Lesson 3
Current sources, conductances and dependent sources
Learn how to describe a current source using the j element. Simplify parallel resistors using the pr tool and its shorthand. Learn to describe conductances and dependent sources.
Last updated 2023-07-08
In this lesson you will learn how to describe a current source with the j element, and a trick to simplify parallel resistors with the pr tool or its shorthand. You will also learn to describe conductance and dependent sources using elements you already know.
3.1How to describe a current source
You can use SI prefixes here as well. The value of a current source is often given in milliamps; Symbulator will interpret any 'm given in the value as a division by a thousand.
Answers for a current source
For each current source you get the same answers as for a voltage source, with the same polarity conventions: the voltage drop in it, the current through it, the power consumed by it (for the delivered power, ask for the negative), and the equivalent resistance of the rest of the circuit as seen by that source.
Let's see an example.
B11's Example 8.1
Open in app ↗Open in split view
Given the circuit, determine the current and voltage drop in R1.
B11's Example 8.1
Solution
We ask Symbulator to run a DC simulation of the circuit described between quotations:
j,0,1,10'm
r1,1,0,20'kTwo lines, two elements. Under Type of analysis leave Solve circuit, set Analysis to DC — direct current, and click Run Symbulator. This is a numerical circuit, so it is worth putting Rounding back to approximate to n significant digits with n = 3 and ticking Use SI prefixes, as in Lesson 1.
The r1 block gives both answers at once: ir1 = 10 mA on the current through line, and vr1 = 200 V on the voltage drop line — that is a 10 mA current and a 200 V voltage drop.
3.2What about conductances?
Conductances are really resistors by another name. So we describe them as resistors, using the element r, and give as the value the inverse of the conductance: one divided by the conductance is the resistor in Ω.
The next example is from Elementary Linear Circuit Analysis (2ed) by Leonard S. Bobrow, which from here on I call Bo2.
Bo2's Example 2.2
Open in app ↗Open in split view
Given the circuit, determine the voltages in the nodes.
Bo2's Example 2.2
Solution
As I explained, all conductances are simulated as resistors: the 4 siemens conductance becomes a 1/4 resistor, and so on. My description:
j10,1,0,2
r12,1,2,1
r20,2,0,1/4
r30,3,0,1/3
r13,1,3,1/2
j32,3,2,3Run it in DC. The answers are the first thing on the page, under Node voltages:
That indicates v1 = –1.3 V, v2 = .34 V, v3 = –1.12 V. This is correct.
3.3How to reduce parallel resistors
Symbulator 9 has the same tool, spelled pr, and you can use it in two places.
Inside a circuit description, write it where a value goes — that is what the re,3,0,[6,3] shorthand below does, and pr(6,3) means the same thing.
On its own, type it into the Evaluate card. To reduce three resistors in parallel with values of 10 Ω, 20 Ω and 30 Ω:
pr(10, 20, 30)You get 60/11 Ω with Rounding on exact, or 5.45 Ω approximately. It nests as deeply as you like — the two problems below lean on that — and it works on symbols as readily as on numbers: pr(r1, r2) evaluates to \(r1 r2/(r1 + r2)\), useful when the resistors in your circuit are still unknowns.
AS7's Example 2.10
Find the equivalent resistance.
AS7's Example 2.10
Solution
No simulation is needed — the whole reduction is one nested expression in the Evaluate card, working from the far end of the ladder outward:
10 + pr(3, 6, 1 + pr(12, 4, 1 + 5))The answer is 11.2 Ω. This is correct.
AS7's Practice Problem 2.10
Find the equivalent resistance.
AS7's Practice Problem 2.10
Solution
Again, no simulation:
16 + pr(18, 9, 2 + pr(20, 1 + pr(5, 20)))The answer is 19 Ω. This is correct.
The [r,r,r…] shorthand for descriptions
Calling pr() separately while describing a circuit is not very practical, so there is a shorthand: Symbulator reads any values inside square brackets, such as [10,20,30] or [r1,r2,r3,r4], as input to pr. The Evaluate card runs what you type through the same shorthand, so the brackets work there too; this book writes pr(...) outside descriptions anyway, where square brackets are easy to misread as a list.
