Examples from Nilsson & Riedel 12e

Solution

The circuit consists of a current source feeding a ladder of seven resistors. The question wants the current in one branch, \(i_o\) in the 24 Ω resistor, and the voltage across another, \(v_o\) across the 30 Ω. We describe the circuit as it is drawn, naming the source j and the resistors r1 to r7 from left to right, and taking the bottom rail as ground, so \(i_o\) is the current through r7 and \(v_o\) the voltage across r6. One run in DC returns every current and voltage in the circuit, those two among them.

j,0,1,8
r1,1,2,36
r2,2,0,44
r3,1,0,10
r4,1,3,40
r5,3,4,10
r6,4,0,30
r7,1,0,24

Set Analysis to DC — direct current.

Symbulator returns i_r7 = 2 A (the book's \(i_o\)) and v_r6 = 18 V (the book's \(v_o\)).

Solution

A Wheatstone bridge measures a resistance nobody knows, \(R_x\), by comparing it with resistances that are known. The adjustable resistor \(R_3\) is turned until the galvanometer between the two arms of the bridge carries no current, and at that setting the unknown can be read off the others. So we describe the circuit with the two resistances the question leaves open as symbols rather than numbers, which we name R_3 for the adjustable one and R_x for the unknown, after the book's. We name the two known resistors r1 and r2, and call the two arms' midpoints a and b. We describe the galvanometer as a short circuit, the element s, which we name sg: the figure gives it no resistance, and what we will need from it is its current, which a short reports as isg. We leave the source as a symbol too, V_s, since its value plays no part at balance. We run this as it stands, in DC, and every answer comes back as a formula in the three symbols.

e,1,0,V_s
r1,1,a,1'k
r2,1,b,4'k
sg,a,b
r3,a,0,R_3
rx,b,0,R_x

Set Analysis to DC — direct current.

With the results on screen, we open the Solve card under them and ask the question the way the bridge is used. The balance condition, no current through the galvanometer, is the equation isg=0. The resistance we want is the unknown, R_x. The setting of the dial is a condition, first at its lowest, 10 Ω:

isg=0

R_x

R_3=10

Press Solve equations.

The card returns R_x = 40 Ω.

For the other end of the dial we keep the equation and the unknown as they are and change only the condition, to the highest setting, 2 kΩ, written with the usual shorthand:

R_3=2'k

Then press Solve equations again.

The card returns R_x = 8000 Ω.

So the smallest resistance the bridge can measure is 40 Ω and the largest 8000 Ω: its range is 40 Ω to 8 kΩ.

Solution

The circuit is a bridge: five resistors in a diamond with a sixth across the middle, fed by one source. The question wants what that source delivers, its current and its power. We write the bridge as it is drawn, naming the source e and the six resistors r1 to r6.

e,1,0,40
r1,1,2,5
r2,2,3,100
r3,2,4,125
r4,3,4,25
r5,3,0,40
r6,4,0,37.5

Set Analysis to DC — direct current.

The source's card reports the power it delivers as -pe, which is the power the question asks for. Its current it reports into the source, the same way as for every other element, so the current it supplies is the negative of that, which we read with a minus sign in the Evaluate card.

a) We type -i_e into Evaluate:

-i_e

It gives 0.5 A (the book's \(i\)).

b) The power it supplies is read off its card: -p_e = 20 W (the book's \(p\)).

Solution

The circuit is resistive, with a dependent voltage source at its far end. Its value is eight times \(i_\phi\), the current through the 5 Ω resistor, and the question wants the power that resistor dissipates. In Symbulator a dependent source needs no special element. It is a source whose value names another answer. We name the 5 Ω resistor r3, so its current is ir3, and write the dependent source as e2 with the value 8*ir3. We number the nodes 1 to 4 from the source rightward, with the bottom rail as ground. The power asked for is then pr3, the power consumed by r3.

e1,1,0,20
r1,1,2,2
r2,2,0,20
r3,2,3,5
r4,3,0,10
r5,3,4,2
e2,4,0,8*ir3

Set Analysis to DC — direct current. Set Rounding in Settings to approx to n digits with n = 4.

Symbulator returns p_r3 = 7.2 W (the book's \(p\)).

Solution

The circuit again has a dependent voltage source, this time worth fifteen times \(i_\phi\), the current in the 20 Ω resistor, and again the question wants one resistor's power, the 4 Ω's. We name the 20 Ω r4, so its current is ir4 and the source e2 has the value 15*ir4. The 4 Ω we name r3, and its power is pr3. We number the nodes 1 to 3 with the bottom rail as ground.

e1,1,0,50
r1,1,3,1
r2,1,2,5
r3,2,3,4
r4,2,0,20
e2,3,0,15*ir4

Set Analysis to DC — direct current.

Symbulator returns p_r3 = 16 W (the book's \(p\)).

Solution

The circuit consists of a voltage source and a current source feeding three resistors, and the question wants the current in each of the three branches. We name the sources e and j and the resistors r1 to r3, and take the bottom rail as ground. A current source is written like a voltage source, name, two nodes, value, and its current flows through it from the first node to the second. The 3 A source's arrow points up from the bottom rail into node 2, so we write it j,0,2,3. We write each resistor's nodes in the direction of the figure's arrow, so that its current is counted as the book counts it: \(i_a\) is ir1, \(i_b\) is ir2 and \(i_c\) is ir3.

e,1,0,50
r1,1,2,5
r2,2,0,10
r3,2,0,40
j,0,2,3

Set Analysis to DC — direct current.

Symbulator returns i_r1 = 2 A (the book's \(i_a\)), i_r2 = 4 A (the book's \(i_b\)) and i_r3 = 1 A (the book's \(i_c\)).

Solution

The circuit consists of two sources, one of voltage and one of current, and six resistors. The question wants the voltage across the 100 Ω, and the power each source develops. We write every element as it stands in the figure, the 125 Ω across the voltage source and the 10 Ω under the current source included, naming the resistors r1 to r6 from left to right and the sources e and j. The current source's arrow points down, from node 2 towards the 10 Ω, so we write its nodes in that order, j,2,9,8, calling the node between the source and the resistor 9. \(v_o\) is the voltage across r4.

e,1,0,250
r1,1,0,125
r2,1,2,25
j,2,9,8
r3,9,0,10
r4,2,0,100
r5,2,3,5
r6,3,0,15

Set Analysis to DC — direct current.

The power a source develops is what it delivers, which each source's card reports as -pe and -pj.

