Transient analysis

Bo2's Example 5.1

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

Solution. Here we have two intervals. In the first, t < 0s, we assume things have been steady for a long time and any transient has passed, so we run a DC simulation to find the initial conditions for the second.

e,1,0,V
r1,1,2,r1
c,2,0,c
r,2,0,r

When the simulation is Done, we ask for vc, the voltage in the capacitor c. We get the expression shown below, which is correct.

For the second interval, when 0s ≤ t, we do a transient simulation. We give the capacitor as initial condition its voltage from the previous interval.

c,2,0,c,r*v/(r+r1)
r,2,0,r

When the simulation is Done, we ask for vc again. We get the expression below, which is correct.

Bo2's Drill Exercise 5.1

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

Solution. Again, we have two intervals: the first for DC, the second for TR.

For the first interval, when t < 0s, the current source is 2A. Simulate in DC.

j,0,1,2
r,1,0,3
c,1,0,1/12

The answers you want are vc and ic, in Results.

{ 6 , 0 }

So the voltage in the capacitor is 6V. We store this value in a variable called vc₀, to use it as initial condition of the capacitor for the second interval.

For the second interval, when 0s ≤ t, we do a transient simulation. The current source is 0A, and the capacitor has an initial condition: voltage vc₀.

j,0,1,0
r,1,0,3
c,1,0,1/12,vc0

Set Analysis to TR — transient / time domain.

The answers you want are vc and ic, in Results.

When the simulation is Done, we get the expressions below. They are right.

Bo2's p224 5.2

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

Solution. Again, we have two intervals: the first for DC, the second for TR.

For the interval when t < 0s, the voltage source is 12 V. Simulate in DC.

e,1,0,12
r3,1,2,3
r6,2,0,6
r4,2,3,4
c,3,0,1/12

The answers you want are vc, ic and v2, in Results.

{ 8 , 0 , 8 }

For the interval when 0s ≤ t, the voltage source is 0V. Simulate in TR, giving the capacitor as initial condition its voltage from the previous interval.

e,1,0,0
r3,1,2,3
r6,2,0,6
r4,2,3,4
c,3,0,1/12,8

Set Analysis to TR — transient / time domain.

The answers you want are vc, ic and v2, in Results.

Bo2's Drill Exercise 5.2

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

Solution. For t < 0s, the 6A source is part of the circuit. Simulate in DC.

j,0,1,6
r10,1,0,3
r12,1,2,3
r20,2,0,3
c,2,0,1/10

The answers you want are vc, ic and v1, in Results.

{ 6 , 0 , 12 }

For 0s ≤ t, the source is no longer relevant and can be excluded from the circuit. Simulate in TR, giving the capacitor its proper initial condition.

r10,1,0,3
r12,1,2,3
r20,2,0,3
c,2,0,1/10,6

Set Analysis to TR — transient / time domain.

The answers you want are vc, ic and v1, in Results.

Bo2's Figure 5.10a

Open in app ↗Open in split view

Look at the simple resistor-inductor (RL) circuit shown, where at time t=0 the inductor current is i_L(0). Determine v_L(t), i_L(t) and v_R(t) for t≥0.

Since we are already given the initial condition, we only run the transient simulation for t≥0. In our circuit description below, the initial condition is il0.

l,1,0,l,il0
r,1,0,r

Set Analysis to TR — transient / time domain.

The answers you want are vl, il and vr, in Results.

We get the right answers:

Bo2's Example 5.3

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

Determine i_L(t) for all t.

Solution. For t < 0s, simulate in DC.

e,1,0,v
r1,1,2,r1
l,2,0,l
r2,2,0,r

The answer you want is il, in Results.

For t ≥ 0s, simulate in TR, giving the inductor its initial condition.

l,2,0,l,v/r1
r2,2,0,r

Set Analysis to TR — transient / time domain.

The answer you want is il, in Results.

Bo2's Drill Exercise 5.3

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

Find i_L(t), v_L(t) and i(t) for all t.

Solution. For t < 0s, simulate in DC.

e,1,0,8
r4,1,2,4
l,2,0,1
r12,2,0,12

The answers you want are il, vl and ir4, in Results.

{ 2 , 0 , 2 }

For t ≥ 0s, simulate in TR, giving the inductor its initial condition of 2A.

e,1,0,0
r4,1,2,4
l,2,0,1,2
r12,2,0,12

Set Analysis to TR — transient / time domain.

The answers you want are il, vl and ir4, in Results.

Bo2's p230 (Dependent source)

Open in app ↗Open in split view

Find i_L(t) for t≥0, given that i_L(0) = 5A.

