0.707<\/p>\n\n\n\n

        = 17.15 V<\/p>\n\n\n\n

        Maximum value of resultant voltage<\/p>\n\n\n\n

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

        RMS value of resultant voltage, \"img\"<\/p>\n\n\n\n

        Example 6.18<\/strong><\/p>\n\n\n\n

        Three sinusoidal voltages acting in series are given by<\/p>\n\n\n\n

         <\/p>\n\n\n\n

        \u03bd<\/em>1<\/sub><\/small> = 10sin 440t<\/em>; \u03bd<\/em>2<\/sub><\/small> = 10\"img\" sin (440t<\/em> \u2013 45\u00b0); \u03bd<\/em>3<\/sub><\/small> = 20 cos 440t<\/em><\/p>\n\n\n\n

          \n
        1. an expression for the resultant voltage,<\/li>\n\n\n\n
        2. the frequency and rms value of the resultant voltage.(U.P.T.U., February 2001)<\/strong><\/li>\n<\/ol>\n\n\n\n

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

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

           <\/p>\n\n\n\n

          \u03c51<\/sub> = 10 sin 440 t<\/em><\/p>\n\n\n\n

          \u03c52 <\/sub>= 10\"img\" sin (440t<\/em> \u2013 45\u00b0);<\/p>\n\n\n\n

          \u03c53<\/sub><\/small> = 20 cos 440t<\/em> \"img\"<\/p>\n\n\n\n

          All the three voltages are shown vectorially in Figure 6.32.<\/p>\n\n\n\n

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

          Fig. 6.32  <\/strong>Phasor diagram as per data<\/p>\n\n\n\n

          Resolving voltage along X-axis and Y-axis, we get,<\/p>\n\n\n\n

          V<\/em>xx<\/sub><\/small> = 10cos 0\u00b0 + 10\"img\" cos(\u201345\u00b0) + 20 cos 90\u00b0<\/p>\n\n\n\n

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

          V<\/em>yy<\/sub><\/small> = 10sin 0\u00b0 + 10\"img\" cos(\u201345\u00b0) + 20 sin 90\u00b0<\/p>\n\n\n\n

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

          Maximum value of resultant voltage,<\/p>\n\n\n\n

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

          <\/a>Phase angle, \"img\"<\/p>\n\n\n\n

            \n
          1. Expression for resultant voltage, \u03c5r<\/sub> = 22.36 sin (440 t<\/em> + 0.1476 \u03c0<\/em>) (Ans.)<\/li>\n\n\n\n
          2. Frequency, \"img\"<\/li>\n<\/ol>\n\n\n\n

            RMS value of resultant voltage, \"img\"<\/p>\n\n\n\n

            Example 6.19<\/strong><\/p>\n\n\n\n

            Three voltages represented by e<\/em>1<\/sub> = 10 sin wt<\/em>, e<\/em>2<\/sub> = 15 sin (\u03c9t<\/em> + \u03c0<\/em>\/4), and e<\/em>3<\/sub> = 20 cos (\u03c9t<\/em> \u2212 \u03c0<\/em>\/6) act together in a circuit. Find an expression for the resulting voltage.<\/p>\n\n\n\n

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

            The three voltages are represented vectorially in Figure 6.33. The third voltage (20 V) makes an angle of lag of 30\u00b0 with the vertical.<\/p>\n\n\n\n

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

            Fig. 6.33  <\/strong>Phasor diagram as per data<\/p>\n\n\n\n

            Resolving the components along X-axis<\/p>\n\n\n\n

             <\/p>\n\n\n\n

            E<\/em>XX<\/sub> = 10 + 15 cos 45\u00b0 + 20 cos 60\u00b0<\/p>\n\n\n\n

            = 10 + 15 \u00d7 0.707 + 20 \u00d7 0.5 = 30.607 V<\/p>\n\n\n\n

            Resolving the components along Y-axis,<\/p>\n\n\n\n

             <\/p>\n\n\n\n

            E<\/em>YY<\/sub> = 0 + 15 sin 45\u00b0 + 20 sin 60\u00b0<\/p>\n\n\n\n

            = 0 + 15 \u00d7 0.707 + 20 \u00d7 0.866 = 27.927 V<\/p>\n\n\n\n

            Maximum value of resultant voltage,<\/p>\n\n\n\n

             <\/p>\n\n\n\n

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

            Let \u0278<\/em> be the angle which the resultant voltage makes with X-axis<\/p>\n\n\n\n

