Alternating current analysis

Solution

Since this is the first AC problem we will solve, I will explain every step, including the manual ones, in full detail.

Step 1: sinusoids to phasors. We have to express the value of the source in terms Symbulator can understand, so we convert it from a sinusoid to a phasor: 10∠0°. We can pass this as (10∠0°) or, since the angle is zero, simply as 10.

Step 2: note the frequency. The circuit has a capacitor with a value given in farads. Because of this, Symbulator will need the frequency, given by the problem as 4 rad/s, to convert that value into an impedance in ohms.

Step 3: describe the circuit.

e,1,0,10
r1,1,2,5
c,2,0,.1

Choose AC — alternating current, and put 4 in the ω — angular frequency box that appears beside it.

Notice that, in our circuit description, the order we give to the nodes of the resistor and the capacitor is chosen to be convenient for the answers we will ask for.

Step 4: read the answers. Once Symbulator is done, you should take a look at what it found:

• the usual voltages in the nodes, voltage drops in the elements, and currents through the elements, in variables that should be familiar by now
• the average power consumed in the source and the resistor, in ape and apr1. None is given for the capacitor, since capacitors and inductors do not consume real power.
• the complex power consumed in all elements, in sc, se and sr1
• the equivalent impedance of the rest of the circuit as seen by the source, in ze

To get the answers we need for this problem in particular, we ask for ir1 and vc. It is likely that the computer will give you the answers in rectangular form, with a real part and an imaginary part.

If we want to see them as an amplitude and angle, open the Mini-Tools card under the results, leave the tool set to aa — amplitude and angle, and give it the answer's name:

ir1

It reads 1.789∠26.57° A.

That is the one-value way of doing it. If you would rather see every answer in polar form, there is a setting for it — see Rectangular or polar, above.

This is correct. You manually convert it to a sinusoid, by putting it back in the same terms the input was given in:

Asking for the capacitor voltage the same way gives 4.472∠-63.43°, that is, \(v(t) = 4.472 \cos(4t - 63.43°)\) V, which is correct.

Future solved problems will not include this level of detail in the solution, only the circuit description and the commands we give.

Solution

e,1,0,(20∠30°)
r1,1,2,4
l,2,0,.2

The source is given in polar form, and version 9 reads it as written — magnitude, the angle sign, then the angle in degrees.

Then AC, with 10 for ω. Mini-Tools with aa reads ir1 as 4.472∠3.43° A and vl as 8.944∠93.43° V. Both are correct.

We ask for the current in the resistor and get 4.472∠3.43°, and for the voltage in the inductor and get 8.944∠93.43°. Both are correct.

Solution

Since this is a passive circuit, to reduce it we will use the er tool.

ca,1,2,2'm
r1,2,3,3
cb,3,0,10'm
l1,2,4,.2
r2,4,0,8

Set Type of analysis to Find equivalent and Type of equivalent to Resistance / impedance, with nodes 1 and 0, in AC at ω 50. The answer is called zeq.

3.22–𝐢11.07

That is 3.22 − j11.07 Ω, which is correct.

Solution

The capacitor is in farads and the inductor in henries, so Symbulator needs the frequency, and the source is entered as its phasor: 50 cos 200t is 50 at an angular frequency of 200.

e,1,0,50
r,1,2,10
c,2,3,5'm
l,3,0,20'm

AC, with ω — angular frequency set to 200.

Read it with Mini-Tools set to aa: aa(ir) gives 4.789∠-16.70° A.

That is 4.789 A at an angle of −16.7°, which is correct.

Solution

Since the values are in ohms, these are all impedances. This is a passive circuit, so we could use the er tool. However, the structure of this circuit is so simple that we can reduce it by hand. The problem asks for the equivalent admittance, so our answer will be the inverse of the equivalent impedance:

1/pr(4,8j,-10j)

0.25-0.025𝐢

That is 0.25 − j0.025 S, which is correct.

Solution

Notice that the question gives you a frequency and that it is entirely superfluous: every value is already in ohms, so there is nothing for a frequency to convert. You will not need it.

Notice also that nothing here asks for a full simulation. The equivalent impedance falls out of one expression, using the shorthand for parallel combination:

4+20j+pr(16,-14j+25j)

9.135+𝐢27.47

That is 9.135 + j27.47 Ω. The current is the source voltage divided by it:

There is no stored zeq to divide by, so put the whole thing in Mini-Tools with aa: aa(12/(4+20j+pr(16,-14j+25j))) gives 0.4145∠-71.60° A.

That is 414.5 mA at an angle of −71.6°, which is correct.

Solution

Eight impedances and no source, so this is a job for the equivalent-impedance tool rather than a solve. The names below carry the two nodes each element bridges, which is only a convenience — any unique names would do.

r10,1,0,6j
r20,2,0,8j
r30,3,0,8j
r40,4,0,12j
r12,1,2,2
r23,2,3,-6j
r34,3,4,4
r14,1,4,-4j

Set Type of analysis to Find equivalent and Type of equivalent to Resistance / impedance, with nodes 1 and 0, in AC. The answer is called zeq.

0.3794+𝐢1.46

That is 0.3794 + j1.46 Ω, which is correct.

Solution

The answer given by the book is 25 µF. But I think this is one of those unfortunately common instances where there is a mistake in the book. I believe they meant to ask "at 2k rad/s", because that is the frequency at which the answer they give is right. Let me tell you how I solved this problem and I will let you be the judge.

c,1,0,c
r1,1,2,10
l,2,0,5'm

Find equivalent, Resistance / impedance, nodes 1 and 0, AC. The problem gives 2000 Hz and the ω box wants radians per second, so type 2*pi*2e3 straight into it — it takes an expression, not only a number.

Once the simulation finishes, solve for the value of c that makes the imaginary part of the equivalent impedance zero:

Now open the Solve card and ask for the capacitance that leaves no imaginary part:

im(zeq) = 0

c

Tick real solutions only before running it. Without it the answer comes back carrying an im(c) term, because nothing has told Symbulator that a capacitance is a real number.

The answer is c = 0.000001235, or 1.235 µF. This is not the answer the book gives.

I first used cSolve here, which also works, but my friend Qifan Wang — who verified my answers — pointed out, correctly, that it is not necessary: focusing on the imaginary part removes every reference to the complex operator, so we are solving an equation in real terms only.

Let's now repeat the process, using what I suspect is the frequency they meant: 2000 rad/s. I also want to show you something cool. Now that we know the answer is in the range of µF, we can declare the value of the capacitor as c'µ, so that the value of c we get will be in that scale:

c,1,0,c'µ
r1,1,2,10
l,2,0,5'm

Find equivalent, Resistance / impedance, nodes 1 and 0, in AC at ω 2000. That gives zeq = \(\frac{500 i}{- c + 25 + 25 i}\), and the condition goes to the Solve card:

im(zeq) = 0

with c as the unknown and real solutions only ticked — the same equation as before, now against the new zeq — which answers c = 25. The calculator spells that function imag; version 9 uses im.

The answer is c = 25, that is to say 25 µF. This is the right answer in the book. So they asked the wrong question — they asked it using the wrong frequency unit.

Solution

e,1,0,vs
c,1,2,c
r1,2,o,r
o,0,2,o

The ω box takes a name as readily as a number: type omega into it and the answers come back as functions of it.

We ask for vo/vs and get −c·ω·r·𝐢, which is correct.