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Exam 4: Circuit Theorems

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In the circuit shown below, let Vs=5 V,R1=200Ω,R2=500Ω,R3=300Ω,R4=450Ω,R5=400Ω,gm=0.0006V _ { \mathrm { s } } = 5 \mathrm {~V} , \mathrm { R } _ { 1 } = 200 \Omega , \mathrm { R } _ { 2 } = 500 \Omega , \mathrm { R } _ { 3 } = 300 \Omega , \mathrm { R } _ { 4 } = 450 \Omega , \mathrm { R } _ { 5 } = 400 \Omega , \mathrm { g } _ { \mathrm { m } } = 0.0006 Find the Norton equivalent current In\mathrm { I } _ { \mathrm { n } } and the Norton equivalent resistance Rn\mathrm { R } _ { \mathrm { n } } between terminals aa and bb . In the circuit shown below, let V _ { \mathrm { s } } = 5 \mathrm {~V} , \mathrm { R } _ { 1 } = 200 \Omega , \mathrm { R } _ { 2 } = 500 \Omega , \mathrm { R } _ { 3 } = 300 \Omega , \mathrm { R } _ { 4 } = 450 \Omega , \mathrm { R } _ { 5 } = 400 \Omega , \mathrm { g } _ { \mathrm { m } } = 0.0006 Find the Norton equivalent current \mathrm { I } _ { \mathrm { n } } and the Norton equivalent resistance \mathrm { R } _ { \mathrm { n } } between terminals a and b .

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V1−VsR1+V1R2+V1−V2R3=0\frac { V _ { 1 } - V _ { s } } { R _ { 1 } } + \frac { V _ { 1 } } { R _ { 2 } } + \frac { V _ { 1 } - V _ { 2 } } { R _ { 3 } } = 0
V2−V1R3−gmV1+V2R4+V2R5=0\frac { V _ { 2 } - V _ { 1 } } { R _ { 3 } } - g _ { m } V _ { 1 } + \frac { V _ { 2 } } { R _ { 4 } } + \frac { V _ { 2 } } { R _ { 5 } } = 0
V1=2.8717 V, V2=1.4022 V\mathrm { V } _ { 1 } = 2.8717 \mathrm {~V} , \mathrm {~V} _ { 2 } = 1.4022 \mathrm {~V}
In=V2/R5=3.5054 mA\mathrm { I } _ { \mathrm { n } } = \mathrm { V } _ { 2 } / \mathrm { R } _ { 5 } = 3.5054 \mathrm {~mA}
Finding open-circuit voltage:
Va−VsR1+VaR2+Va−VbR3=0\frac { V _ { a } - V _ { s } } { R _ { 1 } } + \frac { V _ { a } } { R _ { 2 } } + \frac { V _ { a } - V _ { b } } { R _ { 3 } } = 0
Vb−VaR3−gmVa+VbR4=0\frac { V _ { b } - V _ { a } } { R _ { 3 } } - g _ { m } V _ { a } + \frac { V _ { b } } { R _ { 4 } } = 0
