KVL holds for the supermesh, so we can write a KVL equation to generate the second equation we need to solve for the two unknown mesh currents, ixix and iyiy. Write the KVL equation for the supermesh by summing the voltages around the supermesh in the direction of the mesh currents and setting the sum to zero.

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danialamin danialamin · Answer 1 · 2020-03-13T07:59:33+01:00

Answer:

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

Explanation:

As the complete question is not given the complete question is found online and is attached herewith.

By applying KCL at node 1

[tex]i_x+50mA=i_y\\i_x-i_y=0.05A[/tex]

Also

[tex]V_{\Delta}=1K*i_y[/tex]

Now applying KVL on loop 1 as indicated in the attached figure

[tex]1K*i_y+5K(i_y-i_z)+3K*i_x=0\\3i_x+6 i_y-5i_z=0[/tex]

Similarly for loop 2

[tex]2V_{\Delta}+5K(i_z-i_y)=0\\2*1K*i_y+5K(i_z-i_y)=0\\2K*i_y+5K(i_z-i_y)=0\\3i_y-5i_z=0[/tex]

So the system of equations become

[tex]i_x-i_y+0i_z=0.05\\3i_x+6i_y-5i_z=0\\0i_x-3i_y+5i_z=0[/tex]

Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as

[tex]V_{\Delta}=1K*i_y\\V_{\Delta}=1K*-25 mA\\V_{\Delta}=-25 V[/tex]

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V