When to reduce resistors
A simulation where it makes sense to use pr is B11's Example 7.4, which you saw in the practice problems of Lesson 1. It makes sense to reduce R4 and R5 to an equivalent resistor, since we do not need to know their individual currents or power use:
e,1,0,16.8
r1,1,2,9
r2,1,2,6
r3,2,3,4
re1,3,0,[6,3]
r6,2,0,3Run it in DC, then ask Evaluate for -ie, the current the source delivers. The answer is 3 A.
An example where using pr makes no sense is B11's Example 8.3, because you need to know the value of the current through R1.
Finally, a simulation where you can reduce part of the resistors is B11's Example 6.22. We must leave R1 alone, because we need the current through it, but we can reduce R2 and R3:
jt,0,1,12'm
r1,1,0,1'k
re,1,0,[10'k,22'k]Moving forward, we will use the pr(,,) tool, or its shorthand [,,], in circuit descriptions whenever we feel it is appropriate.
3.4How to describe dependent sources
One of my favourite scenes in cinema comes from The Dark Knight: the Joker, played masterfully by Heath Ledger, is rolling on the floor of a Gotham City prison, taking a bare-knuckle beating from an ever-more-frustrated Batman. Master of the situation and laughing hysterically, the Joker says: "You have nothing! Nothing to threaten me with!"
Even though the movie had not been made yet, I remember feeling something similar — if less hysterical — back in 1999, when I realised that a 100% symbolic implementation let me make any element's value depend on any answer of the circuit. I could simulate voltage or current sources dependent on any voltage, current or combination of them as easily as a 12 V source.
Here is what you need to know for simulating dependent sources in Symbulator: nothing. There is nothing special to it at all. Just write the value as a function of the circuit's answers, using the variables you know by now, and run the simulation like it's nobody's business. For example:
- if the source depends on the current through a resistor called r1, you define its value as
ir1 - if the source depends on the voltage drop in a resistor called r2, you define its value as
vr2 - if the source depends on the current through a short called s3, you define its value as
is3 - if the source depends on the difference between the voltage of two nodes a and b, you define its value as
va-vb
With Symbulator, instead of fearing them, you will laugh in the face of dependent sources, thinking: "You have nothing!" Booyah!
3.5Instructive solved examples
Circuits with E, J and R
B11's Example 8.2
Open in app ↗Open in split view
Determine the values of VS, I1 and I2.
j,0,1,7
e,1,0,12
r1,1,0,4The answers you want are v1, ie and ir1, in Results.
The answers show VS is 12V, I1 is 4A and I2 is 3A. These are the correct answers.
B11's Example 8.15
Open in app ↗Open in split view
Determine the current through each resistor. My solution, with the direction marked in blue:
j6,0,1,6
r2,1,0,2
r6,1,2,6
r8,0,2,8
j8,2,0,8The answers you want are ir2, ir6 and ir8, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
We get these answers: {1.25,4.75,3.25}. So IR2 is 1.25A, IR6 is 4.75A, and IR8 is 3.25A.
B11's Example 8.21
Open in app ↗Open in split view
Determine the voltage in each node and the current through each resistor.
My solution:
j1,0,1,4
r1,1,0,2
r3,1,2,12
r2,0,2,6
j2,2,0,2The answers you want are v1, v2, ir1, ir2 and ir3, in Results.
We get the following answers: {6,-6,3,1,1}. So V1=6V, V2=-6V, IR1=3A, and IR2=IR3=1A.
Bo2's Drill Exercise 2.2 (Conductances)
Open in app ↗Open in split view
Given the circuit, determine the voltages in the nodes.
Below my solution:
r10,1,0,1/3
r12,1,2,1/2
r13,1,3,1/2
r23,2,3,1/6
r20,2,0,1/8
j12,1,2,17
j03,0,3,2The answers you want are v1, v2 and v3, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
The answer, {-2.,1.,.5}, indicates v1=-2V, v2=1V, v3=0.5V. This is correct.
HK5's Figure 1-24b (Expert)
Open in app ↗Open in split view
Determine ix and vx in the following circuit.
My solution:
j6,0,1,6
r5,1,2,5
r2,2,0,2
r1,1,3,1
r3,0,4,3
j10,2,3,10
rx,3,4,rxSet Analysis to DC — direct current. This one needs Enable Expert Mode ticked in the Expert Mode box; the equations and unknowns go in the fields it reveals.
The answers you want are ir1 and vrx, in Results.