Symbulator returns a) v_r4 = 20 V (the book's \(v_o\)), b) -p_e = 2800 W (the book's \(p_{250\,V}\)) and c) -p_j = 480 W (the book's \(p_{8\,A}\)).

Solution

The circuit consists of a source and two resistors with a load \(R_L\) connected across the second, and the question is which load draws the most power from those terminals, and how much. That is a question about the circuit seen from the load, so we describe the source and the two resistors and leave \(R_L\) out, calling its terminals node 2 and ground. The Find equivalent card, with Thévenin / Norton chosen and those two terminals named, reduces the circuit to its Thévenin equivalent and reports beside it the load that would draw the most power and how much that is: by the maximum power theorem that load equals the Thévenin resistance, which the card reports as req, and the power it reports as pmax.

e,1,0,360
r1,1,2,30
r2,2,0,150

Open Find equivalent, choose Thévenin / Norton, and give the two terminals 2 and 0.

Symbulator returns a) req = 25 Ω (the book's \(R_L\)) and b) pmax = 900 W.

Solution

The circuit consists of two independent sources, two dependent ones and three resistors, and the question wants the voltage across the 20 Ω. We name the 5 Ω resistor r1, the 20 Ω r2 and the 10 Ω r3. The dependent current source is worth \(0.4v_\Delta\), and \(v_\Delta\) is the voltage across the 10 Ω, so we write its value as 0.4*vr3. The dependent voltage source is worth \(2i_\Delta\), and \(i_\Delta\) is the current through the 5 Ω, so we write its value as 2*ir1. We write each source's nodes in the order its arrow or its polarity marks give, name the three inner nodes a, b and c, and take the bottom-right node as ground. \(v_o\) is the voltage across r2.

e1,1,c,10
r1,1,a,5
r2,a,c,20
r3,b,0,10
j1,0,b,5
j2,b,a,0.4*vr3
e2,0,c,2*ir1

Set Analysis to DC — direct current.

Symbulator returns v_r2 = 24 V (the book's \(v_o\)).

Solution

The circuit is an op amp with two inputs, \(v_a\) into the inverting side through a 25 kΩ resistor and \(v_b\) straight into the non-inverting side, with a 100 kΩ feedback resistor. The question wants the output for two pairs of input values, and then the range of one input that keeps the amplifier out of saturation. An ideal op amp is the element o, whose three nodes are its non-inverting input, its inverting input and its output, in that order. We choose to write the two inputs as voltage sources with the symbolic values va and vb rather than the numbers in the question, so that one run returns the output as a formula in both, and each part can then be asked of that formula. We name the sources ea and eb, the resistors r1 and r2, and number the nodes from the \(v_a\) input, the output being node 3. Symbulator's ideal op amp has no supplies, so it reports whatever output the inputs demand. Whether that output is within the ±10 V supplies of the figure is something we check afterwards.

ea,1,0,va
r1,1,2,25'k
r2,2,3,100'k
eb,4,0,vb
o,4,2,3

Set Analysis to DC — direct current.

v_3 is the book's \(v_o\).

Part (a) is that formula at \(v_a\) = 1 V and \(v_b\) = 0 V. We type the output's name into Evaluate and the two values into its Conditions box:

v_3

va = 1
vb = 0

It gives -4 V (the book's \(v_o\)).

Part (b) changes only the second condition:

v_3

va = 1
vb = 2

It gives 6 V (the book's \(v_o\)).

Part (c) asks where saturation begins. The op amp is linear while its output lies between the supply rails, which the figure gives as ±10 V, so the question is which \(v_b\) puts the output exactly on a rail. That is a question for the Solve card: the output on the upper rail is the equation, \(v_b\) the unknown, and \(v_a\) = 1.5 V a condition:

v_3=10

vb

va=1.5

Press Solve equations.

The card returns vb = 3.2 V.

For the lower rail we keep the unknown and the condition and change only the equation:

v_3=-10

Then press Solve equations again.

The card returns vb = -0.8 V.

a) So v_o = -4 V, inside the ±10 V supplies: the op amp is in its linear region and that is the answer.

b) Now v_o = 6 V, which is inside the supplies again.

c) The output reaches +10 V at v_b = 3.2 V and −10 V at v_b = -0.8 V, so the op amp avoids saturation for -0.8 V ≤ v_b ≤ 3.2 V.

Solution

The circuit is a summing amplifier, which adds several input voltages, each with its own gain, and inverts the sum. The question asks us to verify a design made for the gains given, then to find the range of one input that keeps it linear. The output of a summing amplifier with feedback resistor \(R_f\) and input resistors \(R_a\), \(R_b\) and \(R_c\) is \(v_o = -(R_f/R_a)v_a - (R_f/R_b)v_b - (R_f/R_c)v_c\), so with \(R_f\) fixed at 20 kΩ the gains of 4, 1 and 5 want \(R_a\) = 20/4 = 5 kΩ, \(R_b\) = 20/1 = 20 kΩ and \(R_c\) = 20/5 = 4 kΩ. That is the design, and the run checks it. We describe the circuit with those four resistors, which we name r1, r2, r3 and rf. We write the three inputs as sources ea, eb and ec with the symbolic values va, vb and vc, so that the output comes back as a formula, and we write the op amp o with its non-inverting input at ground, its inverting input at a node we call n, and its output at node 4.

ea,1,0,va
eb,2,0,vb
ec,3,0,vc
r1,1,n,5'k
r2,2,n,20'k
r3,3,n,4'k
rf,n,4,20'k
o,0,n,4

Set Analysis to DC — direct current.

v_4 is the book's \(v_o\).

Part (b) fixes two of the inputs and asks for the range of the third that keeps the output between the ±12 V rails. In the Solve card we put the output on the lower rail as the equation, name \(v_b\) as the unknown, and give the two fixed inputs as conditions:

v_4=-12

vb

va=2
vc=-1

Press Solve equations.

The card returns vb = 9 V.

For the upper rail we change only the equation:

v_4=12

Then press Solve equations again.

The card returns vb = -15 V.

a) The run returns exactly the formula the design was to produce, which verifies the three resistor values.

b) The output sits on the −12 V rail at v_b = 9 V and on the +12 V rail at v_b = -15 V, so the op amp remains linear for -15 V ≤ v_b ≤ 9 V.