We only run the transient simulation for t≥0, with initial condition 5A.

r,v,0,r
j,v,0,2*ir
l,v,0,l,5

Set Analysis to TR — transient / time domain.

The answer you want is il, in Results.

Bo2's Drill Exercise 5.4 (Dependent source)

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

Find i_L(t) and v_L(t) for all t.

Solution. For t < 0s, simulate in DC. This gives us the initial condition.

ei,1,0,12
r6,1,2,6
l,2,0,16
r8,2,3,8
ed,3,0,3*vr8

The answers you want are il and vl, in Results.

{ 2 , 0 }

For t ≥ 0s, simulate in TR, giving the inductor its initial condition.

l,2,0,16,2
r8,2,3,8
ed,3,0,3*vr8

Set Analysis to TR — transient / time domain.

The answers you want are il and vl, in Results.

Bo2's Example 5.5 (Op Amp)

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

Determine v_C(t), i_C(t) and v_o(t) for all t.

Solution. For t < 0s, simulate in DC. This gives us the initial condition.

e,3,0,4
r2,3,1,2
r5,1,o,5
c,1,o,1/20
o,0,1,o

The answers you want are vc, ic and vo, in Results.

{ 10 , 0 , -10 }

For t ≥ 0s, simulate in TR, giving the capacitor its initial condition.

r2,0,1,2
r5,1,o,5
c,1,o,1/20,10
o,0,1,o

Set Analysis to TR — transient / time domain.

The answers you want are vc, ic and vo, in Results.

Bo2's Drill Exercise 5.5 (Op Amp)

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

Determine v_C(t), i_C(t) and v_o(t) for all t.

Solution. For t < 0s, simulate in DC. This gives us the initial condition.

e,2,0,4
o,2,1,o
c,o,1,1/20
r5,o,1,5
r2,1,0,2

The answers you want are vc, ic and vo, in Results.

{ 10 , 0 , 14 }

For t ≥ 0s, simulate in TR, giving the capacitor its initial condition.

o,0,1,o
c,o,1,1/20,10
r5,o,1,5
r2,1,0,2

Set Analysis to TR — transient / time domain.

The answers you want are vc, ic and vo, in Results.

Bo2's Example 5.6

DC, before the switchOpen in app ↗Open in split viewTR, 0 < t <= 1 sOpen in app ↗Open in split viewTR, t > 1 sOpen in app ↗Open in split view

In the circuit shown, there are two switches: one that opens at time t=0 and one that closes at time t=1 second. Determine v_C(t) and i_C(t) for all t.

As we mentioned before, every time a switch moves marks the end of one interval and the beginning of another interval for Symbulator.

For the first interval, which corresponds to t<0, we simulate in DC.

e,2,0,10
r1,2,1,1
c,1,0,1/4
r4,1,0,4

The answers you want are vc and ic, in Results.

{ 8 , 0 }

The voltage in the capacitor at the end of this interval will serve as the initial condition of the capacitor for the next interval. The second interval runs between 0 and 1 second, i.e. 0 < t ≤ 1 second. We simulate in TR.

c,1,0,1/4,8
r4,1,0,4

Set Analysis to TR — transient / time domain.

The answers you want are vc and ic, in Results.

The capacitor's voltage at the end of this second interval is the initial condition for the third. We can use its exact value, , but the textbook prefers using its approximate value, i.e. 2.943.

The third interval corresponds to t > 1 second. We simulate in TR. To make things easier for Symbulator, we replace the resistors by their equivalent.

c,1,0,1/4,2.943
re,1,0,[4,6]

Ask for {vc,ic}. The expressions we get are equivalent to:

These are the right answers, as can be seen by checking the book's answers.

Bo2's Drill Exercise 5.6

DCOpen in app ↗Open in split viewTR, 0 < t <= 2 sOpen in app ↗Open in split viewTR, t > 2 sOpen in app ↗Open in split view

Let’s see another example where a TR simulation is done for an interval that starts at a time other than t=0. Determine i_L(t) and v_L(t) for all t.

For the first interval, which corresponds to t<0, we simulate in DC.

e,1,0,9
r9,1,0,9
r3,1,2,3
l,2,0,6

The answers you want are il and vl, in Results.

{ 3 , 0 }

The 3A current in the inductor is its initial condition for the second interval, which goes from 0 to 2 seconds, i.e. 0 < t ≤ 2 seconds. We simulate in TR.

r,1,0,9+3
l,1,0,6,3

Set Analysis to TR — transient / time domain.

The answers you want are il and vl, in Results.

