Direct current analysis

I firmly believe one learns best by doing, so I have written this documentation as a tutorial, where you learn Symbulator by working through a series of progressively more diverse and complex examples. In a moment, I'll teach you how to swim. But for now, let me throw you in the deep end!

1.1Run a direct current analysis

Open Symbulator 9 — the local app if you have it installed, otherwise the online app in a browser. Click Clear all inputs, at the top right, so every input field is empty. Then type this into the Circuit Description box, one element per line:

e,1,0,36
r1,1,2,1'k
r2,2,3,3'k
r3,3,0,2'k

Scroll down to Analysis & Settings (if it is collapsed, open it with the blue triangle to its left). Set Type of analysis to Solve circuit and Analysis to DC — direct current. Then click Run Symbulator.

The button reads Solving… for a moment, then Solved!. Scroll down to Results, under OUTPUTS: there is an answer for every node and every element of the circuit.

Congratulations! You have just run your first simulation in Symbulator 9.

How does it work

Let's now go over what we just did, one piece at a time.

The two menus in Analysis & Settings tell Symbulator what to do (in this case, solve the circuit) and which analysis to run (in this case, in direct current). Everything else there is optional; later lessons introduce it, and until then leave it as it is.

The most important input is the description of the circuit you want to analyse. It goes in the Circuit Description area, as a series of lines of text, each describing one element.

e,1,0,36
r1,1,2,1'k
r2,2,3,3'k
r3,3,0,2'k

Using one line per element keeps a long circuit readable, and it makes mistakes easier to find. But it is not the only option: Symbulator also accepts a colon as a separator. For example, this is the same circuit, described in one line:

e,1,0,36:r1,1,2,1'k:r2,2,3,3'k:r3,3,0,2'k

To reuse this circuit later, save it as an entry in an input file. You will be asked to name both. Input files must have the .cir extension.

Look at the elements

Now let's examine the contents of the circuit description. Study it, and you will see it has four lines. Each line describes one element.

At the moment they may look cryptic. But by the end of this lesson you will read them with ease.

1.2How to describe a resistor

Notice that the second, third and fourth elements in the list all start with the letter r. These are resistors.

How to name resistors

Every resistor must have a name. As long as it starts with the letter r and is unique, it can be whatever you want. Unique means no other element or node has the same name.

How to name nodes

Naming the nodes is the first thing we do to solve a circuit. Every node must have a name, and node names must be unique: no other node or element may have the same one. Otherwise they can be whatever you want, with one constraint: at least one node should be called 0 (zero). This will be considered the ground node of your circuit and will have, by definition, a voltage of 0 volts.

Values can use SI prefixes

The values of circuit elements are often given with prefixes of the International System (SI) — kilo, milli, micro. So Symbulator has a shorthand for them: an SI prefix in an element's value, preceded by an apostrophe, multiplies it by the corresponding factor, exactly.

Eleven prefixes are accepted, from peta down to atto. “SI prefixes” lists them all with their factors, and covers the two where case matters — optional reading; 'k and 'm are the two this lesson uses.

For example, an 8 kΩ resistor can be entered as 8000 or as 8'k, and both are exact. 8000. and 8E3 are the same resistance approximately, which is a real difference once an answer is symbolic; “SI prefixes” sets the four side by side.

1.3What answers do you get

After the simulation in DC is complete, Symbulator fills in the Results section under OUTPUTS with every answer it worked out.

The results are arranged in groups: Node voltages first, one per node, and then Results by element, one group per element, headed by its name and its kind — r1 resistor, e voltage source.

Every answer is written the way you would write it by hand, as a named quantity and its value: v₁ = 36 V, i_r1 = 6 mA.

How answers are shown

By default, Symbulator answers exactly: it works symbolically, so a current comes back as 3/500 A rather than a decimal. That is what you want for symbolic results. For numerical work like the example in this lesson, decimals read better. Open the Settings card and set Rounding to approx to n digits with n = 3, then tick Use SI prefixes in answers. The current above then reads 6 mA instead of 3/500 A.

Ticking the prefix box moves Rounding off exact by itself, since a prefixed value is a decimal, and choosing exact again switches the prefixes off. Symbulator says so on screen when it does.

In this tutorial, numerical examples assume approximate answers to three or four figures with prefixes on, and symbolic ones assume exact. The text says so when a change is needed.

The card holds a few more controls than these two, and the lessons reach for them as they come up. “Settings” is the whole card in one place — optional reading, and nothing here waits on it.

Answer for each node

Symbulator calculates for each node a voltage with reference to ground, which is shown under Node voltages as v plus the node's name: node 1's voltage is v1.

