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Symbulator 9 9

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 function 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 function 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

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Given the circuit, determine the current and voltage drop in R₁.

B11's Example 8.1

Solution

The circuit:

Circuit Description

j,0,1,10'm
r1,1,0,20'k

Two lines, two elements. Leave Type of analysis on Solve circuit, set Analysis to DC — direct current, and click Run Symbulator. The circuit is numerical, so set Rounding to approx to n digits with n = 3 and tick Use SI prefixes, as in Lesson 1.

We read:

current through r1

\[i_{r1} = 10\ \mathrm{mA}\]

voltage drop in r1

\[v_{r1} = 200\ \mathrm{V}\]

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

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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:

Circuit Description

j10,1,0,2
r12,1,2,1
r20,2,0,1/4
r30,3,0,1/3
r13,1,3,1/2
j32,3,2,3

Run it in DC. The answers are 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 has a function for reducing resistors (and impedances) in parallel, pr(). You can use it in the circuit description and in the Evaluate area.

To reduce three resistors in parallel with values of 10 Ω, 20 Ω and 30 Ω:

Evaluate

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, as the two problems below show, and it works on symbols: pr(r1, r2) gives

\[\dfrac{r_{1}\,r_{2}}{r_{1} + r_{2}}\]

AS7's Example 2.10

Find the equivalent resistance.

AS7's Example 2.10

Solution

No simulation needed: the whole reduction is one nested expression in Evaluate, from the far end of the ladder outward:

Evaluate

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:

Evaluate

16 + pr(18, 9, 2 + pr(20, 1 + pr(5, 20)))

The answer is 19 Ω. This is correct.

The [r,r,r…] shorthand

There is a shorthand for pr: put the resistor values in square brackets, separated by commas. Expressions like [10,20,30] or [r1,r2,r3,r4] are read as inputs to pr, both in the Evaluate area or in the circuit description.

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.

The [6,3] below is the shorthand at work:

Circuit Description

e,1,0,16.8
r1,1,2,9
r2,1,2,6
r3,2,3,4
re1,3,0,[6,3]
r6,2,0,3

Run it in DC. The current the source delivers is the opposite of ie: 3 A.

An example where using pr makes no sense is B11's Example 6.13, because you need to know the current through each resistor.

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:

Circuit Description

jt,0,1,12'm
r1,1,0,1'k
re,1,0,[10'k,22'k]

Moving forward, we will use pr(,,), 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 answer names 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

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Determine the values of V_S, I₁ and I₂.

Circuit Description

j,0,1,7
e,1,0,12
r1,1,0,4

The answers you want are v1, ie and ir1, in Results.

The answers show V_S is 12V, I₁ is 4A and I₂ is 3A. All correct.

B11's Example 8.15

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Determine the current through each resistor. My solution, with the direction marked in blue:

Circuit Description

j6,0,1,6
r2,1,0,2
r6,1,2,6
r8,0,2,8
j8,2,0,8

The answers you want are ir2, ir6 and ir8, in Results.

Here we use Rounding — approx to n digits, with n = 3.

The answer is I_R2 = 1.25A, I_R6 = 4.75A and I_R8 = 3.25A.

B11's Example 8.21

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Determine the voltage in each node and the current through each resistor.

My solution:

Circuit Description

j1,0,1,4
r1,1,0,2
r3,1,2,12
r2,0,2,6
j2,2,0,2

The answers you want are v1, v2, ir1, ir2 and ir3, in Results.

The answer is V₁=6V, V₂=-6V, I_R1=3A and I_R2=I_R3=1A.

Bo2's Drill Exercise 2.2 (Conductances)

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Given the circuit, determine the voltages in the nodes.

Below my solution:

Circuit Description

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
j03,0,3,2

The answers you want are v1, v2 and v3, in Results.

Here we use Rounding — approx to n digits, with n = 3.

The answer is v₁=-2V, v₂=1V and v₃=0.5V. This is correct.

HK5's Figure 1-24b (Expert)

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Determine i_x and v_x in the following circuit.

My solution:

Circuit Description

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,rx

Set 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.

In DC, with ir2=4 in Add equation(s) and rx in Add unknown(s), run it. The answer is i_x = -8A and v_x = 80V.

HK5's Example 2.2 (Conductances)

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Determine the voltages in the nodes. My solution:

Circuit Description

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
r30,3,0,1/5

The answers you want are v1, v2 and v3, in Results.

The answer is v₁=1V, v₂=2V and v₃=3V. This is correct.