Va=3.1355 V, Voc=Vb=2.2199 V\mathrm { V } _ { \mathrm { a } } = 3.1355 \mathrm {~V} , \mathrm {~V} _ { \mathrm { oc } } = \mathrm { V } _ { \mathrm { b } } = 2.2199 \mathrm {~V}
Rn=Voc/In=633.2776Ω\mathrm { R } _ { \mathrm { n } } = \mathrm { V } _ { \mathrm { oc } } / \mathrm { I } _ { \mathrm { n } } = 633.2776 \Omega clear all;
Vs =5= 5 ;
Rl=200;R2=500;R3=300;R4=450;R5=400;gm=0.0006;\mathrm { R } l = 200 ; \mathrm { R } 2 = 500 ; \mathrm { R } 3 = 300 ; \mathrm { R } 4 = 450 ; \mathrm { R } 5 = 400 ; \mathrm { gm } = 0.0006 ;
syms V1 V2
[V1, V2]−solve((V1−Vs)/R1+V1/R2+(V1−V2)/R3,…[ \mathrm { V } 1 , \mathrm {~V} 2 ] - \mathrm { solve } ( ( \mathrm { V } 1 - \mathrm { Vs } ) / \mathrm { R } 1 + \mathrm { V } 1 / \mathrm { R } 2 + ( \mathrm { V } 1 - \mathrm { V } 2 ) / \mathrm { R } 3 , \ldots
(V2−V1)/R3−gm∗V1+V2/R4+V2/R5,V1,V2)( V 2 - V 1 ) / R 3 - g m * V 1 + V 2 / R 4 + V 2 / R 5 , V 1 , V 2 ) ;
In=V2/R5\mathrm { In } = \mathrm { V } 2 / \mathrm { R } 5 ;
V1=vpa(V1,12)\mathrm { V } 1 = \mathrm { vpa } ( \mathrm { V } 1,12 )
V2=Vpa(V2,12)\mathrm { V } 2 = \mathrm { Vpa } ( \mathrm { V } 2,12 )
In-vpa (In,12)( I n , 12 )
syms Va Vb
[Va,Vb]=s0lve((Va−Vs)/R1+Va/R2+(Va−Vb)/R3,…[ \mathrm { Va } , \mathrm { Vb } ] = s 0 l v e ( ( \mathrm { Va } - \mathrm { Vs } ) / \mathrm { R } 1 + \mathrm { Va } / \mathrm { R } 2 + ( \mathrm { Va } - \mathrm { Vb } ) / \mathrm { R } 3 , \ldots
(Vb−Va)/R3−gm∗Va+Vb/R4,Va,Vb);( V b - V a ) / R 3 - g m * V a + V b / R 4 , V a , V b ) ;
VOc=Vb\mathrm { VOc } = \mathrm { Vb } ;
Rn=Voc/In\mathrm { Rn } = \mathrm { Voc } / \mathrm { In } ;
Va=vpa(Va,12)\mathrm { Va } = \mathrm { vpa } ( \mathrm { Va } , 12 )
Vb=vpa(Vb,12)\mathrm { Vb } = \mathrm { vpa } ( \mathrm { Vb } , 12 )
Rn=vpa(Rn,12)\mathrm { Rn } = \mathrm { vpa } ( \mathrm { R } n , 12 ) Answers: V=2.87166622656 V=1.40216530235In=0.00350541325588\begin{array} { l } \mathrm { V } = \\2.87166622656 \\\mathrm {~V} = \\1.40216530235 \\\mathrm { In } = \\0.00350541325588\end{array} Va=3.13545150502Vb=2.21989966555Rn=633.277591973\begin{array} { l } \mathrm { Va } = \\3.13545150502 \\\mathrm { Vb } = \\2.21989966555 \\\mathrm { Rn } = \\633.277591973\end{array}

