Exam 3: Circuit Analysis Methods

In the circuit shown below, let $I _ { \mathrm { s } 1 } = 5 \mathrm {~mA} , \mathrm { I } _ { \mathrm { s } 2 } = 2 \mathrm {~mA} , \mathrm { R } _ { 1 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 500 \Omega , \mathrm { R } _ { 3 } = 1.5 \mathrm { k } \Omega , \mathrm { R } _ { 4 } = 600 \Omega$ . Use mesh analysis to find mesh currents $\mathrm { I } _ { 1 } , \mathrm { I } _ { 2 }$ , and $\mathrm { I } _ { 3 }$ . Also, find $\mathrm { V } _ { 1 } , \mathrm {~V} _ { 2 }$ , and $\mathrm { V } _ { 3 }$ . $In the circuit shown below, let I _ { \mathrm { s } 1 } = 5 \mathrm {~mA} , \mathrm { I } _ { \mathrm { s } 2 } = 2 \mathrm {~mA} , \mathrm { R } _ { 1 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 500 \Omega , \mathrm { R } _ { 3 } = 1.5 \mathrm { k } \Omega , \mathrm { R } _ { 4 } = 600 \Omega . Use mesh analysis to find mesh currents \mathrm { I } _ { 1 } , \mathrm { I } _ { 2 } , and \mathrm { I } _ { 3 } . Also, find \mathrm { V } _ { 1 } , \mathrm {~V} _ { 2 } , and \mathrm { V } _ { 3 } .$

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Question 1

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$\begin{array} { l } I _ { 1 } = I _ { 1 } 1 \\R _ { 1 } \left( I _ { 2 } - I _ { 1 } \right) + R _ { 3 } I _ { 2 } + R _ { 2 } \left( I _ { 2 } - I _ { 3 } \right) = 0 \\I _ { 3 } = - I _ { 2 } 2 \\V _ { 2 } = R _ { 4 } \left( I _ { 1 } - I _ { 3 } \right) \\V _ { 1 } = V _ { 2 } + R _ { 1 } \left( I _ { 1 } - I _ { 2 } \right) \\V _ { 3 } = V _ { 1 } - R _ { 3 } I _ { 2 } \\I _ { 1 } = 5 \mathrm {~mA} , I _ { 2 } = 1.3333 \mathrm {~mA} , I _ { 3 } = - 2 \mathrm {~mA} , V _ { 1 } = 7.8667 \mathrm {~V} , V _ { 2 } = 4.2 \mathrm {~V} , V _ { 3 } = 5.8667 \mathrm {~V}\end{array}$ % Mesh Analysis clear all
Isl $= 0.005 ; I s 2 = 0.002$ ;
$R 1 = 1000 ; R 2 = 500 ; R 3 = 1500 ; R 4 = 600$ ;
syms I1 I2 I3
$[ I 1 , I 2 , I 3 ] = s 01 v e ( R 1 * ( I 2 - I 1 ) + R 3 * I 2 + R 2 * ( I 2 - I 3 ) , \ldots$
$I 1 = I s 1 , I 3 = - I s 2 , I 1 , I 2 , I 3 )$
$V 2 = R 4 ^ { * } ( I 1 - I 3 \rangle$
$V 1 = V 2 + R 1 * ( I 1 - I 2 )$
$V 3 = V 1 - R 3 * I 2$
IR $4 = I 1 - I 3$
$I R 1 = I 1 - I 2$
$I R 2 = - I 3 + I 2$
IR3 $= I 2$
$S u m I = - I R 1 - I R 2 + I R 4$
$I 1 = \operatorname { vpa } ( I 1,7 )$
$I 2 = \operatorname { vpa } ( \mathrm { I } 2,7 )$
$\mathrm { I } 3 = \mathrm { vpa } ( \mathrm { I } 3,7 )$
$\mathrm { V } 1 = \operatorname { vpa } ( \mathrm { V } , 7 )$
$\mathrm { V } 2 = \mathrm { Vpa } ( \mathrm { V } 2,7 )$
$\mathrm { V } 3 = \mathrm { VPa } ( \mathrm { V } 3,7 )$
IR $4 =$ vpa (IR4, 7)
IR1=vpa (IR1, 7)
IR2=vpa (IR2, T)
IR3=ypa (IR3, 7)
SumI=vpa (SumI, 7) Answers: $\begin{array} { l } I 1 = \\0.005 \\I 2 = \\0.001333333 \\I 3 = \\- 0.002 \\V 1 = \\7.866667 \\V 2 = \\4.2 \\V 3 = \\5.866667\end{array}$

