Find the Thévenin equivalent with respect to the 7k ohm resistor.
Solution Remove the 7k ohm, since it is not part of the circuit we wish to simplify. simplify. Keep the terminals open since we are finding the Thevenin.
Find th, the voltage across the terminals !in this case it is the voltage over the "k ohm#. $om%ine the &k and 'k in parallel. &k (( 'k ) !&k*'k# + !&k'k# ) '-+"k ) '+"k ) 7 ohms
/ow use a voltage divider to compute th across the "k ohm.
th ) 0"k+!7"k#1 * 2 ) 3.&
Find the Thevenin Resistance %y deactivating all sources and computing the total resistance across the terminals. The voltage sources is shorted, as shown4
/ow let5s redraw the circuit, %ringing the &k and 'k into a vertical position !%ut still still keeping them connected the same way wa y electrically#.
They are all in parallel, so4 Rth ) &k (( 'k (( "k ) & + !&+&k &+'k &+"k# ) 232 ohms /ote, as a check, the equivalent resistance for parallel resistors is is always smaller than the smallest resistor in the com%ination. For e6ample, 232 is smaller than &k. The final Thevenin equivalent is then4
Find the /orton quivalent with respect to the '8uF capacitor
Solution Remove the capacitor, since it is not part of the circuit we wish to simplify.
9n order to find the /orton :hort;circuit current, short the terminals where the capacitor used to %e, since we are finding the /orton
/ow the 'k ohm resistor is shorted;out, so we can eliminate it. Then we5ll find the current through the short.
Mesh-Current Solution for Isc -esh;current analysis to find the current through the short.
/otice that all the mesh currents were drawn counter;clockwise. 9" is the current we are particularly interested in. e have trou%le writing the voltage over the current source, so we either must add another varia%le, or simply write4 i3 ) '8m? /ow solve the system of equations. :olving the second equation for i', we get4 i' ) !&2 3k*i"# + 3k
/ow rewrite equation ", plugging in our formulas for i' and i34 ;3k*!&2 3k*i"#+3k &8k*i" ; k*'8m? &' ) 8 :olve for i"4 ;&2 ; 3k*i" &8k*i" ;&'8 &' ) 8 i" ) &'"+k ) '8.2m?
Superposition Solution for Isc :uperposition analysis to find the current through the short means we solve several simpler circuits !one for each source. :ince there are three sources, we have three simpler circuits to solve.
9n each of these circuits, all sources %ut one have %een deactivated !voltage sources are shorted, current sources are opened#. @ur total answer is the sum of the answers we get from each of the circuits4 9sc ) 9sc&2 9sc&' 9sc'8
First, let5s com%ine the Ak, "k, and &k in series4 R ) Ak "k &k ) &'k and also drop the 2k, since no current can flow through it.
The &'k is shorted out %y the horiBontal wire, so it can %e removed.
/ow we can see that the current through the k ohm resistor will %e the current 9sc. Thus, 9sc&2 ) &2+k ) '.2m?
That5s not the final answer thoughC >e have to find the other two parts of 9sc.
The 2k ohm is again left hanging, so we can remove it. The Ak, "k, and &k can %e com%ined in series !&'k#. The 3k ohm is shorted out, so it can %e removed.
9sc&' ) ;&'+k ) ;'m?
>e can remove the 3k and k since they are %oth shorted out. >e can also com%ine the Ak and "k in series !&&k#.
9sc'8 ) '8m?
/ow we add the three su%;answers together to get our final answer4 9sc ) 9sc&2 9sc&' 9sc'8 ) '.2m? ; 'm? '8m? ) '8.2 m? /ow find the /orton resistance !same as Thevenin resistance#. First, open the terminals where the capacitor used to %e4
/ow deactivate all sources !short voltage sources, open current sources#4
The 3k ohm is shorted out, so it can %e removed, and the 2k cannot have any current through it, so it can also %e removed.
>e can com%ine the Ak, "k, and &k in series.
/ow let5s redraw the circuit4 The resistors are in parallel, so the total resistance seen %y the capacitor is Rth ) k (( 'k ) k*'k+!k'k# ) &'-+Ak ) &.2k ohms
Thus the final /orton is shown %elow4
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Dspice :imulation
Find the Thévenin equivalent with respect to the capacitor in the circuit shown. Then replace the capacitor with a resistor chosen for ma6imum power transfer. >hat is the value of the resistorE >hat is the power a%sor%ed %y this resistorE
Solution The Thevenin equivalent has two parts, th and Rth. >e will do the easier one first ;; Rth. To find the Thevenin resistance, deactivate all sources !short voltages and open currents#. :ince we are finding the Thevenin with respect to the capacitor, we also take the cap out of the circuit and consider the resistance seen from the terminals where the cap was.