the circuit shown below, 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) Solve the node equations to find the Thévenin equivalent voltage between a and b , \mathrm { V } _ { \text {th } } = \mathrm { V } _ { 2 } . (d) Find the Thévenin equivalent resistance between a and b . (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) Solve the node equations to find the Thévenin equivalent voltage between aa and bb , Vth =V2\mathrm { V } _ { \text {th } } = \mathrm { V } _ { 2 } . (d) Find the Thévenin equivalent resistance between aa and bb .

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Question 2
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V1−VsR1+V1−V2R3+V1R2=0V2−V1R3+V2R4=0 V1=10.56 V, V2=8.4 V Vth=V2=8.4 VRa=R1∥R2=1.2kΩRb=Ra+R3=3kΩRth=Rb∥R4=2.1kΩ\begin{array} { l } \frac { V _ { 1 } - V _ { s } } { R _ { 1 } } + \frac { V _ { 1 } - V _ { 2 } } { R _ { 3 } } + \frac { V _ { 1 } } { R _ { 2 } } = 0 \\\frac { V _ { 2 } - V _ { 1 } } { R _ { 3 } } + \frac { V _ { 2 } } { R _ { 4 } } = 0 \\\mathrm {~V} _ { 1 } = 10.56 \mathrm {~V} , \mathrm {~V} _ { 2 } = 8.4 \mathrm {~V} \\\mathrm {~V} _ { \mathrm { th } } = \mathrm { V } _ { 2 } = 8.4 \mathrm {~V} \\\mathrm { R } _ { \mathrm { a } } = \mathrm { R } _ { 1 } \| \mathrm { R } _ { 2 } = 1.2 \mathrm { k } \Omega \\\mathrm { R } _ { \mathrm { b } } = \mathrm { R } _ { \mathrm { a } } + \mathrm { R } _ { 3 } = 3 \mathrm { k } \Omega \\\mathrm { R } _ { \mathrm { th } } = \mathrm { R } _ { \mathrm { b } } \| \mathrm { R } _ { 4 } = 2.1 \mathrm { k } \Omega\end{array} clear all;
R1=2000;R2=3000;R3=1800;R4=7000;Vs=20\mathrm { R } 1 = 2000 ; \mathrm { R } 2 = 3000 ; \mathrm { R } 3 = 1800 ; \mathrm { R } 4 = 7000 ; \mathrm { Vs } = 20 ;
Ra=P([R1,R2])R a = P ( [ R 1 , R 2 ] )
Rb=Ra+R3R b = R a + R 3
syms V1 V2
[V1, V2]=s∘1ve((V1−Vs)/R1+V1/R2+(V1−V2)/R3,( V2−V1)/R3+V2/R4, V1, V2);V1=Vpa(V1,7)V2=vpa(V2,7)Vth=V2\begin{array}{l}{[\mathrm{V} 1, \mathrm{~V} 2]=s \circ 1 \mathrm{ve}((\mathrm{V} 1-\mathrm{Vs}) / \mathrm{R} 1+\mathrm{V} 1 / \mathrm{R} 2+(\mathrm{V} 1-\mathrm{V} 2) / \mathrm{R} 3,(\mathrm{~V} 2-\mathrm{V} 1) / \mathrm{R} 3+\mathrm{V} 2 / \mathrm{R} 4, \mathrm{~V} 1, \mathrm{~V} 2) ;} \\\mathrm{V} 1=\mathrm{Vpa}(\mathrm{V} 1,7) \\\mathrm{V} 2=\mathrm{vpa}(\mathrm{V} 2,7) \\\mathrm{Vth}=\mathrm{V} 2\end{array}


Answers:  Ra =Rb=1200 Rth =V=10.56 V2=8.4Vth=8.4\begin{array} { l l } \text { Ra } = & \\\mathrm { Rb } = & 1200 \\\text { Rth } = & \\\mathrm { V } = & \\10.56 & \\\mathrm {~V} 2 = & \\8.4 & \\\mathrm { Vth } = & \\8.4\end{array}

In the circuit shown below, (a) Write a node equation at node 1 by summing the currents away from node 1 . (b) Write a node equation at node 2 by summing the currents away from node 2 . (c) Solve the node equations to find the open circuit voltage Voc=V2V _ { o c } = V _ { 2 } . (d) Find the short-circuit current Isc=InI _ { s c } = I _ { n } and find the Norton resistance RnR _ { n } . In the circuit shown below, (a) Write a node equation at node 1 by summing the currents away from node 1 . (b) Write a node equation at node 2 by summing the currents away from node 2 . (c) Solve the node equations to find the open circuit voltage V _ { o c } = V _ { 2 } . (d) Find the short-circuit current I _ { s c } = I _ { n } and find the Norton resistance R _ { n } .