Solution

The circuit is the same summing amplifier with the three inputs now given as numbers, and the question is turned round: not the output for a given feedback resistor, but the largest feedback resistor for which the output stays within the rails. We describe the circuit as in part (a), the inputs as sources of 2, 3 and −1 V, and leave the feedback resistor as the symbol rf instead of a number, so that the run returns the output as a formula in rf.

ea,1,0,2
eb,2,0,3
ec,3,0,-1
r1,1,n,5'k
r2,2,n,20'k
r3,3,n,4'k
rf,n,4,rf
o,0,n,4

Set Analysis to DC — direct current. Set Rounding in Settings to approx to n digits with n = 4.

v_4 is the book's \(v_o\).

The output is negative for any feedback resistor, so the rail it can reach is −12 V, and the largest feedback resistor is the one that puts the output exactly there. In the Solve card that is one equation and one unknown:

v_4=-12

rf

Press Solve equations.

The card returns rf = 40000 Ω.

c) So the feedback resistor can be as large as 40000 Ω, 40 kΩ. Any larger and the output would have to go beyond −12 V, which it cannot: the op amp saturates.

Solution

The circuit is a difference amplifier, which amplifies the difference between its two inputs. The question asks us to verify a design made for a gain of 8, then to find the range of one input that keeps it linear. Its output is \(v_o = (R_b/R_a)(v_b - v_a)\), provided the four resistors satisfy \(R_a/R_b = R_c/R_d\), so a gain of 8 needs \(R_b\) eight times \(R_a\) and \(R_d\) eight times \(R_c\). One choice, the book's, is \(R_a\) = \(R_c\) = 1.5 kΩ and \(R_b\) = \(R_d\) = 12 kΩ. We describe the circuit with those values, naming the resistors ra to rd after the book's, write the two inputs as sources with the symbolic values va and vb so that the output comes back as a formula, and call the op amp's two input nodes p and n and its output node 3. The run checks the design.

ea,1,0,va
eb,2,0,vb
ra,1,n,1.5'k
rb,n,3,12'k
rc,2,p,1.5'k
rd,p,0,12'k
o,p,n,3

Set Analysis to DC — direct current.

v_3 is the book's \(v_o\).

Part (b) fixes \(v_a\) and asks for the range of \(v_b\) that keeps the output between the ±8 V rails. In the Solve card, the output on the upper rail is the equation, \(v_b\) the unknown and \(v_a\) = 1 V a condition:

v_3=8

vb

va=1

Press Solve equations.

The card returns vb = 2 V.

For the lower rail we change only the equation:

v_3=-8

Then press Solve equations again.

The card returns vb = 0 V.

a) The run returns exactly 8(v_b − v_a), the gain the design was to produce, which verifies it.

b) The output reaches +8 V at v_b = 2 V and −8 V at v_b = 0 V, so the op amp remains linear for 0 V ≤ v_b ≤ 2 V.

Solution

The circuit is a noninverting amplifier built not round an ideal op amp but round a realistic model of one: a dependent voltage source with a large but finite gain, an input resistance between its two inputs, and an output resistance in series with its output. The question wants the gain of the whole amplifier, output over source. We write the model as those three ordinary elements. We call the op amp's two input nodes p and n, write the input resistance as ri between them, the output resistance as ro, and the dependent source as ea with the value 50000*(vp-vn), the open-loop gain times the voltage between the inputs, \(A(v_p - v_n)\). The other three resistors are rg, rs and rf. We leave the source as the symbol vg, so that the output comes back as a multiple of it. The gain is that multiple, which we will read in the Evaluate card.

eg,1,0,vg
rg,1,p,1'k
ri,p,n,100'k
rs,n,0,2'k
rf,n,3,10'k
ro,4,3,7.5'k
ea,4,0,50000*(vp-vn)

Set Analysis to DC — direct current. Set Rounding in Settings to approx (full precision).

Then we type v_3/vg into Evaluate:

v_3/vg

It gives 5.9988 (the book's \(v_o/v_g\)).

Solution

The circuit is a T of three resistors seen as a two-port, and the question wants its z parameters: the four numbers that relate the two port voltages to the two port currents. We take port 1 as node 1 with ground and port 2 as node 2 with ground, write the three resistors between those nodes, naming each after its value, and name the two ports to the Find equivalent card with Two-port parameters chosen and the kind set to z. It returns the four parameters, named z11 to z22.

r5,1,2,5
r20,1,0,20
r15,2,0,15

Open Find equivalent, choose Two-port parameters, kind z, with the ports at 1 and 2.

Symbulator returns z11 = 10 Ω, z12 = 7.5 Ω, z21 = 7.5 Ω and z22 = 9.375 Ω.

Solution

The circuit consists of two identical amplifiers, each known only by its four h parameters, one feeding the other between a source with an internal resistance and a load. The question wants the voltage gain of the pair. A two-port known only by its parameters is an element of its own. Since these are h parameters we use the element h, and write the four values after the two nodes as a bracketed term, [1000,0.0015,100,0.0001], the 100 µS written as 0.0001. We write the two amplifiers as two such lines, h1 and h2, sharing a node we call b, between the source's 500 Ω and the 10 kΩ load. We leave the source as the symbol vg, so that the output comes back as a multiple of it, and the gain is that multiple, which we will read in the Evaluate card.

e,1,0,vg
rs,1,a,500
h1,a,b,[1000,0.0015,100,0.0001]
h2,b,c,[1000,0.0015,100,0.0001]
rl,c,0,10'k

Set Analysis to DC — direct current. Set Rounding in Settings to approx (full precision).

Then we type v_c/vg into Evaluate:

v_c/vg

It gives 33333.33 (the book's \(V_2/V_g\)).

Solution

The circuit consists of an inductor that has been fed by a current source for a long time, and a switch that then cuts the source off and leaves the inductor to discharge through three resistors. The question wants the inductor's current and two other quantities afterwards, as functions of time. There are two intervals, and two runs. Before \(t\) = 0 the switch has been closed for a long time, so any transient has died away and every current is steady. In a steady circuit an inductor carries its current with no voltage across it, which is to say it behaves as a wire. The full current of the source, 20 A, would run through it, and should be the initial condition of the inductor. To verify this intuition, we can run a DC simulation. Describe the circuit as it stands before the switch opens, the 20 A source, the 0.1 Ω resistor, the inductor and the three resistors beyond it, and run it for the inductor's current. We name the source j, the inductor l and the resistors r0 to r3, and write the inductor with no fifth field, since nothing about its past is being told:

j,0,1,20
r0,1,0,0.1
l,1,0,2
r1,1,2,2
r2,2,0,10
r3,2,0,40

Set Analysis to DC — direct current.