B11's Example 8.5

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Determine the current I₂ in the circuit shown. My solution:

Circuit Description

j1,1,0,4
r1,1,0,3
e2,1,2,5
r2,0,2,2

This gives us a value for I₂ of 3.4 A. This is correct.

Bo2's Example 2.5 (Conductances)

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Determine the voltages in the nodes, and the current through the voltage source.

Circuit Description

j,0,1,3
e,3,2,3
r12,1,2,1/7
r20,2,0,1/3
r30,3,0,1/5
r13,1,3,1/2

The answers you want are v1, v2 and v3, in Results; i is the opposite of ie.

Here we use Rounding — approx to n digits, with n = 3.

The answer is v₁=-.5, v₂=-1.5, v₃=1.5 and i=11.5. This is correct.

B11's Example 8.22

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Determine V₁ and V₂. My solution:

Circuit Description

j1,0,1,6
r1,1,0,4
e,1,2,12
r3,1,2,10
r2,2,0,2
j2,2,0,4

The answers you want are v1 and v2, in Results.

Here we use Rounding — approx to n digits, with n = 3.

The answer is V₁ = 10.67V and V₂ = -1.33V.

B11's Example 8.19

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Determine V₁, I₁ and I₂, in the circuit:

My solution:

Circuit Description

e,2,0,24
r1,1,2,6
r2,1,0,12
j,0,1,1

The answer is V₁ = 20V, I₁ = -.667A and I₂ = 1.67 A.

B11's Example 8.14

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Determine I₂ and I₃.

Circuit Description

e1,1,0,20
r1,1,2,6
r2,2,a,4
j,a,0,4
r3,a,3,2
e2,0,3,12

The answers you want are ir2 and ir3, in Results.

Here we use Rounding — approx to n digits, with n = 3.

The answers are:

current through r2

\[i_{r2} = 3.33\ \mathrm{A}\]

current through r3

\[i_{r3} = -.667\ \mathrm{A}\]

This is correct.

B11's Example 8.20

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Determine V₁, V₂, I₁, I₂ and I₃. My solution:

Circuit Description

e,3,0,64
r1,3,1,8
r2,1,2,4
j,1,2,2
r3,2,0,10

The answers you want are v1, v2, ir1, ir2 and ir3, in Results.

Here we use Rounding — approx to n digits, with n = 3.

The answer is V₁ = 37.82V, V₂ = 32.73V, I₁ = 3.27A, I₂ = 1.27A and I₃ = 3.27A.

RM3's Example 9-12 (solve)

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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:

Circuit Description

e,0,b,20
r1,0,a,16
r2,a,b,40
r3,a,b,60

The answer you want is ir3, in Results.

Ask Evaluate for:

Evaluate

va-vb

We 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:

Circuit Description

e,0,b,20
r1,0,a,16
r2,a,b,40
r4,a,b,240
j,a,b,i4

Step 2Open in app ↗Open in split view

Notice 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:

Either equation goes in the Solve card:

Equation(s) to solve in terms of the results

va-vb = 12

with 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

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Solve for the currents through R₂ and R₃ in the circuit shown. My solution:

Circuit Description

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,2,8

The answers you want are ir2 and ir3, in Results.

Here we use Rounding — approx to n digits, with n = 3.

The answer is I_R2 = 1.54 mA and I_R3 = 1.11 mA. This is correct.

B11's Example 6.3 (Hidden source)

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Determine V_S and I₁.

The problem presents a seemingly ‘hanging' node with 20V. To simulate this, imagine a ‘hidden' 20V voltage source connected to the node.

Circuit Description

j,0,1,6
r1,1,2,2
r2,1,2,1
e,2,0,20

The answers you want are v1 and ir1, in Results.

The answer is V_S = 24V and I₁ = 2A. This is correct.

HK5's Figure 1-24c (Expert)

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Determine i_x and v_x in the following circuit.

Circuit Description

e,1,0,60
r8,1,2,8
r10,2,0,10
r4,2,3,4
r2,3,0,2
j,0,3,ix

Set 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.

In DC, with ir8=5 in Add equation(s) and ix in Add unknown(s), run it. The answer is i_x = 1A and v_x = 8V. This is correct.

B11's Example 6.22 (Hidden source)

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Determine I₁.

Although no source is shown, the circuit has a current. I use a ‘hidden' current source.

Circuit Description

jt,0,1,12'm
r1,1,0,1'k
r2,1,0,10'k
r3,1,0,22'k

ir1 gives I₁ = 10.48 mA, which is correct.

Partial reductionOpen in app ↗Open in split view

B11's Example 6.21 (Hidden source, Expert)

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Determine I_S, I₁ and I₃.