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

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

            \u0278<\/em> = tan\u2212<\/sup>1<\/sup> 10.9124 = 42.38\u00b0<\/p>\n\n\n\n

            \u2234    Expression for the resultant voltage,<\/p>\n\n\n\n

             <\/p>\n\n\n\n

            e<\/em>r<\/sub> = E<\/em>mr<\/sub> sin (\u03c9t<\/em> + \u0278<\/em>) = 41.433 sin (\u03c9t<\/em> + 42.38\u00b0) (Ans.)<\/p>\n\n\n\n

            Example 6.20<\/strong><\/p>\n\n\n\n

            Two currents i<\/em>1<\/sub> and i<\/em>2 <\/sub>are given by the expressions: (i) 40 sin (314 t<\/em> + \u03c0<\/em>\/6) (ii) 20 sin (314 t<\/em> \u2212 \u03c0<\/em>\/3)<\/p>\n\n\n\n

            Find i1<\/sub> \u2212 i2<\/sub> and express the answer in the same form as individual currents. Find also the rms value and frequency of the resultant current.<\/p>\n\n\n\n

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

            The two currents are represented by phasors as shown in Figure 6.34. Since current i<\/em>2<\/sub> is to be subtracted from i<\/em>1<\/sub>, reverse the phasor I<\/em>m2<\/sub> (dotted phasor) and find its vector sum with I<\/em>m1<\/sub>.<\/p>\n\n\n\n

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

            Fig. 6.34  <\/strong>Phasor diagram as per data<\/p>\n\n\n\n

            <\/a>Resolving the components along X-axis;<\/p>\n\n\n\n

             <\/p>\n\n\n\n

            I<\/em>XX<\/sub> = 40 cos 30\u00b0 \u2212 20 cos 60\u00b0<\/p>\n\n\n\n

            = 40 \u00d7 0.866 \u2212 20 \u00d7 0.5 = 24.64 A<\/p>\n\n\n\n

            Resolving the components along Y-axis;<\/p>\n\n\n\n

             <\/p>\n\n\n\n

            I<\/em>YY<\/sub> = 40 sin 30\u00b0 + 20 sin 60\u00b0<\/p>\n\n\n\n

            = 40 \u00d7 0.5 + 20 \u00d7 0.866 = 37.32 A<\/p>\n\n\n\n

            Maximum value of resultant current,<\/p>\n\n\n\n

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

            = 44.72 A<\/p>\n\n\n\n

            Let \u0278<\/em> be the angle which resultant current makes with X-axis<\/p>\n\n\n\n

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

            \u2234    Expression for the resultant current,<\/p>\n\n\n\n

             <\/p>\n\n\n\n

            i<\/em>r<\/sub> = I<\/em>mr<\/sub> sin (\u03c9t<\/em> + \u0278<\/em>) = 44.72 sin (314 t<\/em> + 56.56\u00b0) (Ans.)<\/p>\n\n\n\n

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

             <\/p>\n\n\n\n

            i<\/em>r<\/sub> = 44.72 sin (314 \u00d7 t<\/em> + 0.9872) [where \u0278<\/em> is in radians]<\/p>\n\n\n\n

            RMS value of resultant current,<\/p>\n\n\n\n

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

            From the equation, \u03c9<\/em> = 314 or 2 \u03c0<\/em>f<\/em> = 314<\/p>\n\n\n\n