Symbulator returns i_l = 20 A (the book's \(i_L(0)\)).

Opening the switch disconnects the source and the 0.1 Ω resistor and leaves the inductor to release its energy through the three resistors. That is the circuit for the second interval, \(t\) ≥ 0, and we describe it with the same names, dropping j and r0. The inductor now starts with the 20 A just found, which we write into its line as a fifth field, after the inductance. The answers come back as functions of \(t\): \(i_L\) is the current through l, \(i_o\) the current through r3, and \(v_o\) the voltage at node 2.

l,1,0,2,20
r1,1,2,2
r2,2,0,10
r3,2,0,40

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

a) is i_l, b) is i_r3 (the book's \(i_o\)) and c) is v_2 (the book's \(v_o\)).

Solution

The circuit consists of a capacitor charged from one source, then switched over to a set of resistors to discharge. The question wants its voltage and two other quantities after the switch moves. There are two intervals, and two runs. At position x the capacitor has been connected to the 100 V source through the 10 kΩ resistor for a long time, and a capacitor in a steady circuit carries no current, so nothing flows in the 10 kΩ and the capacitor sits at the voltage of the source, 100 V, which would be its initial condition for the next interval. To confirm that intuition, we describe that circuit, the source, the 10 kΩ, which we name r0, and the capacitor, written without a fifth field, and run it in DC to find the capacitor's voltage:

e,1,0,100
r0,1,2,10'k
c,2,0,0.5'u

Set Analysis to DC — direct current.

Symbulator returns v_2 = 100 V (the book's \(v_C(0)\)).

At position y the capacitor is connected instead to the 32 kΩ resistor and the two beyond it, and discharges through them. We describe that second circuit with the 100 V just found as the capacitor's fifth field, naming the three resistors r1 to r3 and the capacitor's node 1, and set the analysis to TR. \(v_C\) is the voltage at node 1, \(v_o\) the voltage at node 2, and \(i_o\) the current through r3.

c,1,0,0.5'u,100
r1,1,2,32'k
r2,2,0,240'k
r3,2,0,60'k

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

a) is v_1 (the book's \(v_C\)), b) is v_2 (the book's \(v_o\)) and c) is i_r3 (the book's \(i_o\)).

Solution

The circuit consists of an inductor that has been carrying a current from one source and is switched, without a break, onto another. The question wants its current afterwards and the voltage across it the instant after the switch moves. There are two intervals, and two runs. With the switch at position a, the inductor has been in parallel with the 10 Ω resistor and the 8 A source for a long time, so the circuit is steady. The inductor behaves as a wire, so the whole 8 A would flow through it and none through the resistor, and that, with its sign, should be its initial condition. To verify this intuition, we can run a DC simulation. The direction matters: the source's arrow points down through the source, so its current comes up through the inductor, against the book's arrow for \(i\). We write the source as j,1,0,8, its current flowing from node 1 down to ground, and the inductor as l,1,0,0.2, so that its current is counted downward like the book's \(i\). A DC run then gives the current with its sign:

j,1,0,8
r,1,0,10
l,1,0,0.2

Set Analysis to DC — direct current.

Symbulator returns i_l = -8 A (the book's \(i(0)\)).

Moving the switch to b connects the inductor, through the 2 Ω resistor, to the 24 V source instead. Because the switch is make-before-break, the inductor's current does not jump at the switching: it starts at the −8 A just found. We describe the second circuit with that −8 as the inductor's fifth field, sign included, naming the source e and the resistor r1, and set the analysis to TR. \(i\) is the current through l.

e,1,0,24
r1,1,2,2
l,2,0,0.2,-8

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

a) is i_l (the book's \(i\)).

b) The inductor is between node 2 and ground, so the voltage across it is v_2, and we read its value the instant after the switch has moved, at \(t\) = 0, with Evaluate:

v_2

t = 0

It gives 40 V (the book's \(v(0^+)\)).

Solution

The circuit consists of two coils wound on one core, both fed from the same node through one resistor, and a switch that connects the source at \(t\) = 0. The question wants the source's current, the voltage across the coils and the current in each, all as functions of time. We write each coil as an inductor line of its own, l1 and l2, and the coupling between them as the m line, which names the two coils and their mutual inductance, 6 H. No energy is stored, so we write neither inductor with a fifth field. The switch closing at \(t\) = 0 needs no element: in TR a source with a plain numerical value is a step that begins at \(t\) = 0, which is exactly what closing the switch on the 120 V source does. We name the source e and the resistor r1, and set the analysis to TR. \(i_o\) is the current through r1, \(v_o\) the voltage at node 2, and \(i_1\) and \(i_2\) the currents through l1 and l2.

e,1,0,120
r1,1,2,7.5
l1,2,0,3
l2,2,0,15
m,l1,l2,6

Set Analysis to TR — transient / time domain.

Here i_r1 is the book's \(i_o\), v_2 is the book's \(v_o\), i_l1 is the book's \(i_1\) and i_l2 is the book's \(i_2\).

Solution

The circuit has an inductor and two switches that open one after the other. The question wants the inductor's current between the two openings. There are three intervals this time, and three runs. In the first, both switches have been closed for a long time and the circuit is steady. We describe the whole circuit, the 60 V source, the 4 Ω, 12 Ω, 6 Ω and 3 Ω resistors, the inductor and the 18 Ω, naming each resistor after its value, and run it in DC for the inductor's current:

e,1,0,60
r4,1,2,4
r12,2,0,12
r6,2,0,6
r3,2,3,3
l,3,0,0.15
r18,3,0,18

Set Analysis to DC — direct current.

Symbulator returns i_l = 6 A (the book's \(i_L(0)\)).

In the second interval switch 1 has opened, which disconnects the 60 V source and the 4 Ω and 12 Ω resistors. We describe what remains, the 6 Ω, the 3 Ω, the inductor and the 18 Ω, with the same names, and give the inductor the 6 A just found as its fifth field. \(i_L\) is the current through l.

r6,2,0,6
r3,2,3,3
l,3,0,0.15,6
r18,3,0,18

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain. Set Rounding in Settings to approx to n digits with n = 4.

a) is i_l.

This run holds until switch 2 opens at 35 ms, and the inductor's current at that instant is where the third interval starts. We read it from the answer with Evaluate, giving the instant in seconds:

i_l

t = 0.035

It gives 1.48 A (the book's \(i_L(35\,\mathrm{ms})\)).

