{"id":2555,"date":"2024-08-24T09:15:10","date_gmt":"2024-08-24T09:15:10","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2555"},"modified":"2024-08-24T09:15:11","modified_gmt":"2024-08-24T09:15:11","slug":"nodal-analysis","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/24\/nodal-analysis\/","title":{"rendered":"\u00a0NODAL ANALYSIS"},"content":{"rendered":"\n

In this method, one of the nodes is taken as the reference node and the other as independent nodes. The voltages at the different independent nodes are assumed and the equations are written for each node as per KCL. After solving these equations, the node voltages are determined. Then, the branch currents are determined.<\/p>\n\n\n\n

Consider a circuit shown in\u00a0Figure 2.49, where D and B are the two independents nodes. Let D be the reference node and the voltage of node B be\u00a0V<\/em>B<\/sub>.<\/p>\n\n\n\n

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

Fig. 2.49<\/strong>  Network with Node B and D<\/p>\n\n\n\n

According to KCL,<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>1<\/sub>+ I<\/em>2<\/sub> = I<\/em>3        <\/sub>(2.40)<\/p>\n\n\n\n

In mesh ABDA, the potential difference across R<\/em>1<\/sub> is E<\/em>1<\/sub>\u2212 V<\/em>B<\/sub><\/p>\n\n\n\n

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

In mesh BCDB, the potential difference across R<\/em>2<\/sub> is E<\/em>2<\/sub>\u2212 V<\/em>B<\/sub><\/p>\n\n\n\n

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

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

Substituting these values in Equation (2.40)<\/a>, we get<\/p>\n\n\n\n

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

Rearranging the terms,<\/p>\n\n\n\n

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

<\/a>Since all other value are known, except V<\/em>B<\/sub>, calculate the value of V<\/em>B<\/sub>. Then, determine the value of I<\/em>1<\/sub>, I<\/em>2<\/sub>, and I<\/em>3<\/sub>. This method is faster as the result are obtained by solving lesser number of equations.<\/p>\n\n\n\n

Example 2.17<\/strong><\/p>\n\n\n\n

Find the current\u00a0I<\/em>1<\/sub>\u00a0and\u00a0I<\/em>2<\/sub>\u00a0in the passive elements of the network shown in\u00a0Figure 2.50.<\/p>\n\n\n\n

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

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

Fig. 2.50<\/strong>  Given network<\/p>\n\n\n\n

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

The independent nodes are A, B, and C. Let C be the reference node and\u00a0V<\/em>A<\/sub>\u00a0and\u00a0V<\/em>B<\/sub>\u00a0be the voltages at node A and B, respectively. Let us assume the direction of flow of current as in\u00a0Figure 2.51.<\/p>\n\n\n\n

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

Fig. 2.51<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

For node A, \"image\" and \"image\", assuming V<\/em>A<\/sub> > V<\/em>B<\/sub><\/p>\n\n\n\n

Similarly for node B, \"image\"   and   \"image\"<\/p>\n\n\n\n

Now, by applying KCL to node A, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>1<\/sub> = I<\/em>4<\/sub>+ I<\/em>3<\/sub><\/p>\n\n\n\n

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

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

By applying KCL <\/em>at node B, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>2 <\/sub>+ I<\/em>4<\/sub> = I<\/em>5<\/sub><\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

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

Solving\u00a0Equations (2.41)\u00a0and\u00a0(2.42), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V<\/em>A<\/sub> <\/em>= 9.25 V   and   V<\/em>B<\/sub> <\/em>= 11 V<\/p>\n\n\n\n

Current \"image\"<\/p>\n\n\n\n

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

<\/a>Example 2.18<\/strong><\/p>\n\n\n\n

Two batteries A and B are connected in parallel to a load of 10 \u03a9. Battery A has an emf of 12 V and an internal resistance of 2 \u03a9 and battery B has an emf of 10 V and internal resistance of 1 \u03a9. Using nodal analysis, determine the currents supplied by each battery and load current.<\/p>\n\n\n\n