Answers for each resistor

Symbulator calculates for each resistor the following answers:

• The voltage drop in the resistor, defined as the voltage in the first node minus the voltage in the second node, in volts. For a resistor called r5, this is given in vr5.
• The current through the resistor, flowing from the first node towards the second, in amperes. For a resistor called rx, this is given in irx.
• The power consumed by the resistor, in watts. For a resistor called r12, this is given in pr12.

You may write the name of any of Symbulator's answers in two ways: with or without an underscore. ir1 and i_r1 are the same current, v2 and v_2 the same voltage; capitals make no difference either. This tutorial uses the shorter spelling throughout. Underscores, and subscripts in your answers says why both exist, and what an underscore does in a value, which is a different thing entirely — optional reading.

By now you should understand the description of the second, third and fourth elements in our example.

1.4How to describe a voltage source

The first element in our example, on the other hand, started with the letter e. This element is a voltage source.

How to name voltage sources

Every voltage source in your circuit must have a unique name that starts with the letter e, and it can be whatever you want.

Values can use SI prefixes

A voltage source's value can use SI prefixes too — they work in the values of every element.

Answers for a voltage source

For each voltage source, the following answers are calculated:

• The voltage drop in the source, defined as the voltage in the first node minus the voltage in the second, in volts. For a source called e5, ve5.
• The current through the source, flowing from the first node towards the second, in amperes. For a source called ex, iex.
• The power consumed — attention: not delivered, but consumed — by the source, in watts. For a source called e12, pe12. If we want the power delivered, we ask for the negative of this value.
• The equivalent resistance of the rest of the circuit, as seen by the source. For a source called e2, re2.

1.5A numerical DC simulation, step by step

Let's go back now to the simulation we ran earlier. That simulation corresponds to the circuit given in Example 5.7 in Boylestad's Introductory Circuit Analysis (11ed). Moving forward, I will refer to that textbook as B11.

The problem statement and the circuit schematic are reproduced exactly as they appear in the textbook, as they will be for every other problem in this tutorial. Since this is for educational purposes, it is my understanding that it falls squarely within the "fair use" doctrine of copyright law. In any event, no copyright infringement is meant.

B11's Example 5.7

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Solution

All the values in this circuit are numbers: no element value is unknown. This is a numerical circuit — one where we know the numerical value of every element in it.

I will now walk you step by step through the solution. This process applies to most numerical simulations in Symbulator: (1) describe the circuit, (2) run the simulation, and (3) get the answers.

Step 1: Describe the circuit

Description starts with naming the nodes. As we said, you can call them anything you want, number or letter, as long as the name is unique — but one node must always be called 0 (zero), the ground node, with a voltage of 0 V. In this circuit the ground node is marked with the ground symbol. That is not always so; when it is not marked, you pick a node to serve as zero.

I labelled the nodes in this circuit, starting in the ground and moving clockwise, as 0, 1, 2 and 3. It helps me to pencil the names in the schematic itself.

After naming the nodes, I am ready to describe the elements of the circuit in Symbulator notation.

Let's start with the source: when I only have one voltage source, like here, I enjoy giving it a short name: e. So the voltage source is e,1,0,36: its name is e, its positive node is 1, its negative node is 0, and its value is 36 volts between them.

Now the resistors. I named the first r1 and described it r1,1,2,1'k: its name is r1, its first node 1, its second node 2, and its value 1 kΩ. The second is r2,2,3,3'k, and the third r3,3,0,2'k.

Those four element descriptions, gathered together, are the whole circuit. Put them in the Circuit description box, one to a line:

Circuit DescriptionCopy

e,1,0,36
r1,1,2,1'k
r2,2,3,3'k
r3,3,0,2'k

Step 2: Run the simulation

Under the box, choose the simulation:

• Type of analysis → Solve circuit
• Analysis → DC — direct current

Then press Run Symbulator.

The page fills in below — this circuit takes under a second on a laptop, a little more on a phone — and a note at the foot of Results says what happened: DC analysis · 16 result(s), and how long it took.

For a DC analysis, these answers are as follows:

• The voltage of each node, so the voltages of nodes 1, 2 and 3 are in v1, v2 and v3, under Node voltages.
• The current through each element. The direction of the current is defined as going from the first node in the element's description to the second.
• The voltage drop in each element, defined as the difference in voltage between the first node in the element's description and the second, in that order.
• The power consumed in each element. An important point is that this is the power consumed, not the power delivered.
• Finally, for each source, the equivalent resistance of the rest of the circuit as seen by that source.

