the circuit shown below, let $$\mathrm { I } _ { \mathrm { s } } = 3 \mathrm {~mA} , \mathrm {~V} _ { \mathrm { s } } = 6 \mathrm {~V} , \mathrm { R } _ { 1 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 1.2 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 1.2 \mathrm { k } \Omega , \mathrm { R } _ { 4 } = 4 \mathrm { k } \Omega$$
Use superposition principle to find $V _ { 1 }$.

Question

Quizplus · Accepted Answer

The Answer is If the current source is deactivated by open-circuiting it, we obtain the circuit shown below. $If the current source is deactivated by open-circuiting it, we obtain the circuit shown below. If the voltage source is deactivated by short-circuiting it, we obtain the circuit shown below. \begin{array} { l } \mathrm { R } _ { \mathrm { c } } = \mathrm { R } _ { 1 } \| \mathrm { R } _ { 3 } = 545.4545 \Omega \ \mathrm { R } _ { \mathrm { d } } = \mathrm { R } _ { 2 } + \mathrm { R } _ { \mathrm { c } } = 1.7455 \mathrm { k } \Omega \ I _ { R _ { 2 } } = I _ { s } imes \frac { R _ { 4 } } { R _ { d } + R _ { 4 } } = 2.08861 \mathrm {~mA} \ V _ { 12 } = R _ { c } I _ { R _ { 2 } } = 1.13924 \mathrm {~V} \ \mathrm {~V} _ { 1 } = \mathrm { V } _ { 11 } + \mathrm { V } _ { 12 } = 4.1013 \mathrm {~V} \end{array} \begin{array} { l } ext { clear all } \ V s = 6 ; I s = 0,003 ; \ R 1 = 1000 ; R 2 = 1200 ; R 3 = 1200 ; R 4 = 4000 \ R a = R 2 + R 4 \ R b = P ( [ R 3 , R a ] ) \ V 11 = V \leqslant * R b / ( R 1 + R b ) \ R c = P ( [ R 1 , R 3 ] ) \ R d = R c + R 2 \ I R 2 = I S ^ { * } R 4 / ( R 4 + R d ) \ V 12 = I s ^ { * } R 4 / ( R 4 + R d ) * R c \ V 1 = V 11 + V 12 \end{array} Answers: \begin{array}{l} \mathrm{Ra}=5200\ \mathrm{Rb}=\ 975\ \mathrm{V} 11=\ 2.962025316455696\ ext { RC }=\ 5.454545454545454 \mathrm{e}+02\ R d=1.745454545454546 \mathrm{e}+03\ ext { IR2 = }\ 0.002088607594937\ \mathrm{V} 12=\ ext { 1.139240506329114 }\ \mathrm{VI}=\ 4.101265822784810 \end{array}$
If the voltage source is deactivated by short-circuiting it, we obtain the circuit shown below. $If the current source is deactivated by open-circuiting it, we obtain the circuit shown below. If the voltage source is deactivated by short-circuiting it, we obtain the circuit shown below. \begin{array} { l } \mathrm { R } _ { \mathrm { c } } = \mathrm { R } _ { 1 } \| \mathrm { R } _ { 3 } = 545.4545 \Omega \ \mathrm { R } _ { \mathrm { d } } = \mathrm { R } _ { 2 } + \mathrm { R } _ { \mathrm { c } } = 1.7455 \mathrm { k } \Omega \ I _ { R _ { 2 } } = I _ { s } imes \frac { R _ { 4 } } { R _ { d } + R _ { 4 } } = 2.08861 \mathrm {~mA} \ V _ { 12 } = R _ { c } I _ { R _ { 2 } } = 1.13924 \mathrm {~V} \ \mathrm {~V} _ { 1 } = \mathrm { V } _ { 11 } + \mathrm { V } _ { 12 } = 4.1013 \mathrm {~V} \end{array} \begin{array} { l } ext { clear all } \ V s = 6 ; I s = 0,003 ; \ R 1 = 1000 ; R 2 = 1200 ; R 3 = 1200 ; R 4 = 4000 \ R a = R 2 + R 4 \ R b = P ( [ R 3 , R a ] ) \ V 11 = V \leqslant * R b / ( R 1 + R b ) \ R c = P ( [ R 1 , R 3 ] ) \ R d = R c + R 2 \ I R 2 = I S ^ { * } R 4 / ( R 4 + R d ) \ V 12 = I s ^ { * } R 4 / ( R 4 + R d ) * R c \ V 1 = V 11 + V 12 \end{array} Answers: \begin{array}{l} \mathrm{Ra}=5200\ \mathrm{Rb}=\ 975\ \mathrm{V} 11=\ 2.962025316455696\ ext { RC }=\ 5.454545454545454 \mathrm{e}+02\ R d=1.745454545454546 \mathrm{e}+03\ ext { IR2 = }\ 0.002088607594937\ \mathrm{V} 12=\ ext { 1.139240506329114 }\ \mathrm{VI}=\ 4.101265822784810 \end{array}$

\begin{array} { l } \mathrm { R } _ { \mathrm { c } } = \mathrm { R } _ { 1 } \| \mathrm { R } _ { 3 } = 545.4545 \Omega \\mathrm { R } _ { \mathrm { d } } = \mathrm { R } _ { 2 } + \mathrm { R } _ { \mathrm { c } } = 1.7455 \mathrm { k } \Omega \I _ { R _ { 2 } } = I _ { s } imes \frac { R _ { 4 } } { R _ { d } + R _ { 4 } } = 2.08861 \mathrm {~mA} \V _ { 12 } = R _ { c } I _ { R _ { 2 } } = 1.13924 \mathrm {~V} \\mathrm {~V} _ { 1 } = \mathrm { V } _ { 11 } + \mathrm { V } _ { 12 } = 4.1013 \mathrm {~V}\end{array}

\begin{array} { l } ext { clear all } \V s = 6 ; I s = 0,003 ; \R 1 = 1000 ; R 2 = 1200 ; R 3 = 1200 ; R 4 = 4000 \R a = R 2 + R 4 \R b = P ( [ R 3 , R a ] ) \V 11 = V \leqslant * R b / ( R 1 + R b ) \R c = P ( [ R 1 , R 3 ] ) \R d = R c + R 2 \I R 2 = I S ^ { * } R 4 / ( R 4 + R d ) \V 12 = I s ^ { * } R 4 / ( R 4 + R d ) * R c \V 1 = V 11 + V 12\end{array}

Answers:

\begin{array}{l}\mathrm{Ra}=5200\\mathrm{Rb}=\975\\mathrm{V} 11=\2.962025316455696\ ext { RC }=\5.454545454545454 \mathrm{e}+02\R d=1.745454545454546 \mathrm{e}+03\ ext { IR2 = }\0.002088607594937\\mathrm{V} 12=\ ext { 1.139240506329114 }\\mathrm{VI}=\4.101265822784810\end{array}

The Circuit Shown Below, Let Use Superposition Principle to Find V1V _ { 1 }V1​

The Circuit Shown Below, Let Use Superposition Principle to Find $V _ { 1 }$