Given circuit with the following phase current, 𝐈p = <...

Given circuit with the following phase current, $\mathbf{I}_p = \begin{bmatrix} I_a \ I_b \ I_c \end{bmatrix} = \begin{bmatrix} 100\angle 10^\circ \ 0 \ 100\angle 190^\circ \end{bmatrix} A$ Determine sequence currents (Hints: Compute $I_{a0}$, $I_{a1}$, and $I_{a2}$)

Transcript

00:01 In this question we are asked to find the current flowing through each register and voltage across each register.

00:08 So let's start to solve this problem.

00:10 Here i am solving this question by using nodal analysis method.

00:14 Why? because in this circuit, if i applied nodal analysis, then i require less equations as compared to mass analysis.

00:25 Here, if i solve this question by using mass analysis, then i require two equations.

00:30 To solve this circuit and here if i solve this circuit by using nodal analysis then i require only one equation so now suppose here this is node a and this is node b here i take node b as a reference node means voltage of node b is equal to zero now after applying nodal analysis at node a we obtain this equation v a minus v1 over r1 plus v a minus v2 over r3 plus v a minus v2 over r3 plus v a minus vb over r2 plus r4 is equal to 0 means algebraic sum of all outgoing currents from node a is equal to zero now after putting the value of v1 r1, v2, r3, vb, r2 and r4, this equation will become equal to va minus v1 is equal to 12 over r1 is equal to 100 plus va minus v2 is equal to 20 over r3, r3 is equal to 270 plus va minus bb is equal to 0 r2 is equal to 120 plus 80 equals to 0 now after taking va common we obtain va 1 over 100 plus 1 over 270 plus 1 over 202 is equal to 20 over 20 20 20 plus 12 over 100 by simplifying this, we obtain va is equal to 10 .40377 volt.

02:47 So now in the circuit, voltage of node a is equal to 10 .4037 volt.

03:01 Now suppose this current is equal to i1, this current is equal to i2, this current is equal to i2, this current is equal to i3.

03:13 Now by applying kirchow voltage load we obtained i1 is equal to 12 minus 10 .4037 over r1.

03:30 Now by simplifying this we obtained i1 is equal to 15 .9623m .m.

03:41 Now next by applying kirchop voltage load we obtained i2 is equal to va.

03:46 Minus vb over r2 plus r4.

03:48 Means we obtain i2 is equal to 10 .40377.

03:54 This is va minus vb is equal to 0 over r2 plus r2 plus r4 is equal to 202.

03:59 And by simplifying this, we obtain i2 is equal to 51 .5038 millie -ampere.

04:06 Now next by applying kirchow voltage load, we obtain i3 is equal to v2 minus va over 270...

Question

Given circuit with the following phase current, $\mathbf{I}_p = \begin{bmatrix} I_a \ I_b \ I_c \end{bmatrix} = \begin{bmatrix} 100\angle 10^\circ \ 0 \ 100\angle 190^\circ \end{bmatrix} A$ Determine sequence currents (Hints: Compute $I_{a0}$, $I_{a1}$, and $I_{a2}$)

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