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Question 3
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Open circuit voltage: V1−VsR1+Vs−V13000+V1−V2R2=0V2−V1R2+V2R3=0 V1=11.63636 V, Voc=V2=4.654545 V\begin{array} { l } \frac { V _ { 1 } - V _ { s } } { R _ { 1 } } + \frac { V _ { s } - V _ { 1 } } { 3000 } + \frac { V _ { 1 } - V _ { 2 } } { R _ { 2 } } = 0 \\\frac { V _ { 2 } - V _ { 1 } } { R _ { 2 } } + \frac { V _ { 2 } } { R _ { 3 } } = 0 \\\mathrm {~V} _ { 1 } = 11.63636 \mathrm {~V} , \mathrm {~V} _ { \mathrm { oc } } = \mathrm { V } _ { 2 } = 4.654545 \mathrm {~V}\end{array} Short circuit current:
Va−VsR1+Vs−Va3000+VaR2=0 Va=9.846154 VIn=Va/R2=3.2821 mARn=Voo/In=1.4181818kΩ\begin{array} { l } \frac { V _ { a } - V _ { s } } { R _ { 1 } } + \frac { V _ { s } - V _ { a } } { 3000 } + \frac { V _ { a } } { R _ { 2 } } = 0 \\\mathrm {~V} _ { \mathrm { a } } = 9.846154 \mathrm {~V} \\\mathrm { I } _ { \mathrm { n } } = \mathrm { V } _ { \mathrm { a } } / \mathrm { R } _ { 2 } = 3.2821 \mathrm {~mA} \\\mathrm { R } _ { \mathrm { n } } = \mathrm { V } _ { \mathrm { oo } } / \mathrm { I } _ { \mathrm { n } } = 1.4181818 \mathrm { k } \Omega\end{array}
clear all;
RI=5000;R2=3000;R3=2000;Vs=16\mathrm { R } I = 5000 ; \mathrm { R } 2 = 3000 ; \mathrm { R } 3 = 2000 ; \mathrm { Vs } = 16 ;
syms V1 V2Va\mathrm { V } 1 \mathrm {~V} 2 \mathrm { Va }
[V1, V2]=solve(⟨V1−Vs)/R1−(Vs−V1)/3000+(V1−V2)/R2,…[ \mathrm { V } 1 , \mathrm {~V} 2 ] = \mathrm { solve } ( \langle \mathrm { V } 1 - \mathrm { Vs } ) / \mathrm { R } 1 - ( \mathrm { Vs } - \mathrm { V } 1 ) / 3000 + ( \mathrm { V } 1 - \mathrm { V } 2 ) / \mathrm { R } 2 , \ldots
(V2−V1)/R2+V2/R3, V1, V2);( \mathrm { V } 2 - \mathrm { V } 1 ) / \mathrm { R } 2 + \mathrm { V } 2 / \mathrm { R } 3 , \mathrm {~V} 1 , \mathrm {~V} 2 ) ;
V1−vpa⁡(V1,7)V 1 - \operatorname { vpa } ( \mathrm { V } 1,7 )
V2=vpa⁡(V2,7)\mathrm { V } 2 = \operatorname { vpa } ( \mathrm { V } 2,7 )
I3=V2/R3I 3 = V 2 / R 3
II=(Vs−V1)/R1\mathrm { I } I = ( \mathrm { Vs } - \mathrm { V } 1 ) / \mathrm { R } 1
IVccs =(Vs−V1)/3000= ( \mathrm { Vs } - \mathrm { V } 1 ) / 3000
Isum-I1+Ivces-I3
Va=s\mathrm { Va } = s solve ( (Va-Vs) /R1−(Vs−Va)/3000+Va/R2,Va)/ \mathrm { R } 1 - ( \mathrm { Vs } - \mathrm { Va } ) / 3000 + \mathrm { Va } / \mathrm { R } 2 , \mathrm { Va } )
ISc=Va/R2
Va-vpa (Va,7)( V a , 7 )
Isc=vpa (Is s,7)( I s \mathrm {~s} , 7 )
Rth=V2/IsC\mathrm { Rth } = \mathrm { V } 2 / \mathrm { IsC } Answers: V=11.63636 V=4.654545I=0.002327272727272728047864802647382I=0.00087272727272727479430614039301872IvCcs=0.0014545454545454579905102339883645Isum=0.0000000000000000047369515717340012391408278325445Va=128/13Isc=16/4875Va=9.846154Isc=0.003282051Rth=I418.1818181818198547224211965441\begin{array} { l } \mathrm { V } = \\11.63636 \\\mathrm {~V} = \\4.654545 \\\mathrm { I } = \\0.002327272727272728047864802647382 \\\mathrm { I } = \\0.00087272727272727479430614039301872 \\\mathrm { I } v \mathrm { Ccs } = \\0.0014545454545454579905102339883645 \\\mathrm { Isum } = \\0.0000000000000000047369515717340012391408278325445 \\\mathrm { Va } = \\128 / 13 \\\mathrm { Isc } = \\16 / 4875 \\\mathrm { Va } = \\9.846154 \\\mathrm { Isc } = \\0.003282051 \\\mathrm { Rth } = \\\mathrm { I } 418.1818181818198547224211965441\end{array}

In the circuit shown below, (a) Write a node equation at node 1 by summing the currents leaving node 1.1 . (b) Write a node equation at node 2 by summing the currents leaving node 2 . (c) Solve the node equations to find the Thévenin equivalent voltage between aa and bb , Vth =V2\mathrm { V } _ { \text {th } } = \mathrm { V } _ { 2 } . (d) Find the Thévenin equivalent resistance Ruth between aa and bb . (e) Find the load resistance R. between aa and bb for the maximum power transfer to the load. Also, find the maximum power delivered to the load. In 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) Solve the node equations to find the Thévenin equivalent voltage between a and b , \mathrm { V } _ { \text {th } } = \mathrm { V } _ { 2 } . (d) Find the Thévenin equivalent resistance Ruth between a and b . (e) Find the load resistance R. between a and b for the maximum power transfer to the load. Also, find the maximum power delivered to the load.