Step 3: Get the answers

We can now answer the six questions in the problem.

Answer to question (a). The equivalent resistance as seen by the source e:

Nothing to compute: in Results by element, the resistance seen by the source reads re = 6 kΩ. Correct.

Answer to question (b). Current I_s is defined in the schematic as the current flowing through the source, in the direction that goes from node 0 to node 1. One way to find this value is switching the sign of the current through the source, which we defined as flowing in the opposite direction:

The current through the source reads ie = −6 mA. The answer we want is its opposite, 6 mA.

Another way, since this is a series circuit where every element carries the same current, is to read the current through any resistor.

Answer to question (c). The voltage drop in resistor R₁ — since its polarity is defined in the schematic in the same way it is defined in our circuit description — is as follows.

The voltage drop vr1 reads 6 V, vr2 reads 18 V and vr3 reads 12 V. All correct.

Answer to question (d). We know that the power consumed is given in pe. We get the power supplied by the source by switching the sign: 216 mW delivered.

Answer to question (e). The power consumed by the resistors is given in pr1, pr2 and pr3: 36 mW, 108 mW and 72 mW, respectively.

Answer to question (f). Let's ask Evaluate whether the sum of the consumed power in the resistors equals the power supplied by the source:

If the resistors consume exactly what the source delivers, the four powers add up to zero. In Evaluate:

EvaluateCopy

pr1 + pr2 + pr3 + pe

The answer is 0.

This is the right answer, and concludes the solution to this, your first ever problem in Symbulator.

1.6Compare it with your own work

You have just watched Symbulator solve a circuit, but not the way your course does it. It writes one equation for every element and one for every node, hands the whole system to its algebra engine at once, and reads the answers back. It never chooses a mesh, never draws a supernode, and never stops to ask which method would be quicker.

You have not seen that system yet, and it is worth looking at once before reading on. Open the Settings card and tick Show equations. An Equations card appears just above Results, open and listing every equation Symbulator assembled for the circuit you have just solved. You do not need to run the analysis again — the system is already there from the last solve, so the card fills in as soon as you tick the box.

Read it beside your own working and the difference is plain: one equation per element and one per node, in no particular order, with not a mesh or a supernode anywhere. That is why many of these equations look different to the ones in your notebook — and it is the reason for the card that comes next.

The By-Hand Equations card, found in the TOOLS section, does things differently. It writes out the system you would write — Kirchhoff's current law at each node, with a supernode where a voltage source sits between two of them, or Kirchhoff's voltage law around each mesh, with a supermesh where a current source is shared. It then solves that system on its own and checks its answers against the ones above, and tells you whether the two agree.

To use it, solve the circuit first, then open the card and press Write the equations. You do not have to pick a method: it works out which ones apply to your circuit, shows the one that needs fewer equations, and offers you the other if both apply. It draws your working on the circuit as well — a ring round every node whose equation is being written, a dashed line round each supernode or supermesh, and the mesh currents I₁, I₂, I₃ curling round their own loops. Every mesh current turns clockwise, as textbooks draw them, and ticking Flip the mesh currents turns them all the other way, which reverses the arrows and the sign of every mesh current.

It is meant for exactly the moment when your homework answer disagrees with Symbulator's and you cannot see why. Both systems use the same names for the same quantities — v2 is still the voltage at node 2, ir3 is still the current through r3 — so you can put your own page beside the card's and find the line where you and it parted company. That is usually a sign somewhere, and it is much easier to spot in one equation than in an answer.

Two things to keep in mind. Not every method suits every circuit: an op-amp has no mesh current to write, and coupled coils belong to mesh rather than to nodal — the card says so, and points you at the one that does apply. And since the By-Hand Equations card is a new feature, should the two ever disagree, I would trust the answers Symbulator gives above: they come from the engine that has been solving these circuits since 1999.

1.7Instructive numerical examples, solved

Numerical problems in DC using e and r

These practice problems are taken from several textbooks, chosen because they apply only the concepts you have learned so far. They let you practise and reinforce them before Lesson 2.

The problem below comes from Figure 1-26 (a) in Hyatt and Kemmerly's Engineering Circuit Analysis (5ed). Moving forward, we will refer to that textbook as HK5.

HK5's Figure 1-26

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We are asked for the current, voltage drop and power consumed in each resistor, the power delivered by each voltage source, and a check that the powers in the circuit add up to zero.

Here is my solution.