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

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

Fig. 2.52<\/strong>  Given network<\/p>\n\n\n\n

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

Considering node Z as reference node and the potentials of nodes X and Y be\u00a0V<\/em>X<\/sub>\u00a0and\u00a0V<\/em>Y<\/sub>, respectively. The assumed current distribution is shown in\u00a0Figure 2.52.<\/p>\n\n\n\n

For node X,<\/p>\n\n\n\n

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

For node Y,<\/p>\n\n\n\n

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

and<\/p>\n\n\n\n

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

By applying KCL <\/em>to node B, we get<\/p>\n\n\n\n

I<\/em>1<\/sub>+ I<\/em>2<\/sub> = I<\/em>3 <\/sub>   or   \"image\"        (2.43)<\/p>\n\n\n\n

Moreover,<\/p>\n\n\n\n

V<\/em>X<\/sub> = V<\/em>Y        <\/sub>(2.44)<\/p>\n\n\n\n

Solving\u00a0Equations (2.43)\u00a0and\u00a0(2.44), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V<\/em>X<\/sub> = V<\/em>Y<\/sub> <\/em>= 10 V<\/p>\n\n\n\n

Thus, current supplied by battery \"image\"<\/p>\n\n\n\n

Current supplied by battery \"image\"<\/p>\n\n\n\n

Load current, \"image\"<\/p>\n\n\n\n

Example 2.19<\/strong><\/p>\n\n\n\n

Using nodal analysis, find current\u00a0I<\/em>\u00a0through 10-\u03a9 resistor in\u00a0Figure 2.53.<\/p>\n\n\n\n

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

Fig. 2.53<\/strong>  Given network<\/p>\n\n\n\n

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

The independent nodes are A, B, and C. Let C be the reference node and\u00a0V<\/em>A<\/sub>\u00a0and\u00a0V<\/em>B<\/sub>\u00a0<\/em>be the voltages at node A and B, respectively. Let us\u00a0assume the direction of flow of current is as marked in\u00a0Figure 2.54. By applying KCL at node A, we get<\/p>\n\n\n\n

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

Fig.2.54<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

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

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

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

 <\/p>\n\n\n\n

\u22125V<\/em>A<\/sub> <\/em>+ 60 \u2212 4V<\/em>A<\/sub> <\/em>= 2V<\/em>A<\/sub> <\/em>\u2212 2V<\/em>B<\/sub>   or   11V<\/em>A<\/sub> <\/em>\u22122V<\/em>B<\/sub> <\/em>= 60         (2.45)<\/p>\n\n\n\n

By applying KCL <\/em>at node B, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I <\/em>= I<\/em>4<\/sub>+ I<\/em>3<\/sub><\/p>\n\n\n\n

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

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

 <\/p>\n\n\n\n

12V<\/em>A<\/sub> <\/em>\u2212 12V<\/em>B<\/sub> <\/em>= 30V<\/em>B<\/sub> \u2212<\/em> 900 + 20V<\/em>B<\/sub><\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

12V<\/em>A<\/sub> <\/em>\u2212 62V<\/em>B<\/sub> <\/em>= \u2212900        (2.46)<\/p>\n\n\n\n

Solving\u00a0Equation (2.45)\u00a0and\u00a0(2.46), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V<\/em>A<\/sub> = 8.39 V   and   V<\/em>B<\/sub> = 16.14 V<\/p>\n\n\n\n

Current,<\/p>\n\n\n\n

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

I <\/em>= <\/strong>0.775 A (from B to A)<\/p>\n\n\n\n

Example 2.20<\/strong><\/p>\n\n\n\n

Calculate currents in all the resistors of the circuit shown in Figure 2.55<\/a> using node analysis method.<\/p>\n\n\n\n