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Question 4

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) Solve the two node equations to find V2V _ { 2 } at node 2.2 . (d) Find the equivalent resistance seen from the terminals aa and bb looking into the circuit after deactivating the voltage source and the current source. 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) Solve the two node equations to find V _ { 2 } at node 2 . (d) Find the equivalent resistance seen from the terminals a and b looking into the circuit after deactivating the voltage source and the current source.

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Question 5

In the circuit shown below, let I5=5 mA,R1=500Ω,R2=300Ω,R3=600Ω,R4=400ΩI _ { 5 } = 5 \mathrm {~mA} , \mathrm { R } _ { 1 } = 500 \Omega , \mathrm { R } _ { 2 } = 300 \Omega , \mathrm { R } _ { 3 } = 600 \Omega , \mathrm { R } _ { 4 } = 400 \Omega Find the Norton equivalent current In\mathrm { I } _ { \mathrm { n } } and the Norton equivalent resistance Rn\mathrm { R } _ { \mathrm { n } } between terminals aa and bb . In the circuit shown below, let I _ { 5 } = 5 \mathrm {~mA} , \mathrm { R } _ { 1 } = 500 \Omega , \mathrm { R } _ { 2 } = 300 \Omega , \mathrm { R } _ { 3 } = 600 \Omega , \mathrm { R } _ { 4 } = 400 \Omega Find the Norton equivalent current \mathrm { I } _ { \mathrm { n } } and the Norton equivalent resistance \mathrm { R } _ { \mathrm { n } } between terminals a and b .

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Question 6

Find the Norton equivalent circuit across a and b for the circuit shown in Figure 2. (a) Short circuit aa and bb by connecting a wire between aa and bb . Write a node equation at node 1 by summing the currents leaving node 1 . Solve the equation to find voltage V1V _ { 1 } . (b) Find the short-circuit current between aa and bb , i.e., Isc=In\mathrm { I } _ { \mathrm { sc } } = \mathrm { I } _ { \mathrm { n } } . (c) Deactivate the current source and find the Norton equivalent resistance Rn\mathrm { R } _ { \mathrm { n } } between aa and bb . Find the Norton equivalent circuit across a and b for the circuit shown in Figure 2. (a) Short circuit a and b by connecting a wire between a and b . Write a node equation at node 1 by summing the currents leaving node 1 . Solve the equation to find voltage V _ { 1 } . (b) Find the short-circuit current between a and b , i.e., \mathrm { I } _ { \mathrm { sc } } = \mathrm { I } _ { \mathrm { n } } . (c) Deactivate the current source and find the Norton equivalent resistance \mathrm { R } _ { \mathrm { n } } between a and b .

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Question 7

 In the circuit shown below, i is the current through R2 and v is the voltage across R2. \text { In the circuit shown below, } i \text { is the current through } \mathrm { R } _ { 2 } \text { and } v \text { is the voltage across } \mathrm { R } _ { 2 } \text {. } (a) Write a mesh equation for mesh 1 (left side) by summing the voltage drops around mesh 1 in a clockwise direction. (b) Write a mesh equation for mesh 2 (right side) by summing the voltage drops around mesh 2 in a clockwise direction. (c) Solve the mesh equations to find I2\mathrm { I } _ { 2 } . (d) Find the Thévenin equivalent voltage Vth, V _ { \text {th, } } , which is the voltage across R3R _ { 3 } . (e) Deactivate the voltage source Vs\mathrm { V } _ { \mathrm { s } } and apply a test voltage of 1 V1 \mathrm {~V} between aa and bb to find the Thévenin resistance Rth. R _ { \text {th. } } . \text { In the circuit shown below, } i \text { is the current through } \mathrm { R } _ { 2 } \text { and } v \text { is the voltage across } \mathrm { R } _ { 2 } \text {. } (a) Write a mesh equation for mesh 1 (left side) by summing the voltage drops around mesh 1 in a clockwise direction. (b) Write a mesh equation for mesh 2 (right side) by summing the voltage drops around mesh 2 in a clockwise direction. (c) Solve the mesh equations to find \mathrm { I } _ { 2 } . (d) Find the Thévenin equivalent voltage V _ { \text {th, } } , which is the voltage across R _ { 3 } . (e) Deactivate the voltage source \mathrm { V } _ { \mathrm { s } } and apply a test voltage of 1 \mathrm {~V} between a and b to find the Thévenin resistance R _ { \text {th. } } .

