{"id":2553,"date":"2024-08-24T09:11:31","date_gmt":"2024-08-24T09:11:31","guid":{"rendered":"https:\/\/workhouse.sweetdishy.com\/?p=2553"},"modified":"2024-08-24T09:11:32","modified_gmt":"2024-08-24T09:11:32","slug":"wheatstone-bridge","status":"publish","type":"post","link":"https:\/\/workhouse.sweetdishy.com\/index.php\/2024\/08\/24\/wheatstone-bridge\/","title":{"rendered":"\u00a0WHEATSTONE BRIDGE"},"content":{"rendered":"\n

For the first time, Wheatstone (an English telegraph engineer) proposed this bridge for measuring the value of an unknown resistance. This bridge consists of four arms AB, BC, AD, and DC having resistances\u00a0P<\/em>,\u00a0Q<\/em>,\u00a0X<\/em>, and\u00a0R<\/em>, respectively (see\u00a0Fig. 2.22). Resistance\u00a0P\u00a0<\/em>and\u00a0Q\u00a0<\/em>are the known (fixed value) resistances and are called ratio arms. While resistance\u00a0R\u00a0<\/em>is a variable resistance of known value and\u00a0X\u00a0<\/em>is an unknown resistance whose value is to be determined.<\/p>\n\n\n\n

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

Fig. 2.22<\/strong>  Wheatstone Bridge<\/p>\n\n\n\n

To determine the value of X<\/em>, connect a battery E <\/em>across A <\/em>and C <\/em>and a galvanometer G <\/em>across B <\/em>and D <\/em>through key K<\/em>. The bridge is said to be balanced, when galvanometer G <\/em>gives zero defection on closing key K<\/em>. The balance is obtained by selecting the values of resistors P<\/em> and Q <\/em>suitably, and finally, adjusting the value of R<\/em>.<\/p>\n\n\n\n

At balance,<\/p>\n\n\n\n

Current flowing through galvanometer = 0<\/p>\n\n\n\n

Current flowing through resistor P <\/em>and Q <\/em>= I<\/em>1<\/sub><\/p>\n\n\n\n

Current flowing through resistor X <\/em>and R <\/em>= I<\/em>2<\/sub><\/p>\n\n\n\n

Voltage drop across AB = Voltage drop across AD<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>1<\/sub>P <\/em>= I<\/em>2<\/sub> X<\/em>        (2.5)<\/p>\n\n\n\n

Voltage drop across BC = Voltage drop across DC<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>1<\/sub>Q <\/em>= I<\/em>2<\/sub>R<\/em>        (2.6)<\/p>\n\n\n\n

Dividing\u00a0Equation (2.5)\u00a0by\u00a0(2.6), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

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

Product of opposite arms = Product of other opposite arms<\/p>\n\n\n\n

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

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

<\/a>Example 2.3<\/strong><\/p>\n\n\n\n

Figure 2.23\u00a0shows two batteries connected in parallel, each represented by an emf along with its internal resistance. A load resistance of 6 \u03a9 is connected across the ends of the batteries.<\/p>\n\n\n\n

Calculate the current through each battery and the load.<\/p>\n\n\n\n

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

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

Fig. 2.23<\/strong>  Given network<\/p>\n\n\n\n

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

Let the current flowing through various branches be as marked in\u00a0Figure 2.24.<\/p>\n\n\n\n

By applying KVL to mesh ABEFA, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u22122I<\/em>1<\/sub> + 4I<\/em>2<\/sub> \u2212 44 + 40 = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

2I<\/em>1<\/sub> \u2212 4I<\/em>2<\/sub> = 40 \u2212 44  or  2I<\/em>2<\/sub> \u2212 I<\/em>1<\/sub> = 2        (2.7)<\/p>\n\n\n\n

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

Fig. 2.24<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

By applying KVL to the mesh BCDEB, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u22124I<\/em>2<\/sub> \u2212 6 (I<\/em>1<\/sub> + I<\/em>2<\/sub>) + 44 = 0<\/p>\n\n\n\n

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

4I<\/em>2<\/sub> + 6 (I<\/em>1<\/sub> + I<\/em>2<\/sub>) = 44   or   5I<\/em>2<\/sub> + 3I<\/em>1<\/sub> = 22         (2.8)<\/p>\n\n\n\n

Multiplying\u00a0Equation (2.7)\u00a0by 3 and adding with\u00a0Equation (2.8), we get<\/p>\n\n\n\n

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

Substituting the value of\u00a0I<\/em>2<\/sub>\u00a0in\u00a0Equation (2.7), we obtain<\/p>\n\n\n\n