The "K and &K are in series, %ut then that com%ination is shorted out %y the wire that replaced the voltage source. ?nother way to think a%out it is that we have &K"K )3K, and then 3K (( 8 ) 8. That is, a 8 ohm resistor in parallel with an ything else is still 8. Thus we have4
Redrawing it slightly !%ut maintaining the same connections#4
>e now see the 3K is in parallel to the 2K, so Rth ) 3K (( 2K ) 3K*2K+!3K2K# ) '.''K ohms /ow we must find th. For this, we must find the open circuit voltage at the terminals4
/ote that there is ' across the 3K and 2K in series. 9t does not matter that the ' is also across the "K and &K in series. >e will use a voltage divider for the 3K and 2K in series with a know total voltage of '4 th ) 2K+!3K2K#*' ) &.&& :o the final Thevenin equivalent is4
The /orton could %e found directly %y computing Rth in the same way and finding 9sc %y shoring the terminals and computing the current. /ow that we have the Thevenin, we can also find the /orton simply %y @hm5s law4 9sc ) th+Rth ) &.&&+'.''Kohm ) 8.2m?. The /orton is then4
Find the Thévenin equivalent with respect to the capacitor in the circuit shown. Then replace the capacitor with a resistor chosen for maximum power transfer. What is the value of the resistor? What is the power absorbed by this resistor?
Solution :ince we are finding the Thevenin with respect to the capacitor, we also take the cap ou t of the circuit and consider the resistance seen from the terminals where the cap was.
The Thevenin equivalent has two parts, th and Rth. >e will do the easier one first ;; Rth. To find the Thevenin resistance, deactivate all sources !short voltages and op en currents#.
From the point of view of the capacitor terminals, the &K and 'K are shorted out. The 7K also is not included %ecause no current can flow through it. 9f current was fed into the top terminal, it would flow through the 3K and 2K and then come %ack through the other terminal. Thus Rth ) 3K (( 2K ) !3K*2K#+!3K2K# ) '.'K ohms /e6t, we5ll find th using node;voltage analysis, with one node !the %ottom wire is the reference node#.
>riting K$= at the node & !current leaving#4
:olve for %y multiplying through %y '8K4 2 ; &2 3 ; &'8 ) 8 ) &"2 ) &"2+ ) &2 :o the final Thevenin quivalent is4
Find the Thévenin equivalent circuit with respect to the capacitor. Gou must use superposition to find th
Solution :ince we are finding the Thevenin with respect to the capacitor, we also take the cap ou t of the circuit and consider the resistance seen from the terminals where the cap was.
The Thevenin equivalent has two parts, th and Rth. >e will do the easier one first ;; Rth. To find the Thevenin resistance, deactivate all sources !make their value 8#.
oltage sources of value 8 can %e modeled as shorts. $urrent sources of value 8 can %e modeled as opens.
9f current was fed into the top terminal, it would split left and right. The current going left would flow through the 3K !skipping the &K and 'K %ecause they are shorted out#, then split again through the AK and K. The current going right would flow through the 2K !skipping the 7k %ecause of the open#. Rth ) 2K (( !3K AK (( K# ) ".&K ohms
/e6t, we5ll find th using superposition. There are three sources, so we5ll have three su%pro%lems to solve. The total th can %e found a s the sum of the individual effects4
Subproblem 1 th due to the " can %e found with this circuit4
Hse a voltage divider to find th4 th ) 2K+!3K 2K AK((K# * " ) &.&"
Subproblem 2 th due to the &8 can %e found with this circuit4
>e can find the voltage across the AK !on our way to finding the voltage across the terminals# %y com%ining the 3K and 2K in series and that in parallel with the AK4 Req ) !2K3K#((AK ) 3.'3K & across Req can %e found %y a voltage divider4 & ) Req+!ReqK#*&8 ) ".'8 Then the voltage over the 2K !which is also the voltage over the terminals# is another voltage divider on the ".'4 th ) 2K+!3K2K#*".' ) &.7A
Subproblem 3 th due to the m? can %e found with this circuit4
/ow com%ine the AK and K in parallel and that in series with the 3K4 Req ) 3K AK((K ) A.'3K Hse a current divider to find the current through the 2K %ranch4 9 ) Req + !Req 2K#*m? ) ".7"m?
Hse @hm5s law to find the voltage over the 2K4 th ) ".7"m? * 2K ) &A.7 /ow add up the su%answers to get the total th4 th ) &.&" &.7A &A.7 ) '&.2A The Thevenin equivalent circuit is then4
Find the Norton quivalent with respect to the ! "ohm resistor in the middle of the circuit# i.e.# the ! "ohm resistor itself should not be part of the equivalent that you compute.
