A) Identify the meshes in the circuit, and using Mesh...

a) Identify the meshes in the circuit, and using Mesh Analysis, determine the mesh currents. Note: Use of a numerical software application such as MATLAB to solve for the Mesh currents is acceptable. b) From part a), determine the voltage, $V_o$. 3 k$\Omega$ 4 k$\Omega$ 2 k$\Omega$ 2 mA $\uparrow$ 12 V + 6 k$\Omega$ 3 k$\Omega$ + 10 V + 6 k$\Omega$ $V_o$ 0 + 8 V

Transcript

00:01 Hello everyone so here we are given that there is the circuit with current i1 i2 and i3 at these loops we have to find their values by mesh analysis.

00:12 So first of all we have in this mesh calling it as mesh 1 we have 24 volts that is the supply vote is equal to in this element we have r that is 2 multiplied by i minus i 2.

00:38 So we have 2 multiplied by i 1 minus i 2.

00:43 Similarly in the other 4 home register we have 4 multiplied by i 1 minus i 3.

00:51 Let us simplify it we have 2 i 1 minus 2 i 2 i 2 plus 4 i 1 minus 4 i 3 minus 24 is equal to 0 or simply we have 6 i1 minus 2 i2 minus 4 i 3 minus 24 is equal to 0 taking out 2 common it will become 0 on rhes we are left with 3 i1 minus i 2 minus 2 i 3 minus 12 is equal to 0 calling it as equation number 1 now next we have here for the super mesh 2 multiplic 2 multiplic 3 by i2 minus i1 plus 8 multiplied by this i2 plus this 2 volt plus 4 multiplied by i 3 minus i 1 so for the super mesh let us write the equation we have twice of i 2 minus i 1 plus 8 i2 plus 2 v0 that's given plus 4 multiplied by i3 minus i1 is equal to 0 from here we know that 2 multiplied by i 1 minus i 2 is equal to v0 that is since 2 multiplied by i 1 minus i 2 is equal to v0 we have therefore simply we have minus of i 1 plus 3 i2 plus 2 i 3 i 3 is equal to 0 calling this as equation number 2 also we know that 3 plus i2 is equal to this current that is i 3 so writing it here we know that i 3 is equal to 3 plus i 2 calling it as equation number 3 let us put equation number 3 in 1 and 2 we get in equation number 1 put 3 we get 3 i 1 minus i 2 i 3 i 3 i 3 is 3 plus i 2 minus 12 is equal to 0 or this is equal to 3 i 1 minus 3 i 2 minus 18 is equal to 0 out 3 common it becomes i1 minus i2 is equal to 18 calling this as equation number 8.

04:14 Now in equation number 2 put equation number 3 we get minus i1 plus 3 i2 plus twice of i 3 is 3 plus i2 and this is equal to 0.

04:32 Now let us solve it we get minus i 2.

04:38 I1 plus 5 i2 is equal to minus 6 or i1 minus 5 i2 is equal to 6 calling this as equation number b.

04:54 Now equation a minus equation b will be i 1 minus i 2 is equal to 18 i 1 minus 5 i 2 is equal to 6...

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