This ‘hidden source' problem is perfect for Expert Mode. My solution:

Circuit Description

js,0,1,is
r1,1,0,6
r2,1,0,3
r3,1,0,1

Choose DC. Open Expert Mode, put ir2 = 2'm in Add equation(s) and is in Add unknown(s), then Run Symbulator.

The answer is I_S = 9mA, I₁ = 1mA and I₃ = 6mA. This is correct.

Bo2's Example 1.9

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Determine I₁, I₂ and v.

This is my solution.

Circuit Description

ji,0,1,2
r1,1,0,3
jd,0,1,4*v1
r2,1,0,5

The answers you want are ir1, v1 and ir2, in Results.

The answer is I₁=-5/26 A, I₂=-3/26 A and v=-15/26 V. This is correct.

AS2's Practice Problem 2.7

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Find v_o and i_o in the circuit. My solution:

Circuit Description

ji,0,o,6
ro,o,0,2
jd,o,0,iro/4
r8,o,0,8

The answers you want are vo and iro, in Results.

The answer is v_o = 8 V and i_o = 4 A. This is correct.

HK5's Example 1-3

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Determine the power delivered by each source and consumed by both resistors.

Circuit Description

ei,1,0,120
r1,1,2,30
ed,2,3,2*vra
ra,0,3,15

The power each source delivers is the opposite of pei and of ped. For the resistors together, ask Evaluate for:

Evaluate

pr1+pra

The independent source delivers 960W, the dependent source delivers 1920W, and the resistors consume 2880W together. This is correct.

AS2's Practice Problem 2.6

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Find v_x and v_o in the circuit. My solution:

Circuit Description

ei,x,1,35
rx,x,0,10
ed,0,2,2*vx
ro,1,2,5

The answers you want are vx and vro, in Results.

The answer is v_x =10 and v_o =-5. This is correct.

Bo2's Example 1.10

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Determine v₁, v₂ and i. My solution:

Circuit Description

ei,1,0,2
r1,1,2,1/3
ed,3,2,4*ir1
r2,3,0,1/5

The answers you want are ir1, vr1 and vr2, in Results.

The answer is v₁=-5/26 V, v₂=-3/26 V and i=-15/26 A. This is correct.

AS2's Example 2.6

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Determine v_o and i in the circuit. My solution is shown below:

Circuit Description

e12,1,o,12
ri,1,2,4
ed,2,3,2*vo
e4,0,3,4
ro,o,0,6

The answers you want are vo and iri, in Results.

The answer is v_o =48 and i = -8. This is correct.

HK5's Drill Problem 1.11

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Find the power absorbed by each element in the circuit. My solution:

Circuit Description

r1,x,0,30
ei,1,x,12
r2,1,2,8
r3,2,3,7
ed,3,0,4*vx

The answers you want are pr1, pr2 and pr3 in Results, and for the two sources the power delivered row of each card, -pei and -ped.

Here we use Rounding — approx to n digits, with n = 3.

The answers are:

power consumed by r1

\[p_{r1} = .768\ \mathrm{W}\]

power delivered by ei

\[-p_{ei} = -1.92\ \mathrm{W}\]

power consumed by r2

\[p_{r2} = .2048\ \mathrm{W}\]

power consumed by r3

\[p_{r3} = .1792\ \mathrm{W}\]

power delivered by ed

\[-p_{ed} = 3.072\ \mathrm{W}\]

This is correct: the source ei delivers a negative power, which is to say it absorbs 1.92 W, and ed delivers 3.072 W.

AS2's Example 3.6

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Determine the value of I_o in the circuit. My solution:

Circuit Description

ei,a,0,24
ro,a,b,10
r12,b,0,12
r4,b,c,4
r24,a,c,24
ed,c,0,4*iro

The answer you want is iro, in Results.

Here we use Rounding — approx to n digits, with n = 3.

The answer, 1.5 A, is correct.

Bo2's Drill Exercise 1.12

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Determine i, v and i_d.

We are given an unnecessary piece of information: the 4V drop in the 2Ω resistor.

My solution:

Circuit Description

ei,1,0,10
r1,1,2,1
r2,2,3,2
r3,2,0,3
r4,3,0,2
ed,2,3,ir1/2

The answers you want are ir1, vr3 and ied, in Results.

The answer is i=4, v=6 and i_d=1. This is correct.

AS2's Example 3.2

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Determine the voltages at the nodes.

My solution:

Circuit Description

ji,0,1,3
jd,3,0,2*ir2
r2,1,2,2
r4a,1,3,4
r8,2,3,8
r4b,2,0,4

The answers you want are v1, v2 and v3, in Results.