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

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

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

Example 2.4<\/strong><\/p>\n\n\n\n

Two batteries A and B are connected in parallel and a load of 10 \u03a9 is connected across their terminals. A has an emf of 12 V and an internal resistance of 2 \u03a9 and B has an emf of 8 V and an internal resistance of 1 \u03a9. Use Kirchhoff\u2019s laws to determine the magnitude of currents and also the directions in each of the batteries. Further, determine the potential difference across external resistance.<\/strong><\/p>\n\n\n\n

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

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

Fig. 2.25<\/strong>  Circuit as per given data<\/p>\n\n\n\n

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

Let the current flowing through various branches be as marked in\u00a0Figure 2.25.<\/p>\n\n\n\n

By applying KVL to mesh ABCDA, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

2I<\/em>1<\/sub> \u2212 I<\/em>2<\/sub> + 8 \u2212 12 = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

2I<\/em>1<\/sub> \u2212 I<\/em>2<\/sub> = 12 \u2212 8<\/p>\n\n\n\n

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

2I<\/em>1<\/sub> \u2212 I<\/em>2<\/sub> = 4        (2.9)<\/p>\n\n\n\n

By applying KVL to mesh ADCEA, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>2<\/sub> + 10 (I<\/em>1 <\/sub>+ I<\/em>2<\/sub>) \u2212 8 = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

10I<\/em>1<\/sub> + 11I<\/em>2<\/sub> = 8        (2.10)<\/p>\n\n\n\n

Solving\u00a0Equations (2.9)\u00a0and\u00a0(2.10), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>1<\/sub> = 1.625 A<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

I<\/em>2<\/sub> = \u22120.75 A<\/p>\n\n\n\n

The negative sign with current I<\/em>2<\/sub> shows that current I<\/em>2<\/sub> is flowing opposite to the assumed direction.<\/p>\n\n\n\n

Current flowing through the load resistance of 10 \u03a9 = I<\/em>1<\/sub> + I<\/em>2<\/sub> = 1.625 + (\u22120.75) = 0.875 A<\/p>\n\n\n\n

Potential difference across load resistance = (I<\/em>1<\/sub> + I<\/em>2<\/sub>) R <\/em>= 0.875 \u00d7 10 = 8.75 V<\/p>\n\n\n\n

Example 2.5<\/strong><\/p>\n\n\n\n

By using Kirchhoff\u2019s laws, find the current in XY in the circuit shown in\u00a0Fig<\/a>ure 2.26<\/a>.<\/p>\n\n\n\n

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

Fig. 2.26<\/strong>  Given network<\/p>\n\n\n\n

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

The assumed direction of flow of current in various sections is marked in Figure 2.27<\/a>.<\/p>\n\n\n\n

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

Fig. 2.27<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

By applying Kirchhoff\u2019s second law, we get<\/p>\n\n\n\n

Circuit XYAX<\/p>\n\n\n\n

\u22120.05I<\/em>1<\/sub> \u22120.05 (I<\/em>1<\/sub> \u2212 40) + 0.1I<\/em>2<\/sub> = 0<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u22120.05I<\/em>1<\/sub> \u2212 0.05I<\/em>1<\/sub> + 2 + 0.1I<\/em>2<\/sub> = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

I<\/em>1<\/sub> \u2013 I<\/em>2<\/sub> = 20        (2.11)<\/p>\n\n\n\n

<\/a>Circuit XYBX<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u22120.1I<\/em>2<\/sub> \u2212 0.1 (I<\/em>1<\/sub> + I<\/em>2<\/sub> \u2212 110) \u2212 0.05 (I<\/em>1<\/sub> + I<\/em>2<\/sub> \u2212 160) = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

\u22120.1I<\/em>2<\/sub> \u2212 0.1I<\/em>1<\/sub> \u2212 0.1I<\/em>2<\/sub> + 11\u2212 0.05I<\/em>1<\/sub> \u2212 0.05I<\/em>2<\/sub> + 8 = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

\u22120.15I<\/em>1<\/sub> \u2212 0.25I<\/em>2<\/sub> + 19 = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

3I<\/em>1<\/sub> + 5I<\/em>2<\/sub> = 380        (2.12)<\/p>\n\n\n\n

Multiplying\u00a0Equation (2.11)\u00a0by 3 and subtracting from\u00a0(2.12), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