In the circuit shown below, let $\mathrm { V } _ { \mathrm { s } } = 9 \mathrm {~V} , \mathrm { R } _ { 1 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 400 \Omega , \mathrm { R } _ { 3 } = 500 \Omega , \mathrm { R } _ { 4 } = 1.2 \mathrm { k } \Omega , \mathrm { g } _ { \mathrm { m } } = 0.0004$ Use mesh analysis to find mesh currents $I _ { 1 } , I _ { 2 }$ , and $I _ { 3 }$ . Also, find $V _ { 1 }$ and $V _ { 2 }$ . $In the circuit shown below, let \mathrm { V } _ { \mathrm { s } } = 9 \mathrm {~V} , \mathrm { R } _ { 1 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 400 \Omega , \mathrm { R } _ { 3 } = 500 \Omega , \mathrm { R } _ { 4 } = 1.2 \mathrm { k } \Omega , \mathrm { g } _ { \mathrm { m } } = 0.0004 Use mesh analysis to find mesh currents I _ { 1 } , I _ { 2 } , and I _ { 3 } . Also, find V _ { 1 } and V _ { 2 } .$

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Question 2

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$\begin{array} { l } \mathrm { R } _ { 1 } \mathrm { I } _ { 1 } + \mathrm { R } _ { 4 } \mathrm { I } _ { 3 } + \mathrm { R } _ { 3 } \left( \mathrm { I } _ { 3 } - \mathrm { I } _ { 2 } \right) + \mathrm { R } _ { 2 } \left( \mathrm { I } _ { 1 } - \mathrm { I } _ { 2 } \right) = 0 \\- \mathrm { V } _ { \mathrm { s } } + \mathrm { R } _ { 2 } \left( \mathrm { I } _ { 2 } - \mathrm { I } _ { 1 } \right) + \mathrm { R } _ { 3 } \left( \mathrm { I } _ { 2 } - \mathrm { I } _ { 3 } \right) = 0 \\\mathrm { I } _ { 3 } - \mathrm { I } _ { 1 } = \mathrm { g } _ { \mathrm { m } } \mathrm { V } _ { 1 } \\\mathrm {~V} _ { 1 } = \mathrm { R } _ { 3 } \left( \mathrm { I } _ { 2 } - \mathrm { I } _ { 3 } \right) \\\mathrm { V } _ { 2 } = \mathrm { R } _ { 4 } \mathrm { I } _ { 3 } \\\mathrm { I } _ { 1 } = 3.0891 \mathrm {~mA} , \mathrm { I } _ { 2 } = 14.1095 \mathrm {~mA} , \mathrm { I } _ { 3 } = 4.9258 \mathrm {~mA} , \mathrm {~V} _ { 1 } = 4.5918 \mathrm {~V} , \mathrm {~V} _ { 2 } = 5.9109 \mathrm {~V}\end{array}$ % Mesh Analysis clear all
$V S = 9 ;$
$R 1 = 1000 ; R 2 = 400 ; R 3 = 500 ; R 4 = 1200 ; g m = 0.0004$
syms I1 I2 I3
$[ I 1 , I 2 , I 3 ] - s 0 l v e ( - V s + R 2 * ( I 2 - I 1 ) + R 3 * ( I 2 - I 3 ) , \ldots$
$R 1 * I 1 + R 4 ^ { * } I 3 + R 3 ^ { * } \{ I 3 - I 2 ) + R 2 ^ { * } ( I I - I 2 ) , \ldots$
$\left. I 3 - I 1 = = g _ { m } * R 3 * ( I 2 - I 3 ) , I 1 , I 2 , I 3 \right)$
$V 1 = R 3 ^ { * } ( I 2 - I 3 )$
$\mathrm { V } 2 = \mathrm { R } 4 ^ { * } \mathrm { I } 3$
IR $4 = I 3$
IRl $= \mathrm { Il }$
IR2 $= I 2 - I 1$
IR $3 = I 2 = I 3$
SumI $= - I R 2 + I R 3 + g m * v 1$
$I 1 = \operatorname { vpa } ( I 1,7 )$
$I 2 = \operatorname { vpa } ( I 2,7 )$
I $3 = \operatorname { vpa } ( I 3,7 )$
$V 1 = \operatorname { vpa } ( \mathrm { V } 1,7 )$
$\mathrm { V } 2 = \mathrm { Vpa } ( \mathrm { V } 2,7 )$
$\operatorname { IR4 } = \mathrm { vpa } ( \operatorname { IR4 } , 7 )$
IR1=Vpa (IR1, 7)
IR2 = VPa (IR2, 7)
IR3=vpa $( \operatorname { IR3 } , 7 )$
SumI=vpa $($ SumI, $7 )$ Answers: $\begin{array} { l } I 1 = \\0.003089054 \\I 2 = \\0.01410946 \\I 3 = \\0.004925788 \\V 1 = \\4.591837 \\V 2 = \\5.910946\end{array}$