Solution

This is the third interval of the same problem: switch 2 has opened too, removing the 18 Ω resistor, so the inductor now discharges through the 3 Ω and 6 Ω alone, and the question wants its current from that moment on. We describe that circuit with the 18 Ω dropped, and give the inductor as its fifth field the current found at the end of the previous entry, \(6e^{-40 \times 0.035} = 6e^{-1.4}\), about 1.48 A. We write it as the expression 6*exp(-1.4), which is exact. We measure time from the second switching, as the book does, so this run's \(t\) = 0 is the instant switch 2 opens. The analysis is TR.

r6,2,0,6
r3,2,3,3
l,3,0,0.15,6*exp(-1.4)

Set Analysis to TR — transient / time domain. Set Rounding in Settings to approx to n digits with n = 4.

b) is i_l.

Solution

The circuit consists of a charged capacitor across two resistors and a dependent current source whose value is seven times the current in one of the resistors. The question wants the capacitor's voltage as a function of time, and then how long it takes to reach 150 V. The 10 V on the capacitor is given, and we write it into the capacitor's fifth field. We name the 20 kΩ resistor r2, so its current is ir2 and the dependent source's value is 7*ir2. The source's arrow points up into node 1, so we write it j,0,1,7*ir2. We set the analysis to TR. \(v_o\) is the voltage at node 1.

c,1,0,5'u,10
r1,1,0,10'k
r2,1,0,20'k
j,0,1,7*ir2

Set Analysis to TR — transient / time domain. Set Rounding in Settings to approx to n digits with n = 4.

a) is v_1 (the book's \(v_o\)).

The exponent is positive, so the voltage grows instead of decaying: the dependent source feeds the capacitor faster than the resistors drain it.

b) The time it takes to reach 150 V is a question for the Solve card, with the time as the unknown:

v_1=150

t

Press Solve equations.

The card returns t = 0.0677 s.

So the capacitor short-circuits after 67.7 ms.

Solution

The circuit consists of a capacitor, an inductor and a resistor in parallel with no source, holding an initial voltage and an initial current. The question wants the voltage across the three as a function of time, and the following example wants the three branch currents. We write each element as a line between node 1 and ground, naming them c, l and r, and put the two initial conditions the question gives in the fifth fields: 12 V on the capacitor and 30 mA, written 0.03, on the inductor. The run is the circuit releasing the energy it holds. \(v\) is the voltage at node 1, and \(i_R\) and \(i_L\) the currents through r and l.

c,1,0,0.2'u,12
l,1,0,50'm,0.03
r,1,0,200

Set Analysis to TR — transient / time domain.

v_1 is the book's \(v\).

Solution

The circuit is the same three elements in parallel with different values, an initial current in the inductor and none on the capacitor. The question wants the voltage across them as a function of time. We describe them as before: \(V_0\) = 0 goes in the capacitor's fifth field and \(I_0\) = −12.25 mA in the inductor's, written −0.01225 with the sign the question gives it. We set the analysis to TR. \(v\) is the voltage at node 1.

c,1,0,125'n,0
l,1,0,8,-0.01225
r,1,0,20'k

Set Analysis to TR — transient / time domain. Set Rounding in Settings to approx (full precision).

v_1 is the book's \(v\).

The book prints the amplitude as 100 and the frequency of the sine as 979.80, both rounded. To the same figures the run gives 100.02 and 979.80.

Solution

The circuit is a charged capacitor discharging round a loop through an inductor and a resistor. The question wants the loop current and the capacitor's voltage as functions of time. We write the capacitor with 100 V as its fifth field, and the inductor and the resistor after it round the loop, naming the three c, l and r. We write the inductor's nodes as 2 then 1 on purpose, so that its current is counted in the direction of the book's arrow for \(i\). Written the other way round, the answer would come back with its sign reversed. We set the analysis to TR. \(i\) is the current through l and \(v_C\) the voltage at node 1.

c,1,0,0.1'u,100
l,2,1,0.1
r,2,0,560

Set Analysis to TR — transient / time domain. Set Rounding in Settings to approx to n digits with n = 4.

a) is i_l (the book's \(i\)) and b) is v_1 (the book's \(v_C\)).

Solution

The circuit is a source switched at \(t\) = 0 onto a loop of an inductor, a resistor and a capacitor, none of them holding any energy. The question wants the capacitor's voltage as a function of time. In TR a source with a plain numerical value is a step beginning at \(t\) = 0, which is what the switch closing does, so the switch needs no element. We write the 48 V source and then the three elements in order round the loop, numbering the nodes 1 to 3, and write the inductor and the capacitor without a fifth field, since neither stores energy. We set the analysis to TR. \(v_C\) is the voltage at node 3, the capacitor's upper end.

e,1,0,48
l,1,2,0.1
r,2,3,1250
c,3,0,0.4'u

Set Analysis to TR — transient / time domain.

v_3 is the book's \(v_C\).

Solution

The circuit consists of two loops sharing a resistor, each with an inductor, switched onto a source at \(t\) = 0 with no energy stored. The question wants the two loop currents as functions of time. No energy is stored, so we write the two inductors without fifth fields, and the switch closing at \(t\) = 0 onto the 336 V source needs no element of its own: in TR a numerical source value is a step that begins at \(t\) = 0. We name the inductors l1 and l2 after the book's \(i_1\) and \(i_2\), which are the currents through them, and the resistors r1 and r2. TR returns the two currents as functions of \(t\). The Laplace transform and its inversion happen inside the solver.

e,1,0,336
l1,1,2,8.4
r1,2,0,42
l2,2,3,10
r2,3,0,48

Set Analysis to TR — transient / time domain.

Here i_l1 is the book's \(i_1\) and i_l2 is the book's \(i_2\).

Solution

The circuit consists of two coupled coils, the primary fed from a source through a switch and the secondary closed on two resistors. The switch takes the source out at \(t\) = 0, and the question wants the secondary's current afterwards. There are two intervals, and two runs. With the switch at position a the primary side has been steady for a long time. We describe the whole circuit, naming the resistors after their values, r9, r3, r2b and r10, the coils l1 and l2, and the coupling as the m line with their mutual inductance, 2 H. We call the secondary's top q and the node between its two resistors c, and take its bottom as ground -- nothing conducts between the two windings, so joining their bottoms changes no current. A DC run gives both currents:

e,1,0,60
r9,1,a,9
r3,a,p,3
l1,p,0,2
m,l1,l2,2
l2,q,0,8
r2b,q,c,2
r10,c,0,10

Set Analysis to DC — direct current.