8I<\/em>2<\/sub> = 320<\/p>\n\n\n\n

Current in section XY, \"image\"<\/p>\n\n\n\n

Example 2.6<\/strong><\/p>\n\n\n\n

In\u00a0Figure 2.28, the potential of point A is \u201330 V. Using Kirchhoff\u2019s laws, find (a) value of\u00a0V<\/em>\u00a0and (b) power dissipated by 5 \u03a9 resistance.<\/p>\n\n\n\n

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

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

Fig. 2.28<\/strong>  Given network<\/p>\n\n\n\n

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

The potential at point A is \u201330 V and potential at point G is zero, being grounded. Potential difference across 12-\u03a9 resistor = 30 V<\/p>\n\n\n\n

Current through 12-\u03a9 resistor \"image\"<\/p>\n\n\n\n

A simplified circuit of the network and assumed current distributed is shown in\u00a0Figure 2.29.<\/p>\n\n\n\n

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

Fig. 2.29<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

By applying KVL <\/em>to mesh CDEC, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u22123 (I<\/em>1 <\/sub>\u2212 I<\/em>2<\/sub>) + 6I<\/em>2<\/sub> = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

\u22123I<\/em>1<\/sub> + 3I<\/em>2<\/sub> + 6I<\/em>2<\/sub> = 0   or   I<\/em>1<\/sub> = 3I<\/em>2        <\/sub>(2.13)<\/p>\n\n\n\n

By applying KVL <\/em>to mesh AGEFA, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

12 \u00d7 2.5 \u2212 4 (I<\/em>1<\/sub> \u2212 2.5) = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

30 \u2212 4I<\/em>1<\/sub> + 10 = 0   or   I<\/em>1<\/sub> = 10 A<\/p>\n\n\n\n

From\u00a0Equation (2.13),<\/p>\n\n\n\n

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

<\/a>By applying KVL <\/em>to mesh ABCGA, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V <\/em>\u2013 5I<\/em>1<\/sub> \u2013 6I<\/em>2<\/sub> \u2013 2.5 \u00d7 12 = 0<\/p>\n\n\n\n

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

\"image\"   or   V <\/em>= 100 V<\/p>\n\n\n\n

Power dissipated in 5-\u03a9 resistor = \"image\" \u00d7 5 = (10)2<\/sup> \u00d7 5 = 500 W<\/p>\n\n\n\n

Example 2.7<\/strong><\/p>\n\n\n\n

In the circuit shown in\u00a0Figure 2.30, find the value of\u00a0I<\/em>S<\/sub>\u00a0for\u00a0I<\/em>\u00a0= 0.<\/p>\n\n\n\n

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

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

Fig. 2.30<\/strong>  Given network<\/p>\n\n\n\n

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

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

Current through branch BE (i.e., 2-\u03a9 resistor)<\/p>\n\n\n\n

 <\/p>\n\n\n\n

= I <\/em>+ I<\/em>S<\/sub> <\/em>= 0 + I<\/em>S<\/sub> <\/em>= I<\/em>S<\/sub> <\/em>(I <\/em>= 0)<\/p>\n\n\n\n

By applying KVL <\/em>to mesh ABEFA, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u20133I<\/em>1<\/sub> \u2013 2I<\/em>S<\/sub> <\/em>+ 4 = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

3I <\/em>+ 2I<\/em>S<\/sub> <\/em>= 4<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

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

Example 2.8<\/strong><\/p>\n\n\n\n

Find the value of\u00a0R<\/em>\u00a0and the current through it in the circuit shown in\u00a0Figure 2.31, when the current is zero in branch OA.<\/p>\n\n\n\n

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

Fig. 2.31<\/strong>  Given network<\/p>\n\n\n\n

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

Since current flowing through branch OA is zero, the same current\u00a0I<\/em>1<\/sub>\u00a0flows through branch BA and AC. The assumed direction of flow of current in other branches is also marked in\u00a0Figure 2.32, according to Kirchhoff\u2019s first law.<\/p>\n\n\n\n

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

Fig. 2.32<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

<\/a>By applying Kirchhoff\u2019s second law to the following meshes, we get<\/p>\n\n\n\n

Mesh BAOB<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u2013I<\/em>1<\/sub> \u00b1 0 + 4I<\/em>2<\/sub> = 0<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I<\/em>1<\/sub> = 4I<\/em>2        <\/sub>(2.14)<\/p>\n\n\n\n

Mesh BACB<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u2013I<\/em>1<\/sub> \u2013 1.511<\/sub> \u2013 2 (I<\/em>1<\/sub> + I<\/em>2<\/sub>) + 10 = 0<\/p>\n\n\n\n