The inductor's current at the end of this second interval is the initial condition for the third. Find its approximate value thus:

.055

The third interval corresponds to t > 2 second. We simulate in TR. To make things easier for Symbulator, we replace the resistors by their equivalent.

r,1,0,[9+3,4]
l,1,0,6,.055

Set Analysis to TR — transient / time domain.

The answers you want are il and vl, in Results.

These are the kind of expressions your book or professor are looking for.

Bo2's Figure 5.19

Open in app ↗Open in split view

In the following circuit, assume that all the initial conditions are zero. The source is a step function u(t) with value V volts. Find i_R, v_R, i_C and v_C.

As discussed before, a step source with symbolic value requires that we use the u(t) nomenclature to describe it. Since the value is V volts, we describe the source as V*u(t).

e,1,0,V*u(t)
r,1,2,r
c,2,0,c,0

Set Analysis to TR — transient / time domain.

The answers you want are vc, vr and ic, in Results.

There is no need to ask for i_R, since i_R=i_C. We get the following expressions:

Bo2's Figure 5.24

Open in app ↗Open in split view

In the following circuit, assume that all the initial conditions are zero. The source is a step function u(t) with value V volts. Find i_R, v_R, i_L and v_L.

e,1,0,V*u(t)
l,1,2,l,0
r,2,0,r

Set Analysis to TR — transient / time domain.

The answers you want are vl, vr and il, in Results.

There is no need to ask for i_R, since in a series circuit it will be identical to i_L.

Bo2's Drill Exercise 5.7

Open in app ↗Open in split view

For the circuit, with a step source of 12 volts, find i_L, v_L, i_L and v.

As discussed before, a step source with numerical value can be described without the u(t) nomenclature. Thus, we describe the source as 12.

e,1,0,12
r4,1,2,4
l,2,3,2,0
r2,3,0,2

Set Analysis to TR — transient / time domain.

The answers you want are il, vl and v2, in Results.

Bo2's Figure 5.26

Open in app ↗Open in split view

In the following circuit, assume that all the initial conditions are zero. The source is a step function u(t) with value I amperes. Find i_C, i_R, and v.

As discussed before, a step source with symbolic value requires that we use the u(t) nomenclature to describe it. Since the value is I amperes, we describe the source as I*u(t). Thus:

j,0,1,i*u(t)
c,1,0,c,0
r,1,0,r

Set Analysis to TR — transient / time domain.

The answers you want are v1, ir and ic, in Results.

Bo2's Example 5.7 (Op Amp)

Open in app ↗Open in split view

Suppose that v_s(t)=u(t). Find v_C, i_C, and v_o.

This source is a step source with a value of 1 volt. Since its value is numerical, the u(t) nomenclature is not needed. We describe the source plainly as 1.

e,1,0,u(t)
o,1,2,o
c,2,o,1/8,0
r2,2,o,2
r1,2,0,1

Set Analysis to TR — transient / time domain.

The answers you want are vc, ic and vo, in Results.

Bo2's Drill Exercise 5.8 (Op Amp)

Open in app ↗Open in split view

Suppose that v_s(t)=u(t). Find v_C, i_C, and v_o.

Again, following the same logic as above, we describe the source plainly as 1.

e,1,0,u(t)
r1,1,2,1
r2,2,o,2
c,2,o,1/8,0
o,0,2,o

Set Analysis to TR — transient / time domain.

The answers you want are vc, ic and vo, in Results.

Bo2's p249 F5.28

TR, first intervalOpen in app ↗Open in split viewTR, second intervalOpen in app ↗Open in split view

For the circuit left, the source is v_s(t) = V u(t) - V u(t-t₀). This is shown in the plot to the right. Given this, find the voltage drop in the capacitor for all t.

Because the source has two steps, this problem has to be analyzed in two intervals. For the first interval, the initial conditions are zero and the source value is V*u(t). Our circuit description for the first interval is given below.

e,1,0,V*u(t)
r,1,2,r
c,2,0,c,0

Set Analysis to TR — transient / time domain.

The answer you want is vc, in Results.

The expression for the capacitor's voltage drop in the first interval is:

The second interval starts at t₀. In it, the initial condition of the capacitor is given by the value of the expression above when t=t₀. Here is how we find it:

Put vc in the Evaluate card and write t = to in its Conditions box. That is what the calculator's | does: evaluate this expression under this condition. The answer comes back \(V - V e^{- \frac{to}{c r}}\).

The circuit description for the second interval uses that expression as the capacitor's initial condition. The source is now 0 volts — another way of saying it becomes a short — so it need not appear in the description.

r,0,2,r
c,2,0,c,v-e^(-to/(c*r))*v

Set Analysis to TR — transient / time domain.