I named the nodes thus: the bottom node 0, and the top nodes, from left to right, 1, 2 and 3. My description of the circuit:

Circuit DescriptionCopy

e1,1,0,120
r1,1,2,30
e2,2,3,30
r2,3,0,15

Run it, and the answers are all on the page already:

• The current through either resistor, ir1 or, ir2 is: 2 A
• vr1 gives the voltage drop in the 30Ω resistor: 60 V
• vr2 is the voltage drop in the 15Ω resistor: 30 V
• pr1 is the power consumed in the 30Ω resistor: 120 W
• pr2 is the power consumed in the 15Ω resistor: 60 W
• Flipping the sign of pe1 gets the power delivered by the 120V source: 240 W
• Flipping the sign of pe2 gets the power delivered by the 30V source: -60 W. The source consumes 60W, so it delivers negative 60W.
• pr1+pr2+pe1+pe2 is the sum of powers: 0 W. As expected.

Wasn't that easy?

B11's Example 5.20

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The practice problems get progressively more complicated, so you build up your 'symbulating' skills with confidence.

I named the nodes clockwise from the ground: 0, 1, 2, 3 and 4. My circuit description:

Circuit DescriptionCopy

e1,1,0,50
r1,1,2,4
e2,2,3,12.5
r2,3,4,7
r3,4,0,4

Run it and read the answers. ir1 is the current I: 2.5 A. vr2 is the voltage drop in the 7Ω resistor: 17.5 V

B11's Example 6.13

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Since the ground is the bottom node, I named it 0. I named the top node 1.

Circuit DescriptionCopy

e,1,0,24
r1,1,0,10
r2,1,0,220
r3,1,0,1.2'k

• re is the total resistance: 9.49 Ω
• Flipping the sign of ie gets us the source current: 2.53 A
• ir1 is I₁: 2.4 A, ir2 is I₂: 109 mA, and ir3 is I₃: 20 mA.

B11's Example 7.2

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Determine I₄, I_S and V₂.

My solution: I named the top node 1, and the other 2:

Circuit DescriptionCopy

e,1,0,12
r1,1,2,6.8'k
r2,2,0,18'k
r3,2,0,2'k
r4,1,0,8.2'k

Answers: v2 is 2.51 V, the opposite of ie (i.e. I_S) is 2.86 mA and ir4 is 1.46 mA.

B11's Example 7.7

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I have labelled the node names I used. My solution:

Circuit DescriptionCopy

e1,0,1,6
e2,0,2,18
r1,1,a,5
r2,a,2,3
r3,1,b,6
r4,b,2,2

Answers: vr1 is 7.5 V, vr3 is 9 V. For V_ba, vb-va is -1.5 V. For I_S, the opposite of ie2 is 3 A.

B11's Figure 7.32

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Determine I₆ and V₆.

My solution:

Circuit DescriptionCopy

e,1,0,240
r1,1,2,5
r2,2,0,6
r3,2,3,4
r4,3,0,6
r5,3,4,1
r6,4,0,2

Answers: ir6 is 10 A, and vr6 is 20 V.

B11's Example 7.10

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Calculate the indicated currents and voltages.

My solution:

Circuit DescriptionCopy

r2,2,3,8'k
r1,3,4,4'k
r3,1,2,12'k
r4,1,4,24'k
r5,1,0,12'k
e,4,0,72
r6,4,5,12'k
r7,5,0,9'k
r8,5,6,3'k
r9,0,6,6'k

Answers: ir5 is 3 mA, the opposite of ie (i.e. I_S) is 7.36 mA, and vr7 is 19.6 V.

B11's Example 7.4

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Determine the currents I₁, I₂, I_A, I_B and I_C, and the voltage drop in areas A, B and C.

My solution:

Circuit DescriptionCopy

e,1,0,16.8
r1,1,2,9
r2,1,2,6
r3,2,3,4
r4,3,0,6
r5,3,0,3
r6,2,0,3

Current I₁ is found via ir1 = 1.2 A, I₂ via ir2 = 1.8 A, I_A, via the opposite of ie = 3 A, I_B via ir3 = 1 A and I_C via ir6 = 2 A. The voltage drop in area A is vr1 = 10.8 V; in both B and C it is v2 = 6 V.

B11's Example 6.15

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My solution:

Circuit DescriptionCopy

e,1,0,28
r1,1,0,1.6'k
r2,1,0,20'k
r3,1,0,56'k

• re is the total resistance: 1.44 kΩ
• ir1 is 17.5 mA, ir2 is 1.4 mA, and ir3 is 0.5 mA
• Flipping the sign of pe gets the power: 543 mW

These are the correct answers.

B11's Figure 7.40

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Determine V_b and V_c.