(U.P.T.U. 2006\u201307<\/strong>)<\/p>\n\n\n\n

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

Fig. 2.55<\/strong>  Given network<\/p>\n\n\n\n

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

The independent nodes are A, B, and C. Let C\u00a0<\/em>be the reference node and\u00a0V<\/em>A<\/sub>\u00a0and\u00a0V<\/em>B<\/sub>\u00a0be the voltages at node A and B, respectively. Let us assume the direction of flow of current is marked as in\u00a0Figure 2.56.<\/p>\n\n\n\n

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

Fig. 2.56<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

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

V<\/em>A<\/sub> = 6 V<\/p>\n\n\n\n

By applying KCL <\/em>at node B, we get<\/p>\n\n\n\n

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

Current in 12-\u03a9 resistor,<\/p>\n\n\n\n

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

Current in 2-\u03a9 resistor,<\/p>\n\n\n\n

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

Current in 3-\u03a9 resistor,<\/p>\n\n\n\n

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

Example 2.21<\/strong><\/p>\n\n\n\n

Use nodal analysis to find the current in various resistors of the circuit shown in Figure 2.57<\/a>.<\/p>\n\n\n\n

(U.P.T.U. 2005\u201306<\/strong>)<\/p>\n\n\n\n

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

Fig. 2.57<\/strong>  Given network<\/p>\n\n\n\n

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

The independent nodes are A, B, C, and D. Let D\u00a0<\/em>be the reference node and\u00a0V<\/em>A<\/sub>,\u00a0V<\/em>B<\/sub>,\u00a0<\/em>and\u00a0V<\/em>C<\/sub>\u00a0<\/em>be the voltages at nodes A, B, and C, respectively, The current flowing through various branches are marked in\u00a0Figure 2.58.<\/p>\n\n\n\n

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

Fig. 2.58<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

By applying KCL at different nodes, different node voltage equations are obtained as follows:<\/p>\n\n\n\n

Node A<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>1 <\/sub>+ I<\/em>2 <\/sub>+ I<\/em>3<\/sub> = I<\/em><\/p>\n\n\n\n

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

15V<\/em>A<\/sub>+ 10(V<\/em>A<\/sub>\u2212V<\/em>B<\/sub>) + 6(V<\/em>A<\/sub>\u2212V<\/em>C<\/sub>) = 300   or   31V<\/em>A<\/sub> \u2212 10V<\/em>B<\/sub> \u2212 6V<\/em>C<\/sub> = 300        (2.47)<\/p>\n\n\n\n

Node B<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>2 <\/sub>\u2212 I<\/em>4 <\/sub>\u2212 I<\/em>5<\/sub> = 0<\/p>\n\n\n\n

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

5(V<\/em>A <\/sub>\u2212 V<\/em>B<\/sub>) \u2212 15(V<\/em>B <\/sub>\u2212 V<\/em>C<\/sub>) \u2212 3V<\/em>B<\/sub>= 0   or   5V<\/em>A<\/sub> \u2212 23V<\/em>B<\/sub> + 15V<\/em>C<\/sub> = 0        (2.48)<\/p>\n\n\n\n

Node C<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>3 <\/sub>+ I<\/em>4 <\/sub>\u2212 I<\/em>6<\/sub>\u2212 I<\/em>7<\/sub> = 0<\/p>\n\n\n\n

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

4(V<\/em>A <\/sub>\u2212 V<\/em>c<\/sub>) + 20(V<\/em>B <\/sub>\u2212 V<\/em>C<\/sub>) \u2212 5V<\/em>c<\/sub> \u2212 40 = 0   or   4V<\/em>A<\/sub> + 20V<\/em>B<\/sub> \u2212 29V<\/em>c<\/sub> = 40        (2.49)<\/p>\n\n\n\n