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Question 8

the circuit shown below, let Is=3 mA, Vs1=2 V, Vs2=3 V,R1=1kΩ,R2=3kΩ,R3=1kΩ,R4=3kΩ,R5=4kΩ\mathrm { I } _ { \mathrm { s } } = 3 \mathrm {~mA} , \mathrm {~V} _ { \mathrm { s } 1 } = 2 \mathrm {~V} , \mathrm {~V} _ { \mathrm { s } 2 } = 3 \mathrm {~V} , \mathrm { R } _ { 1 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 3 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 4 } = 3 \mathrm { k } \Omega , \mathrm { R } _ { 5 } = 4 \mathrm { k } \Omega Use source transformation to find V0V _ { 0 } . the circuit shown below, let \mathrm { I } _ { \mathrm { s } } = 3 \mathrm {~mA} , \mathrm {~V} _ { \mathrm { s } 1 } = 2 \mathrm {~V} , \mathrm {~V} _ { \mathrm { s } 2 } = 3 \mathrm {~V} , \mathrm { R } _ { 1 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 3 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 4 } = 3 \mathrm { k } \Omega , \mathrm { R } _ { 5 } = 4 \mathrm { k } \Omega Use source transformation to find V _ { 0 } .

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Question 9

the circuit shown below, let Is1=5 mA,Is2=2 mA,R1=1kΩ,R2=1.5kΩ,R3=1.2kΩR4=4kΩ,R5=3kΩ\mathrm { 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 } = 1.5 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 1.2 \mathrm { k } \Omega \mathrm { R } _ { 4 } = 4 \mathrm { k } \Omega , \mathrm { R } _ { 5 } = 3 \mathrm { k } \Omega Use source transformation to find V0V _ { 0 } . the circuit shown below, let \mathrm { 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 } = 1.5 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 1.2 \mathrm { k } \Omega \mathrm { R } _ { 4 } = 4 \mathrm { k } \Omega , \mathrm { R } _ { 5 } = 3 \mathrm { k } \Omega Use source transformation to find V _ { 0 } .

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Question 10

the circuit shown below, let Is=1 mA,R1=12kΩ,R2=20kΩ,R3=5kΩ. \mathrm { I } _ { \mathrm { s } } = 1 \mathrm {~mA} , \mathrm { R } _ { 1 } = 12 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 20 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 5 \mathrm { k } \Omega \text {. } Find the Thévenin equivalent voltage Vth \mathrm { V } _ { \text {th } } and the Thévenin equivalent resistance Rth between terminals aa and bb . the circuit shown below, let \mathrm { I } _ { \mathrm { s } } = 1 \mathrm {~mA} , \mathrm { R } _ { 1 } = 12 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 20 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 5 \mathrm { k } \Omega \text {. } Find the Thévenin equivalent voltage \mathrm { V } _ { \text {th } } and the Thévenin equivalent resistance Rth between terminals a and b .

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Question 11

the circuit shown below, let Is=3 mA,R1=300Ω,R2=500Ω,R3=400Ω,R4=600Ω,kr=1000I _ { \mathrm { s } } = 3 \mathrm {~mA} , \mathrm { R } _ { 1 } = 300 \Omega , \mathrm { R } _ { 2 } = 500 \Omega , \mathrm { R } _ { 3 } = 400 \Omega , \mathrm { R } _ { 4 } = 600 \Omega , \mathrm { k } _ { \mathrm { r } } = 1000 Find the Thévenin equivalent voltage Vth V _ { \text {th } } and the Thévenin equivalent resistance Rth R _ { \text {th } } between terminals aa and bb . the circuit shown below, let I _ { \mathrm { s } } = 3 \mathrm {~mA} , \mathrm { R } _ { 1 } = 300 \Omega , \mathrm { R } _ { 2 } = 500 \Omega , \mathrm { R } _ { 3 } = 400 \Omega , \mathrm { R } _ { 4 } = 600 \Omega , \mathrm { k } _ { \mathrm { r } } = 1000 Find the Thévenin equivalent voltage V _ { \text {th } } and the Thévenin equivalent resistance R _ { \text {th } } between terminals a and b .

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Question 12

the circuit shown below, let Is=3 mA, Vs=6 V,R1=1kΩ,R2=1.2kΩ,R3=1.2kΩ,R4=4kΩ\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 V1V _ { 1 } . 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 } .

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Question 13
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