Select DC. Add equation ir2=4. Add unknown rx. Run the simulation. The answer, {-8,80}, means that IX is -8A and that VX is 80V.
HK5's Example 2.2 (Conductances)
Open in app ↗Open in split view
Determine the voltages in the nodes. My solution:
j01,0,1,-8
j30,3,0,-25
j21,2,1,-3
r12,1,2,1/3
r23,2,3,1/2
r13,1,3,1/4
r20,2,0,1
r30,3,0,1/5The answers you want are v1, v2 and v3, in Results.
The answer, {1,2,3}, is correct: v1=1V, v2=2V, v3=3V.
B11's Example 8.5
Open in app ↗Open in split view
Determine the current I2 in the circuit shown. My solution:
j1,1,0,4
r1,1,0,3
e2,1,2,5
r2,0,2,2This gives us a value for I2 of 3.4 A. This is correct.
Bo2's Example 2.5 (Conductances)
Open in app ↗Open in split view
Determine the voltages in the nodes, and the current through the voltage source.
j,0,1,3
e,3,2,3
r12,1,2,1/7
r20,2,0,1/3
r30,3,0,1/5
r13,1,3,1/2The answers you want are v1, v2 and v3, in Results.
Ask Evaluate for:
-ieThe calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
The answer, {-.5,-1.5,1.5,11.5}, is correct: v1=-.5, v2=-1.5, v3=1.5, i=11.5
B11's Example 8.22
Open in app ↗Open in split view
Determine V1 and V2. My solution:
j1,0,1,6
r1,1,0,4
e,1,2,12
r3,1,2,10
r2,2,0,2
j2,2,0,4The answers you want are v1 and v2, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
The answer, {10.67,-1.33}, tells us that V1 is 10.67V and V2 is -1.33V.
B11's Example 8.19
Open in app ↗Open in split view
Determine V1, I1 and I2, in the circuit:
My solution:
e,2,0,24
r1,1,2,6
r2,1,0,12
j,0,1,1The answer, {20.,-.667,1.67}, tells us that V1 is 20V, I1 is -.667A and I2 is 1.67 A.
B11's Example 8.14
Open in app ↗Open in split view
Determine I2 and I3.
e1,1,0,20
r1,1,2,6
r2,2,a,4
j,a,0,4
r3,a,3,2
e2,0,3,12The answers you want are ir2 and ir3, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
Answer: {3.33,-.667}. This is correct.
B11's Example 8.20
Open in app ↗Open in split view
Determine V1, V2, I1, I2 and I3. My solution:
e,3,0,64
r1,3,1,8
r2,1,2,4
j,1,2,2
r3,2,0,10The answers you want are v1, v2, ir1, ir2 and ir3, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
Answer: {37.82,32.73,3.27,1.27,3.27}. You should know how to read these by now, but here it is just in case: V1 = 37.82V, V2 = 32.73V, I1 = 3.27A, I2 = 1.27A, I3 = 3.27A.
RM3's Example 9-12 (solve)
Step 1Open in app ↗Open in split viewStep 2Open in app ↗Open in split view
If R3 is to be replaced with R4 and I4, determine the value and direction of the source.
So that we can keep the name of node b, we will use the top left node as reference.
First, make sure you understand what this problem is asking you to do. The idea is that, despite the change, we keep the same voltage drop and current flow between nodes a and b. We must first know what they are. So we simulate the original circuit:
e,0,b,20
r1,0,a,16
r2,a,b,40
r3,a,b,60The answer you want is ir3, in Results.
Ask Evaluate for:
va-vbWe find that the voltage drop is 12V and the current is 0.2A. These are the currents and voltages that we have to keep once we do the replacement. Simulate the circuit now replacing R3 with a resistor R4 of 240Ω and a source j with value I4. Run this:
e,0,b,20
r1,0,a,16
r2,a,b,40
r4,a,b,240
j,a,b,i4Notice that both the voltage drop (given by va-vb) and the current (given by ir4+ij) are algebraic functions in terms of i4. Now you can find i4 solving by voltage drop:
Both routes go in the Solve card, which solves against the answers the circuit just produced:
va-vb = 12with i4 as the unknown — or ir4+ij = 0.2 for the same answer by current instead.
The result is the same: i4 = .15 A. The required current source is .15A from a to b.