 <\/p>\n\n\n\n

4.5I<\/em>1<\/sub> + 2I<\/em>2<\/sub> = 10<\/p>\n\n\n\n

Substituting the value of\u00a0I<\/em>1<\/sub>\u00a0from\u00a0Equation (2.14), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

4.5 (4I<\/em>2<\/sub>) + 2I<\/em>2<\/sub> = 10<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

I<\/em>2<\/sub> = 0.5 A<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

I<\/em>1<\/sub> = 4 \u00d7 0.5 = 2 A<\/p>\n\n\n\n

Mesh BOCB<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u20134I<\/em>2<\/sub> \u2013 RI<\/em>2<\/sub> \u2013 2 (I<\/em>1<\/sub> + I<\/em>2<\/sub>) + 10 = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

\u20134 \u00d7 0.5 \u2013 R <\/em>\u00d7 0.5 \u2013 2 (2 + 0.5) + 10 = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

R <\/em>= 6 \u03a9<\/p>\n\n\n\n

Example 2.9<\/strong><\/p>\n\n\n\n

Determine the current in the 4-\u03a9 resistance of circuit shown in\u00a0Figure 2.33.<\/p>\n\n\n\n

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

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

Fig. 2.33<\/strong>  Given network<\/p>\n\n\n\n

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

A simplified circuit is redrawn as shown in\u00a0Figure 2.34. Let the current flowing through various branches be as marked in\u00a0Figure 2.34.<\/p>\n\n\n\n

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

Fig. 2.34<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

By applying KVL to various meshes, that is, BCDHB, DEFHD, and BHFGAB, we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u20132 (I<\/em>1 <\/sub>\u2013 I<\/em>2<\/sub>) \u2013 10I<\/em>3<\/sub> + 1 \u00d7 (I<\/em>2<\/sub> \u2013 6) = 0;        (2.15)<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u20132 (I<\/em>1<\/sub> \u2013 I<\/em>2<\/sub> + 6 \u2013 I<\/em>3<\/sub>) + 3 (I<\/em>2<\/sub> \u2013 6 + I<\/em>3<\/sub>) + 10I<\/em>3<\/sub> = +10        (2.16)<\/p>\n\n\n\n

 <\/p>\n\n\n\n

and \u22121 \u00d7 (I<\/em>2<\/sub> \u2212 6) \u2212 3 (I<\/em>2<\/sub> \u2212 6 + I<\/em>3<\/sub>) \u2212 4I<\/em>1<\/sub> = \u221224        (2.17)<\/p>\n\n\n\n

 <\/p>\n\n\n\n

From\u00a0Equation (2.15), 2I<\/em>1<\/sub>\u00a0\u2212 3I<\/em>2<\/sub>\u00a0+ 10I<\/em>3<\/sub>\u00a0= \u22126\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2.18)<\/p>\n\n\n\n

 <\/p>\n\n\n\n

From\u00a0Equation (2.16), 2I<\/em>1<\/sub>\u00a0\u2212 5I<\/em>2<\/sub>\u00a0\u2212 15I<\/em>3<\/sub>\u00a0= \u221240\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2.19)<\/p>\n\n\n\n

 <\/p>\n\n\n\n

From\u00a0Equation (2.17), 4I<\/em>1<\/sub>\u00a0+ 4I<\/em>2<\/sub>\u00a0+ 3I<\/em>3<\/sub>\u00a0= 48\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2.20)<\/p>\n\n\n\n

Solving equations by matrices:<\/strong><\/p>\n\n\n\n

The abovementioned equations can be given in the matrix form as<\/p>\n\n\n\n

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

The common determinant is given as<\/p>\n\n\n\n

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

= 2 [(\u22125) \u00d7 3 \u2212 4 \u00d7 (\u221215)] \u2212 2 [(\u22123) \u00d7 3 \u2212 4 \u00d7 10] + 4 [(\u22123) \u00d7 (\u221215) \u2212 (\u22125) \u00d7 10]<\/p>\n\n\n\n

= 2 [\u221215 + 60] \u2212 2 [\u22129 \u2212 40] + [45 + 50]<\/p>\n\n\n\n

= 90 + 98 + 380 = 568<\/p>\n\n\n\n

The determinant for I<\/em>1<\/sub> is given as<\/p>\n\n\n\n

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

= \u2212 6 [(\u22125) \u00d7 3 \u2212 4 \u00d7 (\u221215)] + 40 [(\u22123) \u00d7 3 \u2212 4 \u00d7 10] + 48 [(\u22123) \u00d7 (\u221215) \u2212 (\u22125) \u00d7 10]<\/p>\n\n\n\n