Here we use Rounding — approx to n digits, with n = 3.

The answers are:

voltage of node 1

\[v_{1} = 4.8\ \mathrm{V}\]

voltage of node 2

\[v_{2} = 2.4\ \mathrm{V}\]

voltage of node 3

\[v_{3} = -2.4\ \mathrm{V}\]

This is correct.

AS2's Example 3.4

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Find the node voltages in the circuit. My solution:

Circuit Description

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
r1,4,0,1

The answers you want are v1, v2, v3 and v4, in Results.

Here we use Rounding — approx to n digits, with n = 3.

The answers are:

voltage of node 1

\[v_{1} = 26.67\ \mathrm{V}\]

voltage of node 2

\[v_{2} = 6.67\ \mathrm{V}\]

voltage of node 3

\[v_{3} = 173.33\ \mathrm{V}\]

voltage of node 4

\[v_{4} = -46.67\ \mathrm{V}\]

This is correct.

Bo2's Drill Exercise 2.6

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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:

Circuit Description

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
r4,3,0,1

The answers you want are in Results:

voltage of node 1

\[v_{1} = -1\ \mathrm{V}\]

voltage of node 2

\[v_{2} = 2\ \mathrm{V}\]

voltage of node 3

\[v_{3} = 5\ \mathrm{V}\]

This is correct.

Bo2's Example 2.7

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Determine the voltages in all nodes. My solution:

Circuit Description

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*vr4

The answers you want are in Results:

voltage of node 1

\[v_{1} = 1.5\ \mathrm{V}\]

voltage of node 2

\[v_{2} = -.5\ \mathrm{V}\]

voltage of node 3

\[v_{3} = -2.5\ \mathrm{V}\]

voltage of node 4

\[v_{4} = -4\ \mathrm{V}\]

This is correct.

Bo2's Example 2.6

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Determine the voltages in all nodes.

My solution:

Circuit Description

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
r2,2,4,1/2

The answers you want are in Results:

voltage of node 1

\[v_{1} = -1\ \mathrm{V}\]

voltage of node 2

\[v_{2} = -2\ \mathrm{V}\]

voltage of node 3

\[v_{3} = 1\ \mathrm{V}\]

voltage of node 4

\[v_{4} = .5\ \mathrm{V}\]

This is correct.

HK5's Drill Problem 1-12

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Find i_A, i_B and i_C.

My solution:

Circuit Description

jl,x,0,5.6
ra,0,x,18
jb,0,x,.1*vx
r9,0,x,9
jr,0,x,2

The answers you want are ira, ijb and ir9, in Results.

Here we use Rounding — approx to n digits, with n = 3.

The answers are:

current through ra

\[i_{ra} = 3\ \mathrm{A}\]

current through jb

\[i_{jb} = -5.4\ \mathrm{A}\]

current through r9

\[i_{r9} = 6\ \mathrm{A}\]

This is correct.

Numerical-from-symbolic examples

Bo2's Drill Exercise 1.10

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Determine i, v, i_s and v_s.

With one unknown value and one known solution, this problem is a job for Expert Mode.

Determine i, v, i_s and v_s.

Circuit Description

es,2,0,vs
jd,0,3,2*ir1
r7,0,1,7
r1,3,1,1
r3,3,2,3
r4,1,2,4

Set 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; i_s is the opposite of ies.

In DC, add vr4=4 to the equations and vs to the unknowns, and run it. The answer is i=2, v=-9, i_s=-3 and v_s=3. This is correct.

Bo2's Drill Exercise 1.11

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Determine i, v and v_d. (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.)

Circuit Description

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,vr10/15

The answers you want are vr10, ir2 and vj, in Results.

Here we use Rounding — approx to n digits, with n = 3.

The answer is i=.5, v=7.5 and v_d=.5. This is correct.

Symbolic examples

TR5's Exercise 4.2

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Find v_O and i_O in terms of i_S.

This is my solution.

Circuit Description

ji,0,x,is
r1,x,0,1'k
r2,x,o,2'k
jd,0,o,vx/500
ro,o,0,500

The answers you want are vo and iro, in Results.

The answer is v_O=1000 i_S and i_O=2 i_S. This is correct.

TR5's Example 4.4

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Find v_O and the equivalent resistance R_IN, in terms of v_S, 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.

Circuit Description

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*vr2

The answers you want are vo and re, in Results.

Here we use Rounding — approx to n digits, with n = 3.

The answer is v_O=.904 v_S and R_IN = 10.95 kΩ. This is correct.

TR5's Figure 4-4

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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 i_S.