The answer you want is vc, in Results.

This is the expression for the capacitor's voltage drop in the second interval.

Bo2's Drill Exercise 5.9

TR, first intervalOpen in app ↗Open in split viewTR, second intervalOpen in app ↗Open in split view

Find the current through the inductor if the source is v_s(t) = V u(t) – V u(t-t_o).

Because the source has two steps, this problem has to be analyzed in two intervals. For the first interval, the initial conditions are zero and the source value is V*u(t). Our circuit description for the first interval is given below.

e,1,0,V*u(t)
l,1,2,l,0
r,2,0,r

Set Analysis to TR — transient / time domain.

The answer you want is il, in Results.

The expression for the inductor's current in the first interval is:

The second interval starts at t₀. In it, the initial condition of the inductor is given by the value of the expression above when t=t₀. Here is how we find it:

Put il in the Evaluate card and write t = to in its Conditions box. The answer comes back \(\frac{V}{r} - \frac{V e^{- \frac{r to}{l}}}{r}\), which is the same expression gathered differently.

The circuit description for the second interval uses that expression as the inductor's initial condition. The source becomes a short, so it need not appear in the description.

l,0,2,l,v/r-e^(-to*r/l)*v/r
r,2,0,r

Set Analysis to TR — transient / time domain.

The answer you want is il, in Results.

To put this in terms of the same t as the first interval, we replace t with t-t_o:

This is the expression for the inductor's current in the second interval.

Bo2's Figure 5.36

DC, first intervalOpen in app ↗Open in split viewTR, second intervalOpen in app ↗Open in split view

For the circuit shown in (a), find the voltage drop in the capacitor, given that the voltage source is as shown in (b).

This problem requires two intervals. In the first interval, a DC analysis of the circuit with a source of 2 V gives us the voltage in the capacitor.

e,1,0,2
r3,1,2,3
r5,2,3,5
c,3,0,1
j,0,2,2*ir3

The answer you want is vc, in Results.

2

This is the initial condition for the capacitor in the second interval, which we analyze using TR and a source value of -4 V.

e,1,0,-4*u(t)
r3,1,2,3
r5,2,3,5
c,3,0,1,2
j,0,2,2*ir3

Set Analysis to TR — transient / time domain.

The answer you want is vc, in Results.

This is the expression for the voltage drop in the capacitor after t=0.

Bo2's Drill Exercise 5.11

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

For the circuit of Bo2's Example 5.7, change the value of the capacitor to ¼ F. Find v_C(t) for the case that the source is v_S(t)=1 V for t<0 and 3V for t≥0.

We let Symbulator know that we are only interested in one variable: vc.

This is optional: Symbulator 9 solves quickly enough that limiting the results rarely saves you anything worth having. If you want to anyway, tick Do you want to limit the results to save time? — it sits under the analysis menus once TR is chosen — and list vc in the field it reveals.

Then we run the simulation for the first interval, just as we did before, but with the new capacitor and source values.

e,1,0,1
o,1,2,o
c,2,o,1/4
r2,2,o,2
r1,2,0,1

The answer you want is vc, in Results.

-2

This is the initial condition for the next interval. Before we run the simulation for the second interval, we let Symbulator know, again, that we are only interested in one variable: vc.

This is optional: Symbulator 9 solves quickly enough that limiting the results rarely saves you anything worth having. If you want to anyway, tick Do you want to limit the results to save time? — it sits under the analysis menus once TR is chosen — and list vc in the field it reveals.

Then we run the simulation for the second interval, just as we did before, but with the new capacitor and source values.

e,1,0,3*u(t)
o,1,2,o
c,2,o,1/4,-2
r2,2,o,2
r1,2,0,1

Set Analysis to TR — transient / time domain.

The answer you want is vc, in Results.

Bo2's Figure 5.38 (Impulse)

Open in app ↗Open in split view

Find the voltage drop and current in the capacitor, given a source of I δ(t) A, where δ(t) is the impulse function and I is a constant.

As we mentioned before, impulse sources must be described using the δ(t) nomenclature. Typing delta(t) spells it without the Greek.

j,0,1,i*δ(t)
c,1,0,c,0
r,1,0,r

Once the simulation completes, we ask for the variables of interest:

They are already in Results: vc is \(\frac{i e^{- \frac{t}{c r}}}{c}\) and ic is \(i \delta\left(t\right) - \frac{i e^{- \frac{t}{c r}}}{c r}\).