Symbulator returns i_l1 = 5 A (the book's \(i_1(0)\)) and i_l2 = 0 A (the book's \(i_2(0)\)).

At \(t\) = 0 the switch moves to b, which takes the source and the 9 Ω out and closes the primary on the 3 Ω alone. We describe that circuit with the same names and the two currents just found as the coils' fifth fields, 5 and 0. \(i_2\) is the current through l2.

r3,0,p,3
l1,p,0,2,5
m,l1,l2,2
l2,q,0,8,0
r2b,q,c,2
r10,c,0,10

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain. Set Rounding in Settings to approx to n digits with n = 4.

i_l2 is the book's \(i_2\).

Solution

The circuit consists of two inductors, one carrying a current from a source and the other idle behind a closed switch that shorts it out. Opening the switch at \(t\) = 0 forces the two into series, and the question wants the voltage across the second and the current in the first afterwards. There are two intervals, and two runs. While the switch is closed it joins the junction after \(L_1\) to the bottom rail, and the circuit has been steady for a long time. We describe that circuit and run it in DC for the two inductor currents. We write the closed switch by giving both its ends the same node, ground, and name the inductors l1 and l2 after the book's \(L_1\) and \(L_2\):

e,1,0,100
r1,1,2,10
l1,2,0,3
r2,0,4,15
l2,4,0,2

Set Analysis to DC — direct current.

Symbulator returns i_l1 = 10 A (the book's \(i_1(0)\)) and i_l2 = 0 A (the book's \(i_2(0)\)).

Opening the switch removes that wire, so the two inductors are now in series with the 15 Ω between them. We describe that circuit with the 10 A and 0 just found as their fifth fields, numbering the nodes 1 to 4 from the source. \(v_o\) is the voltage at node 3, the top of the 15 Ω and \(L_2\), and \(i_1\) the current through l1.

e,1,0,100
r1,1,2,10
l1,2,3,3,10
r2,3,4,15
l2,4,0,2,0

TROpen in app ↗Open in split view

Set Analysis to TR — transient / time domain.

Here v_3 is the book's \(v_o\) and i_l1 is the book's \(i_1\).

The \(\delta(t)\) in \(v_o\) is an impulse. Opening the switch puts the two inductors in series, one carrying 10 A and the other none, and they must carry the same current from then on. An inductor's current cannot jump without an impulse of voltage across it, and the weight of this one, 12, is what it takes to bring \(L_2\)'s current from 0 to the 6 A the pair settle on.

Solution

The circuit consists of a sinusoidal current source at 200,000 rad/s feeding a resistor, a capacitor, and a branch of a resistor and an inductor. The question wants the admittance the source sees, the voltage across it and the current in the inductive branch as phasors, and those two as functions of time. The inductor and the capacitor are given as an inductance and a capacitance, and we write them as they are, l,2,0,40'u and c,1,0,1'u, putting the source's frequency, 200,000 rad/s, in the ω — angular frequency box. The conversion to impedances is done inside the solver, so part (a), constructing the frequency-domain equivalent circuit, is out of scope for Symbulator and has been skipped. We write the current source as j,0,1,8, its arrow pointing up into node 1, with its 8 A as the amplitude, which is what the book's phasors carry too, and name the resistors r1 and r2. In AC each answer is a phasor, printed both as a complex number and as an amplitude with an angle: \(v\) is the voltage at node 1 and \(i\) the current through r2.

j,0,1,8
r1,1,0,10
r2,1,2,6
l,2,0,40'u
c,1,0,1'u

Set Analysis to AC — alternating current. Put 200000 in the ω — angular frequency box. Set Rounding in Settings to approx to n digits with n = 4.

b) Symbulator reports the impedance seen by each source, here z_j, and the admittance the source sees is its reciprocal, so we type into Evaluate:

1/z_j

It gives 0.16 + 0.12j S (0.2000∠36.87°, the book's \(Y\)).

c) and d) are read off the run: v_1 = 32 - 24j V (40.00∠-36.87°, the book's \(v\)) and i_r2 = -4j A (4.000∠-90.00°, the book's \(i\)).

e) The steady-state expressions are the two phasors written back as functions of time at the source's frequency, which is circuit theory rather than a run: \(v\) = 40 cos(200,000\(t\) − 36.87°) V and \(i\) = 4 cos(200,000\(t\) − 90°) A.

Solution

The circuit is a bridge of five impedances fed by one source, every impedance given in ohms and several of them complex. The question wants the current in every branch and the voltage at two nodes. We write each impedance exactly as given, -4j, 63.2+2.4j, 20+60j, as a resistor with a complex value, naming the five r1 to r5 in the order of the book's \(I_1\) to \(I_5\), and we write each one's nodes in the direction of the figure's arrow, so that its current is counted as the book counts it. We keep the figure's letters for the nodes, with d as ground, so \(V_1\) and \(V_2\) are the voltages at b and c.

e,a,0,120
r1,a,b,-4j
r2,a,c,63.2+2.4j
r3,b,c,10
r4,b,0,20+60j
r5,c,0,-20j

Set Analysis to AC — alternating current. Leave omega in the ω — angular frequency box, since nothing here depends on the frequency. Set Rounding in Settings to approx to n digits with n = 4.

The source current \(I_0\) leaves the source's positive terminal, which is the opposite of how Symbulator counts a source's current, so we read it with a minus sign in the Evaluate card.

Symbulator returns i_r1 = 2 + 2.667j A (3.333∠53.13°, the book's \(I_1\)), i_r2 = 0.4 + 0.5333j A (0.6667∠53.13°, the book's \(I_2\)), i_r3 = 1.333 + 4.267j A (4.470∠72.65°, the book's \(I_3\)), i_r4 = 0.6667 - 1.6j A (1.733∠-67.38°, the book's \(I_4\)), i_r5 = 1.733 + 4.8j A (5.103∠70.14°, the book's \(I_5\)), v_b = 109.3 + 8j V (109.6∠4.185°, the book's \(V_1\)) and v_c = 96 - 34.67j V (102.1∠-19.86°, the book's \(V_2\)).

Then we type -i_e into Evaluate:

-i_e

It gives 2.4 + 3.2j A (4.000∠53.13°, the book's \(I_0\)).