= \u2212 6 [\u221215 + 60] + 40 [\u22129 \u2212 40] + 48 [45 + 50]<\/p>\n\n\n\n

= \u2212 270 \u2212 1,960 + 4,560 = 2,330<\/p>\n\n\n\n

As per Cramer\u2019s rule, \"image\"<\/p>\n\n\n\n

Example 2.10<\/strong><\/p>\n\n\n\n

What is the difference of potential between X and Y in the network shown in Figure 2.35<\/a>.<\/p>\n\n\n\n

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

Fig. 2.35<\/strong>  Given network<\/p>\n\n\n\n

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

Refer to\u00a0Figure 2.36. Let the current in branch AX be\u00a0I<\/em>1<\/sub>\u00a0and in branch BY be\u00a0I<\/em>2<\/sub>.<\/p>\n\n\n\n

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

Fig. 2.36<\/strong>  Current in different loops<\/p>\n\n\n\n

Current, \"image\"<\/p>\n\n\n\n

Current, \"image\"<\/p>\n\n\n\n

Rise in potential from X to Y is given as<\/p>\n\n\n\n

 <\/p>\n\n\n\n

V<\/em>XY<\/sub> = 3 \u00d7 0.4 + 5 \u2212 3 \u00d7 0.5 = 4.7 V<\/p>\n\n\n\n

Example 2.11<\/strong><\/p>\n\n\n\n

Find the total power delivered to the circuit by two sources in\u00a0Figure 2.37.<\/p>\n\n\n\n

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

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

Fig. 2.37<\/strong>  Given network<\/p>\n\n\n\n

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

Let the current flowing through different branches be as marked in\u00a0Figure 2.38.<\/p>\n\n\n\n

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

Fig. 2.38<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

By applying KVL to mesh ABCFEA, we get<\/p>\n\n\n\n

\u2212 2I<\/em>1<\/sub> \u2212 (I<\/em>1<\/sub> + 6) \u00d7 3 \u2212 5 = 0   or   \"image\"<\/p>\n\n\n\n

<\/a>By applying KVL to mesh ADCFEA, we get<\/p>\n\n\n\n

\u2212I<\/em>2<\/sub> \u2212 4 (I<\/em>2<\/sub> \u2212 6) \u2212 5 = 0   or   \"image\"<\/p>\n\n\n\n

Total power delivered by the two sources<\/p>\n\n\n\n

 <\/p>\n\n\n\n

= Total power absorbed by various resistors<\/p>\n\n\n\n

= I<\/em>1<\/sub>2<\/sup> \u00d7 2 + (I<\/em>1<\/sub> + 6)2<\/sup> \u00d7 3 + (I<\/em>2<\/sub> \u2212 6)2<\/sup> \u00d7 4 + I<\/em>2<\/sub>2<\/sup> \u00d7 1<\/p>\n\n\n\n

= (\u22122.6)2<\/sup> \u00d7 2 + (\u22122.6 + 6)2<\/sup> \u00d7 3 + (5.8 \u2212 6)2<\/sup> \u00d7 4 + (5.8)2 <\/sup>\u00d7 1<\/p>\n\n\n\n

= 82 W<\/p>\n\n\n\n

Example 2.12<\/strong><\/p>\n\n\n\n

In the circuit shown in\u00a0Figure 2.39, determine the value of\u00a0E<\/em>2<\/sub>\u00a0that will reduce the galvanometer current to zero. The galvanometer resistance is 10 \u03a9.<\/p>\n\n\n\n

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

Fig. 2.39<\/strong>  Given network<\/p>\n\n\n\n

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

By applying KCL, the direction of flow of current in various branches is marked in\u00a0Figure 2.40. Since galvanometer does not carry any current, the current in branch BC is the same as that in branch AB, that is,\u00a0I<\/em>1<\/sub>. Similarly, current flowing through branch DC is the same as that in branch AD, that is,\u00a0I<\/em>2<\/sub>.<\/p>\n\n\n\n

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

Fig. 2.40<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

By applying KVL <\/em>to various meshes, we get:<\/p>\n\n\n\n

Mesh ABCDA<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u221211I<\/em>1<\/sub> + 9I<\/em>2<\/sub>+E<\/em>2<\/sub> = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

11I<\/em>1<\/sub> \u2212 9I<\/em>2<\/sub> = E<\/em>2        <\/sub>(2.21)<\/p>\n\n\n\n

Mesh ABCE<\/em>1<\/sub>A<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u221211I<\/em>1<\/sub> \u2212 2 (I<\/em>1<\/sub>+I<\/em>2<\/sub>) + 2 = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