The impulse survives in the capacitor's current, which is right — the charge arrives all at once, and the delta is that instant.

Bo2's Drill Exercise 5.13 (Impulse)

Open in app ↗Open in split view

For the circuit of Bo2's Example 5.7, find v_C(t), i_C(t) and v_o(t) due to an impulse voltage source of v_S(t)= δ(t) (that is to say, a 1V impulse in t=0.)

Nothing new here. Again we use the δ(t) nomenclature for the source.

e,1,0,δ(t)
o,1,2,o
c,2,o,1/8,0
r2,2,o,2
r1,2,0,1

Set Analysis to TR — transient / time domain.

The answers you want are vc, ic and vo, in Results.

Bo2's Figure 5.39 (Ramp)

Open in app ↗Open in split view

Find iL(t) and vL(t) for a circuit that has in parallel a resistor R, an inductor L and a current source with a ramp value I r(t). Initial conditions are zero.

A ramp value I r(t) means a value of I t. We describe this source simply as i*t.

j,0,1,i*t
r,1,0,r
l,1,0,l,0

Set Analysis to TR — transient / time domain.

The answers you want are il and vl, in Results.

Bo2's Drill Exercise 5.14 (Ramp)

Open in app ↗Open in split view

For the circuit of Bo2's Example 5.7, find v_C(t) and i_C(t) due to a ramp input voltage of v_S(t)= r(t).

Nothing new here. The command below should be clear to you by now.

e,1,0,t
o,1,2,o
c,2,o,1/8,0
r2,2,o,2
r1,2,0,1

Set Analysis to TR — transient / time domain.

The answers you want are vc and ic, in Results.

Bo2's Example 5.12 (Exponential)

Open in app ↗Open in split view

Assume initial conditions zero. Find i_L(t) and v_L(t), given that i_S(t) = 2 e ^-4t A.

Describe the source as given to you by the problem, using an exponential.

j,0,1,2e^(-4t)
r,0,1,6
l,1,0,2,0

Set Analysis to TR — transient / time domain.

The answers you want are il and vl, in Results.

Bo2's p265 (Exponential)

Open in app ↗Open in split view

Repeat Bo2's Example 5.12 if the current source is now i_S(t) = 2 e ^-3t A.

All we do is change one digit in the circuit description: replace 4 with 3.

j,0,1,2e^(-3t)
r,0,1,6
l,1,0,2,0

Set Analysis to TR — transient / time domain.

The answers you want are il and vl, in Results.

Bo2's Example 5.13 (Exponential)

Open in app ↗Open in split view

Find the voltage drop in the capacitor and the current in the 3 Ω resistor, if the source value is v_S(t)=18e^-t/2.

e,1,0,18e^(-t/2)
r3,1,2,3
r5,2,3,5
c,3,0,1,0
j,0,2,2*ir3

Set Analysis to TR — transient / time domain.

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

Bo2's Drill Exercise 5.16 (Exponential)

Open in app ↗Open in split view

For the circuit of Bo2's Example 5.7, find v_C(t) and i_C(t) for the case when the source's value is v_S(t)= 2 e ^-4t. As before, assume initial conditions are zero.

e,1,0,2e^(-4t)
o,1,2,o
c,2,o,1/8,0
r2,2,o,2
r1,2,0,1

Set Analysis to TR — transient / time domain.

The answers you want are vc and ic, in Results.

Bo2's Example 6.1

DC, second orderOpen in app ↗Open in split viewTR, second orderOpen in app ↗Open in split view

Find the current in the inductor i(t) and the voltage drop in the capacitor v(t).

Since we have switches changing in time t=0, this is a two interval problem. The first interval is for t<0. We can analyze the first interval using DC.

e,3,0,9
r2,3,1,5
r1,1,0,5
l,1,2,1/2
c,2,0,1/8
r3,2,0,2

The answers you want are vc and il, in Results.

{2,1}

These are the initial conditions for the second interval. The second is for t≥0. We can analyze the second interval using TR. We only care about vc and il.

r1,1,0,5
l,1,2,1/2,1
c,2,0,1/8,2

Bo2's Drill Exercise 6.1

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

For the circuit shown, find v(t) and i(t) for t≥0.

Nothing new here. This is a two interval problem, as others seen before. The first interval, for t<0, is analyzed in DC to find the initial conditions.

j,0,1,1
r3,1,0,3
r6,1,0,6
l,1,2,1/4
c,2,0,1/3

The answers you want are vc and il, in Results.

{2,0}

These conditions are used in the TR analysis of the second interval, for t≥0.

