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. If you already have it installed as a local app, then just open that app. If you do not have it locally, open the online app in a browser connected to the Internet. Let's click on Clear all inputs, found at the top right of the app page, to make sure there is nothing anywhere in the input fields. Then, in the box labelled Circuit Description, type the following circuit description, 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 the section called Analysis & Settings. If it is collapsed, open it with the little blue triangle to its left; under it are two drop-down menus. Set Type of analysis to Solve circuit and Analysis to DC — direct current. Then click Run Symbulator.

If you typed everything correctly, the word Solving… appears on the button for a moment. When it changes to Solved!, the results are ready. Scroll down to the Outputs part of the page, where the Results section is now active, holding the results for every node and every element of the circuit you just solved.

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 selections you made in Analysis & Settings tell Symbulator what kind of question you are asking (solve a circuit) and which analysis to perform (direct current). Everything else there is optional: you will meet the other menus and cards in later lessons, and until then they can be left exactly as they are.

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. Symbulator also accepts a colon as a separator. This is the same circuit 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.

For example, an 8 kΩ resistor can be entered in many ways: 8000, 8'k, 8000. and 8E3 are all equivalent, except that the first two are treated as exact and the other two as approximate.

1.3What answers do you get

After the simulation in DC is complete, Symbulator fills in the Results section for you, under OUTPUTS. Nothing is hidden away: every answer it worked out is on the page, and reading them is a matter of scrolling and looking.

The Results are arranged in groups. Node voltages first, one line per node, and then Results by element, which gives each element a small block of its own 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

Out of the box, Symbulator answers exactly. It works symbolically, so it would rather tell you a current is 3/500 A than round it off. That is the right default for a machine doing algebra, and exactly what you want for symbolic results: an answer of v_in·r₂/(r₁ + r₂) cannot and should not be rounded.

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.

One thing to expect: ticking the prefix box moves Rounding off exact by itself, because a prefixed value is a decimal. Symbulator says so on screen rather than doing it silently, and the reverse happens too — choose exact again and the prefixes switch off.

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

Answer for each node

For each node, its voltage with reference to ground is shown under Node voltages, labelled v with the node's name below it. The voltage in node 1 appears as v1.

Answers for each resistor

For each resistor, the following answers are calculated:

• 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 the line marked voltage drop, labelled vr5.
• The current through the resistor, flowing from the first node towards the second, in amperes. For a resistor called rx, this is the line marked current through, labelled irx.
• The power consumed by the resistor, in watts. For a resistor called r12, this is the line marked power consumed, labelled pr12.

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, the voltage drop line, ve5.
• The current through the source, flowing from the first node towards the second, in amperes. For a source called ex, the current through line, iex.
• The power consumed — attention: not delivered, but consumed — by the source, in watts. For a source called e12, the power consumed line, 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, the resistance seen line, 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. The process applies to most numerical simulations in Symbulator: first you describe the circuit, then you run the simulation, then you 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 in that order.

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

Circuit Description

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

Step 2: run the simulation. The two menus under the box say what kind of simulation to run. We want a plain solve of the circuit as described, in direct current, so:

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

Then press Run Symbulator.

The page fills in below almost immediately — this circuit takes well under a second — and a line at the foot of the Results section tells you what just happened: DC analysis · 16 result(s), and how long it took. Sixteen answers, found and displayed in one go.

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 the three lines under Node voltages: v1, v2 and v3.
• 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. Find the block headed e voltage source in Results by element, and read the line marked resistance seen:

r_e = 6 kΩ

That is 6 kΩ. Correct.

Answer to question (b). Current Is 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 evaluating the negative of the current through the source, which as you know flows in the opposite direction:

The current through line in the same block reads ie = −6 mA. The answer we want is its opposite, 6 mA.

That is 6 mA. Another way, since this is a series circuit where every element carries the same current, is to read the current through line of any resistor — all three say 6 mA.

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.

It is already on screen. In the block for r1, the voltage drop line reads v_r1 = 6 V. Look at the blocks for r2 and r3 and you will find v_r2 = 18 V and v_r3 = 12 V. These are all the right answers.

Answer to question (d). Since the problem asks for the power supplied by the source, and we know that the power consumed line of the source's block, pe, holds the power consumed by it, we need its opposite, which is 216 mW, that is 216 mW delivered.

Answer to question (e). The power consumed by the resistors is on the power consumed line of each resistor's block: pr1 = 36 mW, pr2 = 108 mW and pr3 = 72 mW, that is 36 mW, 108 mW and 72 mW consumed, respectively.

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

The tidiest way to check a balance like this is to add everything up and expect nothing. If the resistors consume exactly what the source delivers, then all four powers together must come to zero. In Evaluate:

Evaluate

pr1 + pr2 + pr3 + pe

The answer is 0 — not a small number, but zero. If you want to be sure it is genuinely zero and not three digits' worth of rounding, set Rounding back to exact for a moment: Symbulator returns to fractions and still says 0.

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

1.6Instructive 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 Description

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 line of either resistor gives the current in the resistors: 2 A
• The voltage drop line of r1 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. This means this source is actually consuming 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, as an argument of the DC simulation command:

Circuit Description

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

When it's done, ask for the answers we need. 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 Description

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 Description

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 Description

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 Description

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 Description

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 areas A, B and C.

My solution:

Circuit Description

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 Description

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 Description

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 Description

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 line of each of the three resistors: 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 Description

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 Description

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 Description

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 line of the r3 block 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 Description

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.

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 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 Description

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

The r4 block 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 Description

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:

Evaluate

va-vb

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, {.6,.2,.4,-2.}, indicates 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 Description

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

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

Ask Evaluate for:

Evaluate

(v1-v2)/irt
va-vb

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 we get indicates that the equivalent resistance, given by (v₁-v₂)/I_T, is 7.2 kΩ, and that I_T=1.11 mA, I₁=.133 mA, I₂=.444 mA and V_ab=-0.8 V. This is correct.