13I<\/em>1<\/sub> + 2I<\/em>2<\/sub> = 2        (2.22)<\/p>\n\n\n\n

Mesh BCDB<\/p>\n\n\n\n

\u22125I<\/em>1<\/sub> + 4I<\/em>2<\/sub> = 0<\/p>\n\n\n\n

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

I<\/em>1<\/sub> = 0.8I<\/em>2        <\/sub>(2.23)<\/p>\n\n\n\n

 <\/p>\n\n\n\n

13 (0.8I<\/em>2<\/sub>) + 2I<\/em>2<\/sub> = 2<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

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

Substituting the value of\u00a0I<\/em>1<\/sub>\u00a0and\u00a0I<\/em>2<\/sub>\u00a0in\u00a0Equation (2.21), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

11 \u00d7 (4\/31) \u2212 9 (5\/31) = E<\/em>2<\/sub><\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

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

Note: <\/strong>The negative sign shows that the cell E<\/em>2<\/sub> should be connected in reverse direction.<\/p>\n\n\n\n

Example 2.13<\/strong><\/p>\n\n\n\n

Determine the current\u00a0I<\/em>\u00a0in 4-\u03a9 resistance in the circuit shown in\u00a0Figure 2.41.<\/p>\n\n\n\n

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

Fig. 2.41<\/strong>  Given network<\/p>\n\n\n\n

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

A simplified circuit is shown in Figure 2.42<\/a>. By applying KCL, different currents are marked in various sections.<\/p>\n\n\n\n

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

Fig. 2.42<\/strong>  Assumed direction of flow of current in various branches<\/p>\n\n\n\n

By applying KVL to various loops, we get<\/p>\n\n\n\n

Loop ABHGA:<\/p>\n\n\n\n

 <\/p>\n\n\n\n

10I<\/em>2<\/sub> + (I <\/em>\u2212 I<\/em>1<\/sub> \u2212 6) \u2212 2I<\/em>1<\/sub> = 0   or   I <\/em>\u2212 3I<\/em>1<\/sub> + 10I<\/em>2<\/sub> = 6        (2.24)<\/p>\n\n\n\n

Loop BCDHB:<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u22122 (I<\/em>1<\/sub>+ I<\/em>2<\/sub> + 6) \u2212 10 + 3 (I <\/em>\u2212 I<\/em>1<\/sub>\u2212 I<\/em>2<\/sub> \u2212 6) \u2212 10I<\/em>2<\/sub> = 0<\/p>\n\n\n\n

 <\/p>\n\n\n\n

3I <\/em>\u2212 5I<\/em>1<\/sub> \u2212 15I<\/em>2<\/sub> = 40        (2.25)<\/p>\n\n\n\n

Loop GHDEFG:<\/p>\n\n\n\n

 <\/p>\n\n\n\n

\u2212(I <\/em>\u2212 I<\/em>1<\/sub> \u2212 6) \u2212 3 (I <\/em>\u2212 I<\/em>1<\/sub>\u2212 I<\/em>2<\/sub> \u2212 6) \u2212 4I <\/em>+ 24 = 0<\/p>\n\n\n\n

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

 <\/p>\n\n\n\n

8I <\/em>\u2212 4I<\/em>1<\/sub> \u2212 3I<\/em>2<\/sub> = 48        (2.26)<\/p>\n\n\n\n

Eliminating\u00a0I<\/em>1<\/sub>\u00a0from\u00a0Equations (2.24)\u00a0and\u00a0(2.25), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

4I <\/em>\u2212 95I<\/em>2<\/sub> = 90        (2.27)<\/p>\n\n\n\n

Eliminating\u00a0I<\/em>1<\/sub>\u00a0from\u00a0Equations (2.25)\u00a0and\u00a0(2.26), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

28I <\/em>+ 45I<\/em>2<\/sub> = 80        (2.28)<\/p>\n\n\n\n

Eliminating\u00a0I<\/em>2<\/sub>\u00a0from\u00a0Equations (2.27)\u00a0and\u00a0(2.28), we get<\/p>\n\n\n\n

 <\/p>\n\n\n\n

568I <\/em>= 2330<\/p>\n\n\n\n

Therefore, current in 4-\u03a9 resistor, I <\/em>= 4.102 A<\/p>\n\n\n\n

Alternatively, the three equation are as follows:<\/p>\n\n\n\n

 <\/p>\n\n\n\n

I <\/em>\u2212 3I<\/em>1<\/sub> + 10I<\/em>2<\/sub> = 6<\/p>\n\n\n\n

3I <\/em>\u2212 5I<\/em>1<\/sub> \u2212 15I<\/em>2<\/sub> = 40<\/p>\n\n\n\n

8I <\/em>\u2212 4I<\/em>1<\/sub> \u2212 3I<\/em>2<\/sub> = 48<\/p>\n\n\n\n

<\/a>The three equations can be solved by the method of determinants, <\/em>that is, by applying Cramer\u2019s rule. The matrix from of the abovementioned equation is<\/p>\n\n\n\n