The answers are vc = \(3 e^{- 2 t} - e^{- 6 t}\) and il = \(- 2 e^{- 2 t} + 2 e^{- 6 t}\), both in Results.

r3,1,0,3
r6,1,0,6
l,1,2,1/4,0
c,2,0,1/3,2

Bo2's Example 6.2

DC, the left halfOpen in app ↗Open in split viewDC, the right halfOpen in app ↗Open in split viewTR, both halves joinedOpen in app ↗Open in split view

Find i(t) and v(t) for the circuit shown.

I decided to simulate this using only fractions, not decimals.

So write the value as the fraction it is: 7/2 rather than 3.5. Version 9 needs no conversion step — a fraction stays exact, and it is the decimal point that would make the arithmetic approximate.

7/2

Because of the switches, this is a two interval problem. The first interval is to be analyzed using DC. Notice that for t<0 the circuit is such that we basically have two separate circuits. We simulate each one of them separately:

e,0,1,7/2
r1,1,2,4
c,2,0,2
r2,2,0,1/6

The answer you want is vc, in Results.

-7/50

r3,3,0,1/6
l,3,0,1/50
r4,3,4,3
e2,4,0,3

The answer you want is il, in Results.

1

The second interval, for t≥0, is analyzed using TR. I could have described the circuit again from scratch, but out of laziness I copied the descriptions from the DC simulations. To avoid renaming nodes, I simulated the right switch as a short between nodes 2 and 3.

vc comes back \(\frac{\left(- \sin{\left(4 t \right)} - 7 \cos{\left(4 t \right)}\right) e^{- 3 t}}{50}\) and il \(\left(- \sin{\left(4 t \right)} + \cos{\left(4 t \right)}\right) e^{- 3 t}\).

c,1,0,2,-7/50
r,1,0,[1/6,1/6]
l,1,0,1/50,1

These are the right answers.

Bo2's Example 6.3

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

Find i(t) and v(t) for the circuit shown.

Since the sources change value, this is a two interval problem. For t<0, in DC:

e,1,0,5
r3,1,2,3
l,2,3,1
c,3,0,1
r1,3,0,1
j,0,3,1

The answers you want are vc and il, in Results.

{2,1}

Using these as initial conditions, now we use TR for t≥0:

r3,0,2,3
l,2,3,1,1
c,3,0,1,2
r1,3,0,1

Bo2's Drill Exercise 6.4

DCOpen in app ↗Open in split viewTR, the book's 1 HOpen in app ↗Open in split viewTR, 1/2 H insteadOpen in app ↗Open in split viewFDOpen in app ↗Open in split view

My copy of Bo2 is second-hand, and this problem had scribbles on it. The previous owner evidently struggled with it and concluded that the inductor had to be ½ H, not the 1 H the schematic says. Symbulator can show he is right and the book wrong. First, solve the circuit as printed, with a 1 H inductor.

e,2,0,2
r,1,2,1
l,1,0,1
c,1,0,1/8

The answers you want are il and vc, in Results.

{2,0}

r,1,0,1
l,1,0,1,2
c,1,0,1/8,0

The answer we get doesn't match the answer provided by the book. Now let's repeat the TR simulation using the ½ H value for the inductor:

r,1,0,1
l,1,0,1/2,2
c,1,0,1/8,0

Now the answer matches the book's. An expert user can reach for the fd gate and the t2s tool, which we have not discussed yet, to find the inductor's value from the book's answer. This is how I found the ½ H myself, working backwards:

t2s(-16*t*e^(-4*t))

which gives \(- \frac{16}{\left(s + 4\right)^{2}}\).

-16/(s+4)^2

2*(s+8)/(s+4)^2

r,1,0,1
l,1,0,l,2
c,1,0,1/8,0

Set Analysis to FD — complex frequency domain.

The answer you want is vc, in Results.

-16*l/(l*s^2+8*l*s+8)

That is what the Solve card is for — a system that is not a circuit:

-16/(s+4)^2 = -16*l/(l*s^2+8*l*s+8)

with l as the unknown. It answers l = 1/2 H.

l=1/2

The fd gate and t2s tool are used for frequency domain analysis, and will be discussed separately in a subsequent part of Symbulator's documentation.

Bo2's Drill Exercise 6.5

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

In the circuit, suppose that v_S(t)= 12 – 12 u(t) V. Find i₂(t) for t≥0.

e,3,0,12
r6,3,1,6
l1,1,0,6
r3,1,2,3
l2,2,0,4

The answers you want are il1 and il2, in Results.