<\/a>The three equations in matrices form are:<\/p>\n\n\n\n

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

Current in various resistors:<\/p>\n\n\n\n

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

Example 2.22<\/strong><\/p>\n\n\n\n

Using nodal analysis, determine current in each branch of the network as shown in Figure 2.59<\/a>. Further, find total power loss in the network.<\/p>\n\n\n\n

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

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

Fig. 2.59<\/strong>  Given network<\/p>\n\n\n\n

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

Redraw the circuit and mark the arbitrary assumed values of currents in various branches as shown in\u00a0Figure 2.60. Let G (or D) be the reference node.<\/p>\n\n\n\n

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

Fig. 2.60<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

By applying KCL at different nodes, we get<\/p>\n\n\n\n

At node A<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>1<\/sub> = I<\/em>2<\/sub> + I<\/em>3<\/sub><\/p>\n\n\n\n

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

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

At node B<\/p>\n\n\n\n

I<\/em>3<\/sub> = I<\/em>4<\/sub>+ I<\/em>5<\/sub>   or   \"image\"<\/p>\n\n\n\n

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

2V<\/em>A<\/sub> \u2212 4V<\/em>B<\/sub> + V<\/em>C<\/sub> = 10         (2.51)<\/p>\n\n\n\n

 <\/p>\n\n\n\n

At node C<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>5<\/sub> + I<\/em>7<\/sub> = I<\/em>C<\/sub><\/p>\n\n\n\n

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

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

<\/a>Solving Equations (2.50)<\/a>, (2.51)<\/a>, and (2.52)<\/a>, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V<\/em>A<\/sub> = 6 V, V<\/em>B<\/sub> = 2 V, and V<\/em>C<\/sub> = 6 V<\/p>\n\n\n\n

Current through different branches:<\/p>\n\n\n\n

Current through current source, I<\/em>1<\/sub> = 1 A<\/p>\n\n\n\n

Current through branch AG (10-\u03a9 resistor), \"image\"<\/p>\n\n\n\n

Current through branch AB (10-\u03a9 resistor), \"image\"<\/p>\n\n\n\n

Current through branch BG (20-\u03a9 resistor), \"image\"<\/p>\n\n\n\n

Current through branch BC (20-\u03a9 resistor), \"image\"<\/p>\n\n\n\n

Current through branch CG (20-\u03a9 resistor), \"image\"<\/p>\n\n\n\n

Current through current source, I<\/em>7<\/sub> = 0.5 A<\/p>\n\n\n\n

Total power loss = (0.6)2<\/sup> \u00d7 10 + (0.4)2<\/sup> \u00d7 10 + (0.6)2<\/sup> \u00d7 20 + (0.2)2<\/sup> \u00d7 20 + (0.3)2<\/sup> \u00d7 20<\/p>\n\n\n\n

= 3.6 + 1.6 + 7.2 + 0.8 + 1.8 = 15 W<\/p>\n\n\n\n

Example 2.23<\/strong><\/p>\n\n\n\n

Use the node voltage method to solve the mesh currents in the network shown in Figure 2.61<\/a>.<\/p>\n\n\n\n

(U.P.T.U. June 2001<\/strong>)<\/p>\n\n\n\n

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

Fig. 2.61<\/strong>  Given network<\/p>\n\n\n\n

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

Redraw the circuit and mark the arbitrary assumed values of currents in various branches, as shown in\u00a0Figure 2.62. Let C be the reference node.<\/p>\n\n\n\n

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

Fig. 2.62<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

By applying KCL at different nodes, we get<\/p>\n\n\n\n

At node A: \"image\" considering V<\/em>A<\/sub> > V<\/em>B<\/sub><\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

8V<\/em>A<\/sub> \u2212 V<\/em>B<\/sub> = 50        (2.53)<\/p>\n\n\n\n

At node B: \"image\"<\/strong><\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

2V<\/em>A<\/sub> \u2212 17 V<\/em>B<\/sub> = \u2212500        (2.54)<\/p>\n\n\n\n