\"image\"\/<\/figure>\n\n\n\n
2.7  MAXWELL\u2019S MESH CURRENT METHOD (LOOP ANALYSIS)<\/h5>\n\n\n\n

In this method, mesh or loop currents are taken instead of branch currents (as in Kirchhoff\u2019s laws). The following steps are taken while solving a network by this method:<\/p>\n\n\n\n

    \n
  1. The whole network is divided into number of meshes. Each mesh is assigned a current having continuous path (current is not split at a junction). These mesh currents are preferably drawn in clockwise direction. The common branch carries the algebraic sum of the mesh currents flowing through it.<\/li>\n\n\n\n
  2. Write KVL equation for each mesh using the same signs as applied to Kirchhoff\u2019s laws.<\/li>\n\n\n\n
  3. Number of equations must be equal to the number of unknown quantities. Solve the equations and determine the mesh currents.<\/li>\n<\/ol>\n\n\n\n

    Example 2.14<\/strong><\/p>\n\n\n\n

    Using loop current method, find the current I<\/em>1<\/sub> and I<\/em>2<\/sub> as shown in Figure 2.43<\/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. 2.43<\/strong>  Given network<\/p>\n\n\n\n

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

    Let the current flowing through the two loops be\u00a0I<\/em>1\u00a0<\/sub>and\u00a0I<\/em>2<\/sub>, as shown in\u00a0Figure 2.44.<\/p>\n\n\n\n

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

    Fig. 2.44<\/strong>  Loop currents in various sections<\/p>\n\n\n\n

    By applying KVL to different loops, we get<\/p>\n\n\n\n

    Loop ABEFA<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    \u22122I<\/em>1<\/sub> \u2212 6 (I<\/em>1<\/sub>\u2212 I<\/em>2<\/sub>) \u2212 6 + 10 = 0<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    8I<\/em>1<\/sub> \u2212 6I<\/em>2<\/sub> = 4<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    4I<\/em>1<\/sub> \u2212 3I<\/em>2<\/sub> = 2        (2.29)<\/p>\n\n\n\n

    Loop BCDEB<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    \u22123I<\/em>2<\/sub> \u2212 2 + 6 \u2212 6 (I<\/em>2<\/sub>\u2212 I<\/em>1<\/sub>) = 0<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    \u22126I<\/em>1<\/sub> + 9I<\/em>2<\/sub> = 4        (2.30)<\/p>\n\n\n\n

    Multiplying\u00a0Equation (2.29)\u00a0by 3 and\u00a0Equation (2.30)\u00a0by 2, we get<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    12I<\/em>1<\/sub> \u2212 9I<\/em>2<\/sub> = 6        (2.31)<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    \u221212I<\/em>1<\/sub> + 18I<\/em>2<\/sub> = 8        (2.32)<\/p>\n\n\n\n

    Adding\u00a0Equations (2.31)\u00a0and\u00a0(2.32), we get<\/p>\n\n\n\n

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

    Example 2.15<\/strong><\/p>\n\n\n\n

    Using mesh equation method, find current in the resistance\u00a0R<\/em>1<\/sub>\u00a0of the network shown in\u00a0Figure 2.45.<\/p>\n\n\n\n

    (U.P.T.U. 2004\u201305<\/strong>)<\/p>\n\n\n\n

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

    Fig.2.45<\/strong>  Given network<\/p>\n\n\n\n

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

    Converting current source of 1 A and internal resistance 5 \u03a9 into voltage source,<\/p>\n\n\n\n

    emf of voltage source, V <\/em>= I <\/em>\u00d7 R <\/em>= 1 \u00d7 5 = 5 V<\/p>\n\n\n\n

    Internal resistance of voltage source, R <\/em>= R <\/em>= 5 \u03a9<\/p>\n\n\n\n

    The circuit is shown in Figure 2.46<\/a>.<\/p>\n\n\n\n

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

    Fig. 2.46<\/strong>  Loop currents in various sections<\/p>\n\n\n\n

    In mesh ABEFA, I<\/em>1<\/sub> (5 + 5) + 10 (I<\/em>1<\/sub>+ I<\/em>2<\/sub>) = 5   or   20I<\/em>1<\/sub> + 10I<\/em>2<\/sub> = 5<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    4I<\/em>1<\/sub> + 2I<\/em>2<\/sub> = 1        (2.33)<\/p>\n\n\n\n