{2,0}

r6,0,1,6
l1,1,0,6,2
r3,1,2,3
l2,2,0,4,0

Bo2's Drill Exercise 6.6 (Op Amp)

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

In the circuit, suppose that v_S(t)= 2 – 2 u(t) V. Find v_o(t) for t≥0.

Simulate the first interval in DC to find the initial conditions of the capacitors.

e,3,0,2
r1,3,1,1
r2,1,2,2
r3,1,o,2
ca,1,0,1
cb,2,o,1/4
o,0,2,o

The answers you want are vca and vcb, in Results.

{0,4}

Then simulate the second interval in TR, to find the voltage in node o.

r1,0,1,1
r2,1,2,2
r3,1,o,2
ca,1,0,1,0
cb,2,o,1/4,4
o,0,2,o

Bo2's Example 6.5 (Plot)

Open in app ↗Open in split view

In the circuit, use v_S=2/5V, R=12Ω, L=2H and C=1/50F. Find v(t) and i(t), and plot them for time 0<t<1.5 seconds.

e,1,0,2/5*u(t)
r,1,2,12
l,2,3,2,0
c,3,0,1/50,0

-(2/5)e^-3tcos(4t)-(3/10)e^-3tsin(4t)+2/5

We look in the results and see that vc is the first of those and il the second.

These are the right answers. To plot them, run the plot tool, thus: s\plot(). Once the plot window opens, enter vc as function, 0 as minimal time and 1.5 as maximal time. Compare to the graph from the book. Repeat the plot for il, and compare that one to the book's graph too.

Bo2's Drill Exercise 6.7

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For the circuit in Bo2's Example 6.5, use v_S=3V, R=5Ω, L=1/2H and C=1/8F. Find v(t) and i(t).

e,1,0,3*u(t)
r,1,2,5
l,2,3,1/2,0
c,3,0,1/8,0

Bo2's Drill Exercise 6.8 (Ramp)

Open in app ↗Open in split view

For the circuit in Bo2's Drill Exercise 6.7, find the voltage drop in the capacitor v(t) if the source has a value v_S(t) = 3 r(t).

e,1,0,3t
r,1,2,5
l,2,3,1/2,0
c,3,0,1/8,0

Bo2's Example 6.6 (Plot)

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

For the circuit, plot the voltage drop in the inductor for 0 < t < 5ms.

First, find the initial conditions for t<0 by running a DC simulation.

e,1,0,12
r,1,2,3
l,2,3,1
c,3,0,1'µ
s1,3,0

The answers you want are il and vc, in Results.

{4,0}

Then, find the vl for t≥0 by running a TR simulation.

e,1,0,12*u(t)
r,1,2,3
l,2,3,1,4
c,3,0,1'µ,0

Finally, run s\plot() and enter vl as function, 0 as minimal time and 0.005 as maximal time. In the resulting plot you will see something outstanding: around t = 1.57 ms, the voltage drop across the inductor is 3991 volts!

Bo2's Drill Exercise 6.9

Open in app ↗Open in split view

For the circuit, R=1Ω, L=2H, C=1/2F and i_S(t)=u(t). Find i(t) and v(t).

j,0,1,u(t)
r,1,0,1
l,1,0,2,0
c,1,0,1/2,0

Bo2's Figure 6.23 (Op Amp)

Open in app ↗Open in split view

For the circuit, find v_o(t) if v_S(t) = u(t).

e,3,0,u(t)
r1,3,1,1
r2,1,2,1
ca,2,0,1/5,0
cb,1,o,1,0
o,2,o,o

Bo2's Drill Exercise 6.10 (Op Amp)

Open in app ↗Open in split view

In the previous circuit, change the capacitor's value from 1/5 F to 25/16 F. Find v_o(t) if v_S(t) = 3 u(t).

e,3,0,3*u(t)
r1,3,1,1
r2,1,2,1
ca,2,0,25/16,0
cb,1,o,1,0
o,2,o,o

-4e^-2t/5+e^-8t/5+3

Bo2's Drill Exercise 6.11 (p307)

Open in app ↗Open in split view

For the circuit, suppose that R₁ = R₂ = 1Ω, L = 1H, C = 1F and v_S(t)= 2e^-2t u(t). Find i(t) if all initial conditions are zero.

e,3,0,2e^(-2t)
r2,3,2,1
r1,2,1,1
c,2,0,1,0
l,1,0,1,0

-(e^t cos(t)-e^t sin(t)-1) e^-2t

A more complex problem

DCOpen in app ↗Open in split viewTROpen in app ↗Open in split view

The circuit comes from the circuits analysis class of Professor Eliane Boulet de Cabrera, from Universidad Tecnologica de Panama. Find v1 and v.