Solution

This is an AC circuit with a dependent source inside it, and the question wants its Thévenin equivalent, a voltage and an impedance, seen from two terminals. The dependent voltage source is worth ten times \(V_x\), the voltage across the 60 Ω resistor. We name that resistor r2 and place it between node 2 and ground, so that \(V_x\) is the voltage at node 2 and the source's value is 10*v2. The figure names the terminals a and b. Terminal b is the bottom rail, which we take as ground, so we keep a as the top terminal's node and name a and 0 to the Find equivalent card with Thévenin / Norton chosen, which returns the equivalent's voltage and impedance. A dependent source in the circuit is no obstacle to it.

e1,1,0,120
r1,1,2,12
r2,2,0,60
r3,2,a,-40j
e2,3,0,10*v2
r4,3,a,120

Open Find equivalent, choose Thévenin / Norton, and give the two terminals a and 0. Set Analysis to AC — alternating current. Leave omega in the ω — angular frequency box, since nothing here depends on the frequency. Set Rounding in Settings to approx to n digits with n = 4.

Symbulator returns vth = 784 - 288j V (835.2∠-20.17°) and zeq = 91.2 - 38.4j Ω (98.95∠-22.83°, the book's \(Z_{Th}\)).

Solution

This is an AC circuit of two loops with a dependent voltage source in the far one, worth 39 times \(I_x\), the current down through the middle branch. The question wants three voltages, each marked across a pair of impedances in series: \(V_1\) across the 1 Ω and j₂ Ω on the left, \(V_2\) across the 12 Ω and −j₁₆ Ω in the middle, \(V_3\) across the 1 Ω and j₃ Ω on the right. Nothing else connects inside a pair, so we write each pair as one impedance, r1 worth 1+2j, r2 worth 12-16j and r3 worth 1+3j, and each voltage asked is then that element's own drop. \(I_x\) is the current through r2, so the dependent source's value is 39*ir2. We name the source's top node a, the top of the middle branch b and the dependent source's top c.

e1,a,0,150
r1,a,b,1+2j
r2,b,0,12-16j
r3,b,c,1+3j
e2,c,0,39*ir2

Set Analysis to AC — alternating current. Leave omega in the ω — angular frequency box, since nothing here depends on the frequency. Set Rounding in Settings to approx to n digits with n = 4.

Symbulator returns v_r1 = 78 - 104j V (130.0∠-53.13°, the book's \(V_1\)), v_r2 = 72 + 104j V (126.5∠55.30°, the book's \(V_2\)) and v_r3 = 150 - 130j V (198.5∠-40.91°, the book's \(V_3\)).

Solution

The circuit consists of a linear transformer, two coupled coils each with its own winding resistance, between a source with an internal impedance and a load. The question wants the Thévenin equivalent seen from the load's terminals. Of the book's seven parts, (a) to (f) are the steps of its own method -- the frequency-domain equivalent circuit, the two self-impedances, the reflected impedance, its scaling factor and the impedance looking into the primary -- and are skipped. Only (g) asks about the circuit itself. The coils are given as inductances, so we write them as they are, l1,p,0,9 and l2,q,0,4, and put the frequency, 400 rad/s, in the ω — angular frequency box. The coupling the book gives as \(k\), and the m line takes it as it is, m,l1,l2,k=0.5, naming the two coils. The mutual inductance is worked out inside. The winding resistances \(R_1\) and \(R_2\) we name r3 and r6, and the source's internal impedance, given in ohms, r1 and r2. We leave the load out, because the question asks for the equivalent seen from its terminals: we call the top one node c and take the bottom one as ground -- nothing conducts between the two windings, so joining their bottoms changes no current -- and name c and 0 to the Find equivalent card with Thévenin / Norton chosen.

e,1,0,300
r1,1,2,500
r2,2,a,100j
r3,a,p,200
l1,p,0,9
m,l1,l2,k=0.5
l2,q,0,4
r6,q,c,100

Open Find equivalent, choose Thévenin / Norton, and give the two terminals c and 0. Set Analysis to AC — alternating current. Put 400 in the ω — angular frequency box. Tick RMS phasors in Settings, since the book's source is given in rms. Set Rounding in Settings to approx to n digits with n = 4.

Symbulator returns vth = 93.94 + 17.77j V (95.60∠10.71°) and zeq = 171.1 + 1224j Ω (1236∠82.04°, the book's \(Z_{Th}\)).

Solution

This entry takes the circuit of Example 9.14, which we describe the same way, and asks a question about power: how much average and reactive power each of its three impedances takes, how much each source supplies, and whether what is delivered balances what is absorbed. One run answers all three parts.

e1,a,0,150
r1,a,b,1+2j
r2,b,0,12-16j
r3,b,c,1+3j
e2,c,0,39*ir2

Set Analysis to AC — alternating current. Leave omega in the ω — angular frequency box, since nothing here depends on the frequency. Set Rounding in Settings to approx to n digits with n = 4.

Every card of an AC run carries two power rows, average (real) power, the answer p, and reactive power, the answer q. On an impedance they read consumed, and each impedance in the figure is one element of our description, so part (a) is read straight off the three cards. On a source they read delivered, -p and -q, the negatives of the p and q answers, so part (b) is read off the two source cards the same way. For part (c), if every watt delivered is absorbed somewhere, the five p answers add to zero, and likewise the five q, which makes two sums in the Evaluate card.

a) The three impedances' cards read p_r1 = 1690 W (the book's \(P_1\)), q_r1 = 3380 var (the book's \(Q_1\)), p_r2 = 240 W (the book's \(P_2\)), q_r2 = -320 var (the book's \(Q_2\)), p_r3 = 1970 W (the book's \(P_3\)) and q_r3 = 5910 var (the book's \(Q_3\)).

b) The two sources' cards read -p_e1 = -1950 W, -q_e1 = 3900 var, -p_e2 = 5850 W and -q_e2 = 5070 var.

c) We type the sum of the five p answers into Evaluate:

p_e1+p_e2+p_r1+p_r2+p_r3

It gives 0 W.

Then we do the same with the five q:

q_e1+q_e2+q_r1+q_r2+q_r3

It gives 0 var.

A negative reading is the same power the other way: the \(12 - j16\) Ω impedance's q_r2 = −320 var means it delivers 320 var, and the independent source's -p_e1 = −1950 W means it absorbs 1950 W. Both sums being zero, the average power delivered equals the average power absorbed, and the reactive power likewise.