Multiplying\u00a0Equation (2.54)\u00a0by 4 and subtracting it from\u00a0Equation (2.53), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V<\/em>A<\/sub> = 10.0746 V<\/p>\n\n\n\n

Substituting the value of\u00a0V<\/em>A<\/sub>\u00a0in\u00a0Equation (2.53), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V<\/em>B<\/sub> = 30.597 V<\/p>\n\n\n\n

Various currents of the network<\/p>\n\n\n\n

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

and<\/p>\n\n\n\n

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

Example 2.24<\/strong><\/p>\n\n\n\n

By applying KCL, determine current\u00a0I<\/em>s<\/sub>\u00a0in the electric circuit at\u00a0Figure 2.63. Take\u00a0V<\/em>0<\/sub>\u00a0= 16 V<\/p>\n\n\n\n

(U.P.T.U. Feb. 2001<\/strong>)<\/p>\n\n\n\n

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

Fig. 2.63<\/strong>  Given network<\/p>\n\n\n\n

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

Let the current flowing through the various branches of the circuit be as shown in\u00a0Figure 2.64.<\/p>\n\n\n\n

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

Fig. 2.64<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

By applying KCL to node B, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>2<\/sub> + I<\/em>S<\/sub> = I<\/em>1<\/sub>        (2.55)<\/p>\n\n\n\n

By applying KCL to node C, we get<\/p>\n\n\n\n

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

Voltage at node C = V<\/em>0<\/sub> = 16 V<\/p>\n\n\n\n

 <\/p>\n\n\n\n

4I<\/em>2<\/sub> + V<\/em>1<\/sub> = 16        (2.57)<\/p>\n\n\n\n

 <\/p>\n\n\n\n

In branch BG, \"image\"   or   V1<\/sub> = 6I<\/em>1<\/sub>        (2.58)<\/p>\n\n\n\n

In branch DE, \"image\"<\/p>\n\n\n\n

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

Substituting the value of V<\/em>1<\/sub> = 6I<\/em>1<\/sub>, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

4I<\/em>2<\/sub> + 6I<\/em>1<\/sub> = 16   or   3I<\/em>1<\/sub> + 2I<\/em>2<\/sub> = 8         (2.60)<\/p>\n\n\n\n

Solving\u00a0Equations (2.59)\u00a0and\u00a0(2.60), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

6I<\/em>1<\/sub> = 12   or   I<\/em>1<\/sub> = 2 A   and   I<\/em>2<\/sub> = 1 A<\/p>\n\n\n\n

From\u00a0Equation (2.55), we get\u00a0I<\/em>S<\/sub>\u00a0=\u00a0I<\/em>1<\/sub>\u00a0\u2013\u00a0I<\/em>2<\/sub>\u00a0= 2 \u2013 1 = 1 A<\/p>\n\n\n\n

whereas V<\/em>1<\/sub> = 6I<\/em>1<\/sub> = 6 \u00d7 2 = 12 V<\/p>\n","protected":false},"excerpt":{"rendered":"

In this method, one of the nodes is taken as the reference node and the other as independent nodes. The voltages at the different independent nodes are assumed and the equations are written for each node as per KCL. After solving these equations, the node voltages are determined. Then, the branch currents are determined. Consider […]<\/p>\n","protected":false},"author":1,"featured_media":2556,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[404],"tags":[],"class_list":["post-2555","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-dc-circuit-analysis-and-network-theorems"],"jetpack_featured_media_url":"https:\/\/workhouse.sweetdishy.com\/wp-content\/uploads\/2024\/08\/download-12.jpeg","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2555","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=2555"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2555\/revisions"}],"predecessor-version":[{"id":2557,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2555\/revisions\/2557"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/2556"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=2555"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2555"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2555"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

Substituting the value of I<\/em>3<\/sub> and V<\/em>1<\/sub> in Equation (2.56)<\/a>, we get<\/p>\n\n\n\n