I have not labeled the nodes in the figure, so you can practice doing it. My solution:

Circuit Description

js,0,1,is
r50,1,0,50
rx,1,0,25
jd,o,0,48*irx
r3,o,0,300
ro,o,0,500

The answers you want are iro, vo and pro, in Results.

Ask Evaluate for:

Evaluate

pro/(-pjs)

The answers we get are correct: i_O=-12is, v_O=-6000is, p_O=72000is², and p_O/p_S=4320.

TR5's Example 4.1 (Symbolic)

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Find v_O.

I did not label the nodes, so you can practice.

Circuit Description

ei,1,0,vs
rs,1,2,rs
rx,2,0,rp
ed,0,3,r*irx
rc,3,o,rc
rl,o,0,rl

The answer you want is vo, in Results.

The answer, shown first below, is correct. The textbook's answer follows it.

voltage of node o

\[v_{o} = -\dfrac{r\,rl\,vs}{rc\,rp + rc\,rs + rl\,rp + rl\,rs}\]

Bo2's Example 1.11 (Symbolic)

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Determine v₂.

My solution. I named the source's value v₁, as in the book, so no node may be called 1: its voltage would be v1 too. Using r1 as a value is fine.

Circuit Description

e,a,0,v1
r1,a,3,r1
rg,3,0,rg
j,2,0,gm*vrg
rd,2,0,rd
rl,2,0,rl

The answer you want is v2, in Results.

The answer I got is shown first below (the textbook's follows it.)

voltage of node 2

\[v_{2} = -\dfrac{gm\,rd\,rg\,rl\,v_{1}}{r_{1}\,rd + r_{1}\,rl + rd\,rg + rg\,rl}\]

TR5's Example 4-7 (Symbolic)

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Find R_IN, i.e. the resistance as seen by the current source. My solution:

Circuit Description

ji,0,a,is
re1,a,0,re1
jd,b,a,β*is
rl,b,0,rl

The answer you want is rji, in Results.

The answer we get — re1*(β+1) — is correct, as can be seen by comparing it to the textbook's answer.

TR5's Exercise 4.3 (Symbolic)

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Find v_O, 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.

Circuit Description

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/gl

The answer you want is vo, in Results.

The answer is correct; the textbook's follows it.

voltage of node o

\[v_{o} = \dfrac{vs\,(g_{2}\,\mu + gx)}{g_{2}\,\mu + g_{2} + gl + gx}\]

TR5's Example 4.5 (Symbolic)

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Find i_B.

My solution is shown below.

Circuit Description

e1,1,0,vcc
rb,1,b,rb
e2,e,b,vγ
re1,e,0,re1
rc,1,c,rc
j,c,e,β*irb

The answer you want is irb, in Results.

Compare my answer, first below, to the book's answer beneath it.

current through rb

\[i_{rb} = \dfrac{vcc + v\gamma}{r_{e1}\,\beta + r_{e1} + rb}\]

The answer given by the book uses variables with subscripts, which look very elegant. Symbolic answers can be made to read that way too, and the trick is one character: an underscore in a symbolic variable name puts what follows into a subscript. Underscores, and subscripts in your answers shows it on this very circuit — optional reading, and it changes no answer, only its look.

The last four problems show Symbulator at its DC best.

The Showing-off Problem (Expert)

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In 2013, I invented a problem to showcase how easy it is to describe dependent sources in Symbulator.

Question: Find positive values for V_s and I_s that will result in 80W delivered by the VCCS and 0W dissipated in the CCVS.

Use this circuit description and equations:

Circuit Description

es,e,0,vs
js,0,d,is
r1,e,m,10
r2,a,e,20
r3,m,0,30
r4,b,m,40
r5,n,m,50
r6,c,d,60
r7,n,d,70
jd1,a,b,0.2*v_r7
ed2,c,b,0.1*i_r5
jd3,n,c,2*i_r1
ed4,0,n,0.7*v_r6

Symbulator reports the power consumed by each element, so delivering 80W is consuming −80W. In Expert Mode:

Add equation(s)

pjd1 = -80 and ped2 = 0

Add unknown(s)

vs, is

The problem asks for positive values, which is what the conditions are for: the two power equations are quadratic, so four solutions fit, and the conditions pick the one meant:

Add condition(s)

is > 0 and vs > 0

Run it in DC, with Rounding at 4 significant digits. Everything comes back numerical: v_s = 17.61 V and i_s = 0.3973 A.

The answers also show how the CCVS dissipates nothing: i_R5 comes back exactly 0 A, and its voltage is 0.1 i_R5, so the source itself is zero.