RM3's Example 8-13
Open in app ↗Open in split view
Solve for the currents through R2 and R3 in the circuit shown. My solution:
r1,a,0,10'k
r2,1,0,5'k
r3,b,a,6'k
r4,0,2,16'k
j,a,b,2'm
e1,1,b,10
e2,b,2,8The answers you want are ir2 and ir3, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
The answer, {.00154,.00111} is correct: IR2 = 1.54 mA and IR3 = 1.11 mA.
HK5's Figure 1-24c (Expert)
Open in app ↗Open in split view
Determine ix and vx in the following circuit.
e,1,0,60
r8,1,2,8
r10,2,0,10
r4,2,3,4
r2,3,0,2
j,0,3,ixSet Analysis to DC — direct current. This one needs Enable Expert Mode ticked in the Expert Mode box; the equations and unknowns go in the fields it reveals.
The answers you want are ix and v3, in Results.
Select DC. Add equation ir8=5. Add unknown ix. Run the simulation, and you will get: {1,8}. This is correct: IX is 1A and that VX is 8V.
Bo2's Example 1.9
Open in app ↗Open in split view
Determine I1, I2 and v.
This is my solution.
ji,0,1,2
r1,1,0,3
jd,0,1,4*v1
r2,1,0,5The answers you want are ir1, v1 and ir2, in Results.
The answer, {-5/26,-15/26,-3/26}, is correct: I1=-5/26 A, I2=-3/26 A and v=-15/26 V.
AS2's Practice Problem 2.7
Open in app ↗Open in split view
Find vo and io in the circuit. My solution:
ji,0,o,6
ro,o,0,2
jd,o,0,iro/4
r8,o,0,8The answers you want are vo and iro, in Results.
The answer, {8,4}, is correct: vo = 8 V and io = 4 A.
HK5's Example 1-3
Open in app ↗Open in split view
Determine the power delivered by each source and consumed by both resistors.
ei,1,0,120
r1,1,2,30
ed,2,3,2*vra
ra,0,3,15Ask Evaluate for:
-pei
-ped
pr1+praThe answer, {960,1920,2880}, is right: the independent source delivers 960W, the dependent source delivers 1920W, and the resistors consume 2880W together.
AS2's Practice Problem 2.6
Open in app ↗Open in split view
Find vx and vo in the circuit. My solution:
ei,x,1,35
rx,x,0,10
ed,0,2,2*vx
ro,1,2,5The answers you want are vx and vro, in Results.
The answer, {10,-5}, is correct: vx =10 and vo =-5.
Bo2's Example 1.10
Open in app ↗Open in split view
Determine v1, v2 and i. My solution:
ei,1,0,2
r1,1,2,1/3
ed,3,2,4*ir1
r2,3,0,1/5The answers you want are ir1, vr1 and vr2, in Results.
The answers, {-15/26,-5/26,-3/26}, is correct: v1=-5/26 V, v2=-3/26 V and i=-15/26 A.
AS2's Example 2.6
Open in app ↗Open in split view
Determine vo and i in the circuit. My solution is shown below:
e12,1,o,12
ri,1,2,4
ed,2,3,2*vo
e4,0,3,4
ro,o,0,6The answers you want are vo and iri, in Results.
The answer, {48,-8}, is correct: vo =48 and i = -8.
HK5's Drill Problem 1.11
Open in app ↗Open in split view
Find the power absorbed by each element in the circuit. My solution:
r1,x,0,30
ei,1,x,12
r2,1,2,8
r3,2,3,7
ed,3,0,4*vxThe answers you want are pr1, pei, pr2, pr3 and ped, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
The answer, {.768,1.92,.2048,.1792,-3.072}, is correct.
AS2's Example 3.6
Open in app ↗Open in split view
Determine the value of Io in the circuit. My solution:
ei,a,0,24
ro,a,b,10
r12,b,0,12
r4,b,c,4
r24,a,c,24
ed,c,0,4*iroThe answer you want is iro, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
The answer, 1.5 A, is correct.
Bo2's Drill Exercise 1.12
Open in app ↗Open in split view
Determine i, v and id.
We are given an unnecessary piece of information: the 4V drop in the 2Ω resistor.
My solution:
ei,1,0,10
r1,1,2,1
r2,2,3,2
r3,2,0,3
r4,3,0,2
ed,2,3,ir1/2The answers you want are ir1, vr3 and ied, in Results.