Because of the switches, this is a two interval problem. The first is run in DC.

e,3,0,18
r8,3,1,8
ca,1,0,1/6
r12,1,0,12
r18,1,2,18
r6,2,0,6
cb,2,0,1/3

The answers you want are vca and vcb, in Results.

{9,9/4}

These serve as initial conditions for the second interval, which is run in TR. Since we only want v1 and v2, it pays off to let Symbulator know this:

ca,1,0,1/6,9
r12,1,0,12
r18,1,2,18
r6,2,0,6
cb,2,0,1/3,9/4
j,0,2,10e^(-t)*sin(2t+30*pi/180)

Set Analysis to TR — transient / time domain.

Version 9 answers this one in about a second. v1 and v2 are in Results; they are long enough that they are easier read on screen than in print.

These are the correct answers, which explains why Professor Boulet – who made the problem up - is a living legend among circuit students at UTP.

Bo2's Drill Exercise 4.5 (Expert)

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For this circuit, we know that the capacitor’s initial condition is zero, that v_s is an unknown step source (i.e. of the form A u(t), where A is a constant value in volts) and that the voltage drop in the capacitor for t>0 is found to be 1-e^(-t/2). Find the voltage drop in the resistor, the current through the capacitor and the value of the source.

This problem is a match made in heaven for the expert tool, because we have one unknown value in the circuit (i.e. the value of the step source) and we have one known answer (i.e. the voltage drop in the capacitor.) So, the game plan here is to run this circuit through Symbulator’s expert mode, add one new equation and one new unknown, and then solve. First, let’s generate the new equation.

The non-expert user is right here: the new equation is simply 1-e^(-t/2) = vc. Everything you type into expert mode in version 9 — equations, conditions, expressions — is read in the time domain, the same domain as the answers on screen. You do not convert, and you do not have to remember which side of the transform you are standing on.

If you would rather convert by hand, t2s is still there and an equation already written in s is left alone rather than transformed twice:

t2s(1-e^(-t/2))

which gives \(\frac{1}{s \left(2 s + 1\right)}\) — the same statement, one domain over.

We have our new equation. Copy this equation into the clipboard, since we will want to paste it in the Expert window. Now let’s run the Expert simulation of the circuit. Let’s define the value of the source as a*u(t), since we know it’s a step source; the variable a will serve as the unknown value, for which we will solve in the Expert mode.

e,1,0,a*u(t)
r,1,2,2
c,2,0,1,0

Choose TR, then open Expert Mode. Add equations takes one per line, so there is no and to prefix — the word exists on the calculator only because the equation is being appended to a list:

1-e^(-t/2) = vc

Add unknowns likewise takes the name on its own, with no leading comma:

a

Let Symbulator solve the equations now. Once it is done, we want to do a quick sanity check. If the system solved correctly, we should see that Symbulator has as voltage drop in the capacitor the expression we already know from the problem statement:

vc in Results reads \(1 - e^{- \frac{t}{2}}\), which it does.

This is correct, and the answers we want are on screen with it: ic is \(\frac{e^{- \frac{t}{2}}}{2}\), and the resistor's voltage drop line gives \(e^{- \frac{t}{2}}\).

These are the correct answers. The value of the step source is 1 volt.

Bo2's Drill Exercise 4.13 (Expert)

Open in app ↗Open in split view

This problem is solved just like the one we already saw. First, find the new equation:

No transform this time either — write what you know:

vr = e^(-t)

Then run the Expert simulation. I decided to use vs as the step value of the source:

e,1,0,vs*u(t)
r,1,2,1
c,2,0,1,0

Choose TR, and put vs in Add unknowns beside the equation above.

Run it. The sanity check is already on screen: vr reads \(e^{- t}\), which is what we told it.

And the answers with it: ir is \(e^{- t}\) and the source's value vs is 1 V.

Ain’t that a beauty? Yes, ma’am! Atta boy, Symbulator! Atta boy!

Bo2's 164 (Expert)

Open in app ↗Open in split view

This is the same thing as the previous two, so there is no need for commentary.

e,1,0,vs*u(t)
r,1,2,2
l,2,0,1,0

Choose TR, put vs in Add unknowns, and give Add equations the answer you know — in time, as always here:

1-e^(-2*t) = il

Run it, then put the sanity check into Evaluate:

It does add up — il reads \(1 - e^{- 2 t}\) — and the answers are on screen with it: vl is \(2 e^{- 2 t}\), vr is \(2 - 2 e^{- 2 t}\), and the source's value vs is 2 V.

This is the end of the TR section.