My solution:

Circuit DescriptionCopy

e,a,0,120
r1,a,b,10
r2,b,c,20
r3,c,0,30
rl1,a,0,20
rl2,b,0,20
rl3,c,0,20

Answers: vb is 66.21 V, and vc is 24.83 V.

B11's Example 8.10

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Determine the current through each resistor.

My solution:

Circuit DescriptionCopy

e1,1,0,15
r1,1,a,4
e3,3,0,20
r3,3,a,10
e2,0,2,40
r2,a,2,5

Reading the current through each of the three resistors, ir1, ir2 and ir3: I_R1 =4.77A, I_R2 =7.18A and I_R3 =2.41A. These are the correct answers.

B11's Example 7.6

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My solution:

Circuit DescriptionCopy

e,1,0,24
r1,1,2,6
r2,1,2,6
r3,1,2,2
r4,2,0,8
r5,2,0,12

Answer: The opposite of ie is I_S=4 A, ir2 is I₂ = .8 A, ir4 is I₄=2.4 A, vr1 is V₁=4.8 V, vr5 is V₅=19.2 V.

B11's Example 7.11

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My solution:

Circuit DescriptionCopy

e1,a,0,20
e2,a,b,5
e3,c,0,8
r1,a,c,10
r2,b,c,4
r3,b,0,5

Answers: va=20 V, vb=15 V, vc=8 V, va-vc= 12 V, vb-vc=7 V, ir2=1.75 A, I_S via the opposite of ie3=-2.95 A

B11's Example 8.24

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Find the voltage drop in the 3Ω resistor.

My solution:

Circuit DescriptionCopy

e8,1,0,8
r2,1,2,2
r4,2,0,4
r6,2,3,6
r3,3,0,3
r10,3,4,10
e1,0,4,1

The voltage drop vr3 shows that V_3Ω is 1.1 V. This is correct.

B11's Example 8.18

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Find the current through the 10Ω resistor in the network shown.

My solution:

Circuit DescriptionCopy

e15,1,0,15
r10,1,2,10
r8,1,3,8
r5,3,2,5
r3,3,0,3
r2,2,0,2

The answer you want is ir10, in Results.

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

We find that ir10 = 1.22 A.

B11's Example 8.26

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Find the voltage drop in the 2Ω resistor.

My solution is below:

Circuit DescriptionCopy

e,1,0,240
r1,1,2,3
r2,2,3,4
r3,3,4,1
r4,4,5,2
r5,3,5,6
r6,2,5,6
r7,5,0,9

vr4 gives the voltage drop in R₄ (the 2Ω resistor): it is 10.67 V

Circuits with ‘hidden source'

Sometimes the schematics of circuits are presented in such a way that sources of voltage are not shown explicitly, yet their voltage is provided. These are what I call ‘hidden source' problems. Below I offer two examples of these types of problems. Both are taken from the textbook Circuit Analysis: Theory and Practice (3ed) by Allan H. Robbins and Wilhelm C. Miller, to which from this point on we will refer as RM3.

RM3's Example 7-5 (Hidden source)

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Find the indicated currents and voltages.

In my solution below, notice I introduced two sources, one for 12V and one for -6V:

Circuit DescriptionCopy

e1,1,0,12.
r1,1,b,10
r2,b,a,10
r3,a,2,50
r4,b,2,30
e2,2,0,-6

The answers you want are ir1, ir2 and ir4, in Results.

Ask Evaluate for:

EvaluateCopy

va-vb

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

The answer is I₁=.6 A, I₂=.2 A, I₃=.4 A and V_ab=-2 V. This is correct.

RM3's Figure 7-16 (Hidden source)

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Find the total circuit resistance, and the indicated currents and voltages.

My solution:

This is how I named the nodes: the top, 1; the bottom, 2, and a and b as in the figure. In my solution below, notice I introduced two sources, one for -10V and one for -2V:

Circuit DescriptionCopy

r1,1,b,4'k
r2,1,a,3'k
r3,b,a,2'k
r4,b,a,3'k
r5,1,b,1'k
rt,a,2,6'k
e1,1,0,-2
e2,2,0,-10

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

Three answers are given directly in Results. The values of irt, ir1 and ir2 tell us that I_T=1.11 mA, I₁=.133 mA, and I₂=.444 mA.

The other two answers can be found using Evaluate. The equivalent resistance, given by the expression:

\[\dfrac{v_1 - v_2}{I_T}\]

is found by evaluating:

EvaluateCopy

(v1-v2)/irt

which gives a value of 7.2 kΩ.

And the value of V_ab is found by evaluating:

EvaluateCopy

va-vb

which gives the value of -0.8 V. This is correct.