The answer, {4,6,1}, is correct: i=4, v=6 and id=1.
AS2's Example 3.2
Open in app ↗Open in split view
Determine the voltages at the nodes.
My solution:
ji,0,1,3
jd,3,0,2*ir2
r2,1,2,2
r4a,1,3,4
r8,2,3,8
r4b,2,0,4The answers you want are v1, v2 and v3, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
The answer, {4.8,2.4,-2.4}, is correct.
AS2's Example 3.4
Open in app ↗Open in split view
Find the node voltages in the circuit. My solution:
r2,1,0,2
e1,1,2,20
j,0,2,10
r6,2,3,6
rx,1,4,3
r4,3,0,4
ed,3,4,3*vrx
r1,4,0,1The answers you want are v1, v2, v3 and v4, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
The answer, {26.67,6.67,173.33,-46.67}, is correct.
Bo2's Drill Exercise 2.6
Open in app ↗Open in split view
Determine the voltage in each node.
Notice that in the schematic, the resistors' values are given in siemens. Symbulator has to be fed the resistors with values in ohms. This means that, when we simulate them, we have to convert them from siemens to ohms by dividing 1 over the siemens value.
My solution:
j,0,1,6
ei,3,1,6
ed,2,3,3*v1
r5,1,0,1/5
r2,1,2,1/2
r3,2,0,1/3
r1,2,3,1
r4,3,0,1The answers you want are v1, v2 and v3, in Results.
The answer, {-1,2,5}, is correct.
Bo2's Example 2.7
Open in app ↗Open in split view
Determine the voltages in all nodes. My solution:
j,0,1,1
r3,0,1,3
r4,2,1,4
r1,2,0,1
r2,2,3,2
r5,3,0,5
ei,3,4,1.5
ed,4,0,2*vr4The answers you want are v1, v2, v3 and v4, in Results.
The answer, {1.5,-.5,-2.5,-4.}, is correct.
Bo2's Example 2.6
Open in app ↗Open in split view
Determine the voltages in all nodes.
My solution:
e1,0,1,1
e2,3,4,.5
ed,3,2,3*vr4
j,0,4,2
r4,1,2,1/4
r1,2,0,1
r8,3,0,1/8
r2,2,4,1/2The answers you want are v1, v2, v3 and v4, in Results.
The answer, {-1,-2.,1.,.5}, is correct.
HK5's Drill Problem 1-12
Open in app ↗Open in split view
Find iA, iB and iC.
My solution:
jl,x,0,5.6
ra,0,x,18
jb,0,x,.1*vx
r9,0,x,9
jr,0,x,2The answers you want are ira, ijb and ir9, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
The answer, {3.,-5.4,6.}, is correct.
Numerical-from-symbolic examples
Bo2's Drill Exercise 1.10
Open in app ↗Open in split view
Determine i, v, is and vs.
With one unknown value and one known solution, this problem is a job for Expert.
Determine i, v, is and vs.
es,2,0,vs
jd,0,3,2*ir1
r7,0,1,7
r1,3,1,1
r3,3,2,3
r4,1,2,4Set Analysis to DC — direct current. This one needs Enable Expert Mode ticked in the Expert Mode box; the equations and unknowns go in the fields it reveals.
The answers you want are ir1, vjd and vs, in Results.
Ask Evaluate for:
-iesSelect DC, Add vr4=4 to the equations and vs to the unknowns. Run the simulation. The answer, {2,-9,-3,3}, is correct: i=2, v=-9, is=-3 and vs=3.
Bo2's Drill Exercise 1.11
Open in app ↗Open in split view
Determine i, v and vd. (Since all element values are known, the tip we are given by the book – namely, that the voltage drop in the 6Ω resistor is 1.5V – is totally superfluous.)
e,1,0,12
r1,1,2,1
r4,2,0,4
r10,2,3,10
r6,3,0,6
r2,3,4,2
j,4,0,vr10/15The answers you want are vr10, ir2 and vj, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
The answer, {7.5,.5,.5} is correct: i=.5, v=7.5 and vd=.5.
Symbolic examples
TR5's Exercise 4.2
Open in app ↗Open in split view
Find vO and iO in terms of iS.