Solution :ince we are finding the /orton with respect to the " Kohm, we take the " Kohm out of the circuit and consider the resistance seen from the terminals where the "K was.
The circuit to the left of the "K is already a /orton equivalent, where the /orton current is ;&8 m? !%ecause it is facing down#. The resistance is infinite. That is, when you open the current source to deactivate it, the &K and 'K are left disconnected. The circuit to the right of the "K is already a Thevenin, where the voltage is and the equivalent resistance is Kohms. $onverting to a /orton, we get /orton current of 8.7 m? and a resistance of Kohms. /ow com%ine the two /ortons. The total curent will %e ;&8m? 8.7m? ) ;."" m?. The total resistance is infinite in parallel with K, which is simply K.
Find the Thévenin equivalent with respect to the $nF capacitor. %ou must use super& position to 'nd (th# the Thévenin volta)e.
Solution * Thévenin equivalent is a circuit# li+e the one shown here.
. 9t has two parts, th and Rth. >e5ll find them each %elow. First, let5s remove the capa citor, since we are finding the equivalent with respect to the capacitor !and thus it is not included in the
circuit we are reducing#. :ince we are finding the Thévenin, we leave a gap !an open# %ecause we will %e finding the open;circuit voltage for th. The terminals shown in the circuit %elow are the connections from the removed capacitor to the rest of the circuit !sliding the 88K resistor to the left a %it, %u t keeping its electrical connections the same#.
Find Vth The question requires that we use superposition to 'nd (th. There are two sources in the circuit# so we will have a reduced circuit for each source ,with all other sources deactivated-. The total (th will be the sum ,super&imposin)- of the two subcircuit answers (total / (!0u* 1 (23(
V35uA = Vth due to 35 microamp source
We deactivate the 23( source by shortin) it. The resultin) circuit is
The current source of !0u* will 4ow down throu)h the !33"# then split between two branches ,a- the 233" and ,b- the $5 and 633" in series. These two branches ,a- and ,b- are in parallel because they are connected electrically at the head ,where the 233"# and $5 are conected- and the tail ,where the 233" and 633" are connected-. We can use a current divider to 'nd how much of the !0u* )oes down the ,b- branch
Now we can use the 7u* in branch ,b- to 'nd the volta)e across the 633" ,which is also the open&circuit volta)e across the terminals of the capacitor-. 8sin) 9hm:s law# we )et (!0u* / 7u* ; 633" / 2.<( Note that the volta)e has polarity with the =1= at the bottom of the 633" and the =&= at the top of the 633"# because the current must 4ow in the =1= terminal for the passive si)n convention.
V40V Vth due to 40V source
We deactivate the !0u* source by openin) it. The resultin) circuit is
>n this reduced circuit# the !33" is not connected on the left side# so we can safely i)nore it. The 23( source now forms two independent volta)e divider circuits •
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*bove it the series combination of the <33" and then the combined parallel 033" and 733" elow it the series combination of the 233"# $5# and 633"
These are independent# @ust li+e mountain climbers climbin) up to the 23 thousand foot pea+ of 5t. Aimalaya on the north face and another )roup of climbrs on the south face. The fraction of the total hei)ht for one )roup has no eBect on the other )roup. So we will use a volta)e divider @ust for the combination of 233"# $5# and 633"# which )oes across the entire volta)e ,hei)ht of the mountain- of 23(. The volta)e across @ust the 633" ,which is also the open circuit volta)e across the
capacitor- is Notice that the $<( has polarity with the =1= at the bottom of the 633" and the =&= at the top of the 633"# because the volta)e is hi)her at the =1= side of the volta)e source and lower at the =&= side of the volta)e source ,where the =&= of the volta)e source is at the top of the 633"-. Vtotal by Superposition
8sin) the answers to the subcircuits above# we now have (total / (!0u* 1 (23( We computed the volta)e in each subcircuit with the =1= at the bottom of the 633"
and the =&= at the top of the 633"# so we can add them directly now. (total / 2.<( 1 $<( (total / $6.<(
Find Rth To 'nd Cth# we deactivate all the sources# so open the current source and short the volta)e source. The resultin) circuit is
The 288K is in parallel with the 788K and that com%ination is in series with the '88K.
Cth / 633" DD ,233" 1 $5Cth / 633" DD $.25 Cth / ,633" ; $.25- E ,633" 1 $.25Cth / 2<3"
Dra !ircuit The 'nal equivalent circuit is then