            \u2234    Frequency, f<\/em> = 314\/2 \u03c0<\/em> = 50 Hz (Ans.)<\/p>\n\n\n\n

            Example 6.21<\/strong><\/p>\n\n\n\n

            The instantaneous values of two alternating voltages are represented by \u03c51<\/sub> = 60 sin \u03b8<\/em> and \u03c52<\/sub> = 40 sin (\u03b8<\/em> \u2212 \u03c0<\/em>\/3). Derive expression for the instantaneous values of (i) the sum and (ii) the difference of these voltages.<\/p>\n\n\n\n

            (U.P.T.U. July, 2002)<\/strong><\/p>\n\n\n\n

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

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

            \u03c51<\/sub> = 60 sin \u03b8<\/em> and \"img\"<\/p>\n\n\n\n

            The two voltages are shown vectorially in Figure 6.35(a)<\/p>\n\n\n\n

              \n
            1. When the two voltages are to be added, V<\/em>xx<\/sub> = V<\/em>1 <\/sub>cos 0\u00b0 + V<\/em>2 <\/sub>cos (\u221260\u00b0) = 60 \u00d7 1 + 40 \u00d7 0.5 = 80 VV<\/em>yy<\/sub> = V<\/em>1<\/sub> sin 0\u00b0 + 40 sin (\u221260\u00b0) = 60 \u00d7 0 + 40 \u00d7 (\u22120.866) = \u221234.64 V\"img\"<\/a>Phase angle, \"img\"Expression for the instantaneous value of the resultant voltage, V<\/em>r<\/sub> = V<\/em>r(m)<\/sub> sin (\u03b8<\/em> \u2212 \u0278<\/em>r<\/sub>) = 87.2 sin (\u03b8<\/em> \u2212 0.13 \u03c0<\/em>) volt (Ans.) <\/li>\n\n\n\n
            2. When one of the voltage (say V<\/em>2<\/sub>) is to be subtracted from the other (say V<\/em>1<\/sub>)The vector representing V<\/em>2<\/sub> is reversed as shown in Figure 6.35(b) and then added to V<\/em>1<\/sub>. V<\/em>xx<\/sub> = V<\/em>1<\/sub>cos \u03b8<\/em> \u00b0 \u2212 V<\/em>2<\/sub> sin (\u221260\u00b0) = 60 \u00d7 1 \u221240 \u00d7 0.5 = 40 VV<\/em>yy<\/sub> = V<\/em>1<\/sub> sin \u03b8<\/em> \u00b0\u2212 V<\/em>2<\/sub> sin (\u221260\u00b0) = 60 \u00d7 0 \u221240 \u00d7 (\u22120.866) = 34.64 V\"img\"Phase angle, \"img\"Expression for the instantaneous value of resultant voltage, V<\/em>r<\/sub> = V<\/em>r(<\/sub>m<\/sub><\/em>)<\/sub> sin (\u03b8<\/em> + \u0278<\/em>r<\/sub>) = 52.9 sin (\u03b8<\/em> + 0.227 \u03c0<\/em>) (Ans.)<\/li>\n<\/ol>\n\n\n\n
              \"img\"\/<\/figure>\n\n\n\n

              Fig. 6.35  <\/strong>(a) Phasor diagram for addition (b) Phasor diagram for subtraction<\/p>\n","protected":false},"excerpt":{"rendered":"

              In AC circuits, it is required to add or subtract the alternating quantities. In such cases, it should be proceed as follows: 6.19.1 Addition of Alternating Quantities The given alternating quantities are represented as phasor, and then, they are added in the same manner as forces are added. Only phasors of the similar quantities are […]<\/p>\n","protected":false},"author":1,"featured_media":2474,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[408],"tags":[],"class_list":["post-2681","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-ac-fundamentals"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/08\/air-conditioner.png","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2681","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=2681"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2681\/revisions"}],"predecessor-version":[{"id":2682,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2681\/revisions\/2682"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/2474"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=2681"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2681"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2681"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}