This is my solution.
ji,0,x,is1
r1,x,0,1'k
r2,x,o,2'k
jd,0,o,vx/500
ro,o,0,500The answers you want are vo and iro, in Results.
The answer, {1000*is,2*is}, is correct: vO=1000 iS and iO=2 iS. Version 9's panel names the source's value is1, so its answers read 1000*is1 and 2*is1.
TR5's Example 4.4
Open in app ↗Open in split view
Find vO and the equivalent resistance RIN, in terms of vS, when R1 is 50 Ω, R2 is 1 kΩ, R3 is 100 Ω, R4 is 5 kΩ and g is 100 mS (i.e. 100'm).
This is my solution.
e,1,0,vs
r1,1,2,50
r2,2,o,1'k
r3,o,0,100
r4,o,0,5'k
j,0,o,100'm*vr2The answers you want are vo and re, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
The answer, {.904*vs,10952.}, is correct: vO=.904 vS and RIN = 10.95 kΩ.
TR5's Figure 4-4
Open in app ↗Open in split view
Find the voltage drop, current, and power consumed by the 500Ω resistor, and the ratio of that power to that delivered by the independent source, all in terms of iS.
I have not labeled the nodes in the figure, so you can practice doing it. My solution:
js,0,1,is1
r50,1,0,50
rx,1,0,25
jd,o,0,48*irx
r3,o,0,300
ro,o,0,500The answers you want are iro, vo and pro, in Results.
Ask Evaluate for:
pro/(-pjs)The answers we get are correct: iO=-12is, vO=-6000is, pO=72000is2, and pO/pS=4320. Version 9's panel names the source's value is1, so its answers carry is1 where these carry is.
TR5's Example 4.1 (Symbolic)
Open in app ↗Open in split view
Find vO.
I did not label the nodes, so you can practice.
ei,1,0,vs
rs,1,2,rs
rx,2,0,rp
ed,0,3,r*irx
rc,3,o,rc
rl,o,0,rlThe answer you want is vo, in Results.
The answer, shown first below, is correct. The textbook's answer follows it.
Bo2's Example 1.11 (Symbolic)
Open in app ↗Open in split view
Determine v2.
My solution: In my solution I named the value of the source v1, to keep it similar to the book. This required avoiding naming any node as 1: if there was a node 1, Symbulator would store in v1 the voltage of the node, creating trouble. There is no problem with using r1 as a value, since nothing will be stored in that r1 value.
e,a,0,v1
r1,a,3,r1
rg,3,0,rg
j,2,0,gm*vrg
rd,2,0,rd
rl,2,0,rlThe answer you want is v2, in Results.
The answer I got is shown first below (the textbook's follows it.)
TR5's Example 4-7 (Symbolic)
Open in app ↗Open in split view
Find RIN, i.e. the resistance as seen by the current source. My solution:
ji,0,a,is1
re1,a,0,re1
jd,b,a,β*is1
rl,b,0,rlThe answer you want is rji, in Results.
The answer we get — re1*(β+1), version 9's panel having renamed the element — is correct, as can be seen by comparing it to the textbook's answer.
TR5's Exercise 4.3 (Symbolic)
Open in app ↗Open in split view
Find vO, in terms of the value in the circuit. For resistors, use their conductance value.
This is my solution. We use µ as a constant in the dependent source. It is not confused with the SI prefix for micro because the prefix has an apostrophe.
ei,1,0,vs
ed,2,0,μ*(vrx)
r1,1,2,1/g1
r2,2,o,1/g2
rx,1,o,1/gx
rl,o,0,1/glThe answer you want is vo, in Results.
This is the answer we get. It is correct. Compare it with the textbook's answer below it.
TR5's Example 4.5 (Symbolic)
Open in app ↗Open in split view
Find iB.
My solution is shown below.
e1,1,0,vcc
rb,1,b,rb
e2,e,b,vγ
re1,e,0,re1
rc,1,c,rc
j,c,e,β*irbThe answer you want is irb, in Results.
Compare my answer, first below, to the book's answer beneath it. Note re1: as in the previous problem, version 9's panel renamed the element.
The last four problems show Symbulator at its DC best. I don't know of any calculator-based program that was able to provide this kind of purely symbolic answer to a circuit simulator back in 1999 when I made Symbulator. As a matter of fact, even today – a quarter of a century later - I know of no other calculator-based simulator that can do this.