    In mesh BCDEB, 5I<\/em>2<\/sub> + 10 (I<\/em>1<\/sub>+ I<\/em>2<\/sub>) = 10   or   10I<\/em>1<\/sub> + 15I<\/em>2<\/sub> = 10<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    2I<\/em>1<\/sub> + 3I<\/em>2<\/sub> = 2        (2.34)<\/p>\n\n\n\n

    Solving Equations (2.33)<\/a> and (2.34)<\/a>, we get<\/p>\n\n\n\n

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

    Current through \"image\"<\/p>\n\n\n\n

    = 0.625 A (from B to E)<\/p>\n\n\n\n

    <\/a>Example 2.16<\/strong><\/p>\n\n\n\n

    Using mesh current method, determine current\u00a0I<\/em>x<\/sub>\u00a0in the circuit shown in\u00a0Figure 2.47.<\/p>\n\n\n\n

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

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

    Fig. 2.47<\/strong>  Given network<\/p>\n\n\n\n

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

    Let the circuit be as shown in\u00a0Figure 2.48. Suppose voltage across 2 A current source is\u00a0V<\/em>x<\/sub>,<\/p>\n\n\n\n

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

    Fig. 2.48<\/strong>  Loop currents in various sections<\/p>\n\n\n\n

    By applying KVL <\/em>in mesh 1; 3I<\/em>1<\/sub> + (I<\/em>1<\/sub>\u2212 I<\/em>2<\/sub>) = 2<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    4I<\/em>1<\/sub>\u2212 I<\/em>2<\/sub> = 2        (2.35)<\/p>\n\n\n\n

    By applying KVL <\/em>in mesh 2; (I<\/em>2<\/sub>\u2212 I<\/em>1<\/sub>) + V<\/em>x<\/sub> <\/em>= 0<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    I<\/em>1<\/sub> \u2212 I<\/em>2<\/sub> = V<\/em>x        <\/sub>(2.36)<\/p>\n\n\n\n

    By applying KVL <\/em>in mesh 3; 2I<\/em>3<\/sub> = 5 + V<\/em>x<\/sub>        (2.37)<\/p>\n\n\n\n

    Further,<\/p>\n\n\n\n

     <\/p>\n\n\n\n

    I<\/em>3<\/sub>\u2212 I<\/em>2<\/sub> = 2        (2.38)<\/p>\n\n\n\n

    From\u00a0Equations (2.36)\u00a0and\u00a0(2.37)\u00a02I<\/em>3<\/sub>\u00a0= 5 + (I<\/em>1<\/sub>\u2212\u00a0I<\/em>2<\/sub>)<\/p>\n\n\n\n

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

     <\/p>\n\n\n\n

    \u2212 I<\/em>1<\/sub>+ I<\/em>2<\/sub> + 2I<\/em>3<\/sub> = 5        (2.39)<\/p>\n\n\n\n

    From\u00a0Equations (2.35),\u00a0(2.38), and\u00a0(2.39),<\/p>\n\n\n\n

    In matrix form<\/p>\n\n\n\n

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

    Current, \"image\"<\/p>\n\n\n\n

    <\/h5>\n","protected":false},"excerpt":{"rendered":"

    For the first time, Wheatstone (an English telegraph engineer) proposed this bridge for measuring the value of an unknown resistance. This bridge consists of four arms AB, BC, AD, and DC having resistances\u00a0P,\u00a0Q,\u00a0X, and\u00a0R, respectively (see\u00a0Fig. 2.22). Resistance\u00a0P\u00a0and\u00a0Q\u00a0are the known (fixed value) resistances and are called ratio arms. While resistance\u00a0R\u00a0is a variable resistance of known […]<\/p>\n","protected":false},"author":1,"featured_media":2547,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[404],"tags":[],"class_list":["post-2553","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-11.jpeg","_links":{"self":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2553","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=2553"}],"version-history":[{"count":1,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2553\/revisions"}],"predecessor-version":[{"id":2554,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/posts\/2553\/revisions\/2554"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media\/2547"}],"wp:attachment":[{"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/media?parent=2553"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/categories?post=2553"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/workhouse.sweetdishy.com\/index.php\/wp-json\/wp\/v2\/tags?post=2553"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

Substituting the value of I<\/em>1<\/sub> in Equation (2.23)<\/a>, we get<\/p>\n\n\n\n