In the circuit shown below, (a) Write a mesh equation by summing the voltage drops around mesh 1 (left side). (b) Write a mesh equation by summing the voltage drops around mesh 2 (right side). (c) Solve the two equations to find $I _ { 1 }$ and $I _ { 2 }$ . (d) Find voltage $V _ { 3 }$ across $R _ { 2 }$ . $In the circuit shown below, (a) Write a mesh equation by summing the voltage drops around mesh 1 (left side). (b) Write a mesh equation by summing the voltage drops around mesh 2 (right side). (c) Solve the two equations to find I _ { 1 } and I _ { 2 } . (d) Find voltage V _ { 3 } across R _ { 2 } .$

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Question 3

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$\begin{array} { l } - \mathrm { V } _ { 1 } + \mathrm { R } _ { 1 } \mathrm { I } _ { 1 } + \mathrm { R } _ { 2 } \left( \mathrm { I } _ { 1 } - \mathrm { I } _ { 2 } \right) = 0 \\\mathrm { R } _ { 2 } \left( \mathrm { I } _ { 2 } - \mathrm { I } _ { 1 } \right) + \mathrm { R } _ { 3 } \mathrm { I } _ { 2 } + \mathrm { V } _ { 2 } = 0 \\\mathrm {~V} _ { 3 } = \mathrm { R } _ { 2 } \left( \mathrm { I } _ { 1 } - \mathrm { I } _ { 2 } \right) \\\mathrm { I } _ { 1 } = 1 \mathrm {~mA} , \mathrm { I } _ { 2 } = - 1 \mathrm {~mA} , \mathrm {~V} _ { 3 } = 3 \mathrm {~V}\end{array}$ % Mesh Analysis clear all
$\mathrm { V } 1 = 5 ; \mathrm { V } 2 = 6$ ;
$\mathrm { R } 1 = 2000 ; \mathrm { R } 2 = 1500 ; \mathrm { R } 3 = 3000 ;$

$\begin{array}{l}\text { Syms II I2 } \\{[I 1, I 2]=\text { Solve }(-\mathrm{V} 1+\mathrm{R} 1 * I 1+\mathrm{R} 2 *(I 1-I 2), \mathrm{R} 2 *(I 2-I 1)+\mathrm{R} 3 * I 2+\mathrm{V} 2, I 1, I 2)} \\\mathrm{V} 3=\mathrm{R} 2 *(I 1-I 2) \\I 1=\operatorname{vpa}(I 1,7) \\\text { I2 }=\operatorname{vpa}(I 2,7) \\\mathrm{V} 3=\operatorname{vpa}(\mathrm{V} 3,7)\end{array}$
Answers: $\begin{array} { l } 11 = \\0.001 \\12 = \\- 0.001 \\V 3 = \\3.0\end{array}$