the circuit shown below, (a) Write a node equation at node 1 by summing the currents leaving node 1 .
(b) Write a node equation at node 2 by summing the currents leaving node $2 .$
(c) Find $V _ { 1 }$ and $V _ { 2 }$ by solving the two node equations.
(d) Find the currents $\mathrm { I } _ { \mathrm { R } 1 } , \mathrm { I } _ { \mathrm { R } 2 } , \mathrm { I } _ { \mathrm { R } 3 } , \mathrm { I } _ { \mathrm { R } 4 } , \mathrm { I } _ { 5 }$.
(e) Find the power absorbed by $\mathrm { R } _ { 1 } , \mathrm { R } _ { 2 } , \mathrm { R } _ { 3 } , \mathrm { R } _ { 4 }$, and power released by $\mathrm { V } _ { \mathrm { s } }$ and $\mathrm { I } _ { \mathrm { s } }$.

Question

Quizplus · Accepted Answer

The Answer is

\begin{array} { l } - I _ { s } + \frac { V _ { 1 } - V _ { s } } { R _ { 1 } } + \frac { V _ { 1 } - V _ { 2 } } { R _ { 2 } } + \frac { V _ { 1 } } { R _ { 4 } } = 0 \\frac { V _ { 2 } - V _ { 1 } } { R _ { 2 } } + \frac { V _ { 2 } } { R _ { 5 } } + \frac { V _ { 2 } - V _ { s } } { R _ { 3 } } = 0 \\mathrm {~V} _ { 1 } = 12 \mathrm {~V} , \mathrm {~V} _ { 2 } = 6 \mathrm {~V}\end{array}

% Nodal Analysis clear all;

R 1 = 2000 ; R 2 = 3000 ; \mathrm { R } 3 = 500 ; \mathrm { R } 4 = 1000 ;

R 1 = 2000 ; R 2 = 3000 ; R 3 = 500 ; R 4 = 1000

\mathrm { Vs } = 8 ; \mathrm { Ig } = 4 \mathrm { e } - 3 ;

\begin{array}{l}\text { syms } \mathrm { V } 1 \text { V2 }\[ V 1 , V 2 ] = s o l v e ( - I s + ( V 1 - V s ) / R 1 + ( V 1 - V 2 ) / R 2 , \ldots\( V 2 - V 1 ) / R 2 + V 2 / R 4 + ( V 2 - V s ) / R 3 , V 1 , V 2 ) \text {; }\\mathrm { IR } 1 = ( \mathrm { Vl } - \mathrm { Vs } ) / \mathrm { Rl } \text {; }\I R 2 = ( V 1 - V 2 ) / R 2 \text {; }\I R 3 = ( V s - V 2 ) / R 3 \text {; }\\text { I } R 4 = V 2 / R 4 \text {; }\\text { I } 5 = \text { IR3-IRI; }\\mathrm { PRI } = \mathrm { IRl } { } ^ { \wedge } 2 \text { * } \mathrm { R } 1 \text {; }\\mathrm { PR } 2 = \mathrm { IR } 2 ^ { \wedge } 2 * \mathrm { R } 2 \text {; }\\mathrm { PR } 3 = \mathrm { IR } 3 \wedge 2 * \mathrm { R } 3 \text {; }\\mathrm { PR } 4 = \operatorname { IR } 4 { } ^ { \wedge } 2 ^ { * } \mathrm { R } 4 \text {; }\\text { PVs } = - I 5 ^ { * } \mathrm { Vs } \text {; }\\text { PIs } = - I s * V 1 \text {; }\\mathrm { Pa } = \mathrm { PR } 1 + \mathrm { PR } 2 + \mathrm { PR } 3 + \mathrm { PR } 4 \text {; }\\mathrm { Ps } = \mathrm { PVs } + \mathrm { PIs } \text {; }\\text { Psum=PR1 +PR2 +PR3+PR4+PVs+PIs; }\V 1 = \operatorname { vpa } ( \mathrm { V } 1,7 )\\mathrm { V } 2 = \mathrm { vpa } ( \mathrm { V } 2,7 )\\text { IR } 1 = \text { vpa (IR1, } 7 )\\text { IR2=vpa (IR2, 7) }\\text { IR3=vpa (IR3, } 7 )\\text { IR4=vpa (IR4, 7) }\15 \text {-vpa } ( 15,7 )\\mathrm { PR } 1 = \mathrm { vpa } ( \mathrm { PR } 1,7 )\\text { PR2-vpa (PR2, 7) }\\text { PR3 =vpa } ( \text { PR3, } 7 )\\text { PR4 =vpa } ( \mathrm { PR } 4,7 )\\text { PVs=vpa } ( \mathrm { PVs } , 7 )\\text { PIs=vpa (PIs, } 7 \text { ) }\\mathrm { Pa } = \mathrm { vpa } ( \mathrm { Pa } , 7 )\\text { Ps=vpa } ( \mathrm { Ps } , 7 )\\text { Psum=vpa } \{ \text { Psum, 7) }\end{array}

Answers:

\begin{array} { l } \mathrm { V } 1 = \12.0 \\mathrm {~V} 2 = \6.0 \\text { IR1 = } \0.002 \\text { IR2 = } \0.002 \\text { IR3 = } \0.004 \\text { IR4= } \0.006 \\text { I5 = } \0.002 \\text { PR1 = } \0.008 \\text { PR2 = } \0.012 \\text { PR3 = } \0.008 \\text { PR4 = } \0.036 \\text { PVs = } \- 0.016 \\text { PIs = } \- 0.048 \\text { Pa = } \0.064 \\text { Ps = } \- 0.064 \\text { Psum = } \0.0\end{array}

The Circuit Shown Below, (A) Write a Node Equation at Node