Determine (1) the mesh current I2 and (2) the voltage at node X in the circuit of Fig. 7.14.

Question

Determine (1) the mesh current [latex]I_{2}[/latex] and (2) the voltage at node X in the circuit of Fig. 7.14.

Accepted Answer

There are three meshes in total but since there is a current source common to meshes 1 and 2, these can be considered to be a supermesh. Applying the threestep method to the supermesh:

1 we multiply the mesh current [latex]I_{1}[/latex] by the resistance through which it flows [latex](R_{1})[/latex] and add this to the product of the mesh current [latex]I_{2}[/latex] and the resistance through which it flows [latex](R_{2})[/latex];

2 there is one adjacent mesh whose current is [latex]I_{3}[/latex]. We multiply this by the resistance [latex](R_{2})[/latex]common to the two meshes and subtract the product;

3 there is a voltage source [latex](V_{s1})[/latex] which acts in the opposite direction to the mesh current flowing through it so the right-hand side of the supermesh equation is [latex]- V_{s1}[/latex]

The supermesh equation is therefore

[latex]R_{1}I_{1} + R_{2}I_{2} - R_{2}I_{3} = -V_{s1}[/latex]

Now [latex]I_{1}- I_{2} = I_{s} ⇒ I_{1} = I_{s} + I_{2}[/latex], so

[latex]R_{1}(I_{s} + I_{2}) + R_{2}I_{2}- R_{2}I_{3} = - V_{s1}[/latex]

Putting in the values,

[latex]6(5 + I_{2}) + 3I_{2}- 3I_{3} = - 12[/latex]

[latex]9I_{2}- 3I_{3} = - 42[/latex]

Dividing through by 3 we get

[latex]3I_{2}- I_{3} = - 14[/latex] (7.32)

For mesh 3:

1 we multiply the mesh current by the resistance through which it flows [latex](R_{2} + R_{3})[/latex];

2 there is one adjacent mesh whose current [latex](I_{2})[/latex] is multiplied by the resistance common to both meshes [latex](R_{2})[/latex];

3 the voltage source [latex]V_{s2}[/latex] acts in the opposite direction to the current in the mesh so the right-hand side of the mesh equation is [latex]-V_{s2}[/latex].

THe mesh equation is therefore

[latex](R_{2} + R_{3})I_{3}- R_{2}I_{2} = -V_{s2}[/latex]

Putting in the numbers and rearranging we have

[latex]-3I_{2} + 18I_{3} = - 3[/latex] (7.33)

Adding Equations (7.32) and (7.33) we have

[latex]17I_{3}=-17 and I_{3}=-1 A[/latex]

The minus sign indicates that the mesh current [latex]I_{3}[/latex] flows in the opposite direction to that shown in the diagram (i.e. anticlockwise rather than clockwise).

(1) To find the mesh current [latex]I_{2}[/latex]

Substituting for [latex]I_{3}[/latex] = - 1 A in Equation (7.32) we have [latex]3I_{2}- ( - 1 ) = - 1 4[/latex]
and

[latex]I_{2} = - 15 /3 = - 5 A[/latex]

Again the minus sign indicates that the mesh current [latex]I_{2}[/latex] flows in an anticlockwise direction around the mesh.

(2) To find the voltage at node X
The voltage at node X is given by

[latex]V_{x} = V_{s2 }+ I_{3}R_{3} = 3 + ( - 1 ) × 15 = 3 - 15 = -12 V [/latex]

Question 7.16: Determine (1) the mesh current I2 and (2) the voltage at nod...

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