Solution

The circuit consists of a source feeding a load through an ideal transformer whose windings share a node. The question wants the load that draws the most average power and how much that is. An ideal transformer is the element t. Because its primary and secondary here share a node, we write each winding as a bracketed pair of terminals, top node then bottom, [p,x] for the primary and [a,x] for the secondary, followed by the turns. The dots sit at opposite ends of the two windings, the primary's at the top and the secondary's at the bottom, and that polarity is a minus sign on one of the turn counts, so we write the ratio as [-4,1]. We name the two resistors after their values, r60 and r20. The same transformer can also be written with the secondary's pair the other way round and the sign dropped, t,[p,x],[x,a],[4,1], since reversing a pair reverses its polarity too, and the answers are the same. We leave the load \(R_L\) out of the description and name its terminals, node a and ground, to the Find equivalent card with Thévenin / Norton chosen, which reports the load that would draw the most average power from those terminals and how much: by the maximum power theorem that load is the Thévenin impedance, reported as zeq, and the power is pmax.

e,1,0,840
r60,1,p,60
t,[p,x],[a,x],[-4,1]
r20,x,0,20

Open Find equivalent, choose Thévenin / Norton, and give the two terminals a and 0. Set Analysis to AC — alternating current. Leave omega in the ω — angular frequency box, since nothing here depends on the frequency. Tick RMS phasors in Settings, since the book's source is given in rms. Set Rounding in Settings to approx to n digits with n = 4.

Symbulator returns a) zeq = 35 Ω (the book's \(R_L\)) and b) pmax = 315 W.

Solution

The circuit consists of a three-phase generator, a three-phase line and a three-phase load, all Y-connected and all balanced. The question wants the three line currents and the phase and line voltages at the load and at the generator, first for a positive phase sequence and then for a negative one. We describe the whole circuit, one phase at a time. There are three sources at the generator's internal nodes, which we call ga, gb and gc, each written as a phasor, its magnitude and its angle in degrees: (120∠0°), (120∠-120°) and (120∠120°). Then come the generator's, the line's and the load's impedance in each phase, all in ohms, which we name rga, rla and rfa for the a phase and likewise for b and c. We take the generator's neutral as ground and call the load's neutral nn. The generator's terminals we call a, b and c and the load's pa, pb and pc.

ea,ga,0,(120∠0°)
eb,gb,0,(120∠-120°)
ec,gc,0,(120∠120°)
rga,ga,a,0.2+0.5j
rgb,gb,b,0.2+0.5j
rgc,gc,c,0.2+0.5j
rla,a,pa,0.8+1.5j
rlb,b,pb,0.8+1.5j
rlc,c,pc,0.8+1.5j
rfa,pa,nn,39+28j
rfb,pb,nn,39+28j
rfc,pc,nn,39+28j

Set Analysis to AC — alternating current. Leave omega in the ω — angular frequency box, since nothing here depends on the frequency. Set Rounding in Settings to approx (full precision).

Every quantity the question asks for is the current through one element, the voltage at one node or the difference between two node voltages, so each part is read off the run or typed into the Evaluate card. The book prints each answer as a phasor, and the same value in that form follows each answer in parentheses.

b) The three line currents are the currents through the line impedances rla, rlb and rlc, read off their cards: i_rla = 1.92 - 1.44j A (2.4000∠-36.870°, the book's \(I_{aA}\)), i_rlb = -2.2071 - 0.94277j A (2.4000∠-156.87°, the book's \(I_{bB}\)) and i_rlc = 0.28708 + 2.3828j A (2.4000∠83.130°, the book's \(I_{cC}\)).

c) The phase voltages at the load are the voltages of its three nodes above its neutral nn. We type v_pa-v_nn into Evaluate:

v_pa-v_nn

It gives 115.2 - 2.4j V (115.22∠-1.1935°, the book's \(V_{AN}\)).

Likewise v_pb-v_nn:

v_pb-v_nn

It gives -59.678 - 98.566j V (115.22∠-121.19°, the book's \(V_{BN}\)).

And v_pc-v_nn:

v_pc-v_nn

It gives -55.522 + 100.97j V (115.22∠118.81°, the book's \(V_{CN}\)).

d) The line voltages at the load are the differences between pairs of its nodes. We type v_pa-v_pb:

v_pa-v_pb

It gives 174.88 + 96.166j V (199.58∠28.807°, the book's \(V_{AB}\)).

Likewise v_pb-v_pc:

v_pb-v_pc

It gives -4.1569 - 199.53j V (199.58∠-91.193°, the book's \(V_{BC}\)).

And v_pc-v_pa:

v_pc-v_pa

It gives -170.72 + 103.37j V (199.58∠148.81°, the book's \(V_{CA}\)).

e) The phase voltages at the generator's terminals are the voltages at nodes a, b and c, read off the run: v_a = 118.9 - 0.672j V (118.90∠-0.32383°, the book's \(V_{an}\)), v_b = -60.03 - 102.63j V (118.90∠-120.32°, the book's \(V_{bn}\)) and v_c = -58.866 + 103.3j V (118.90∠119.68°, the book's \(V_{cn}\)).

f) The line voltages at the generator's terminals are the differences between those three. We type v_a-v_b:

v_a-v_b

It gives 178.93 + 101.96j V (205.94∠29.676°, the book's \(V_{ab}\)).

Likewise v_b-v_c:

v_b-v_c

It gives -1.1639 - 205.93j V (205.94∠-90.324°, the book's \(V_{bc}\)).

And v_c-v_a:

v_c-v_a

It gives -177.76 + 103.97j V (205.94∠149.68°, the book's \(V_{ca}\)).

g) A negative phase sequence is the same description with the angles of eb and ec exchanged, (120∠120°) and (120∠-120°). Running it returns the same fifteen answers with every b-phase value and c-phase value exchanged, and the line voltages now lag the phase voltages by 30° instead of leading them.

Solution

This is the circuit of Example 7.3, a capacitor charged from one source and switched over to a set of resistors, asked again with the answer wanted as a Laplace transform. Its first interval is the same: before the switch moves the capacitor has sat across the 100 V source through the 10 kΩ for a long time, so the circuit is steady. We describe that circuit as before and run it in DC for the capacitor's voltage:

e,1,0,100
r0,1,2,10'k
c,2,0,0.5'u

Set Analysis to DC — direct current.

Symbulator returns v_2 = 100 V (the book's \(v_C(0)\)).

The second interval is the same description as Example 7.3's second run, with the 100 V in the capacitor's fifth field. The analysis is the one difference. This chapter of the book works every problem by the Laplace method, which transforms the circuit into the \(s\) domain, solves it there and inverts the result, so we set the analysis to FD. FD does the first two of those, returning each answer as a function of \(s\) with the initial condition already inside it. \(V_o(s)\) is the voltage at node 2.

c,1,0,0.5'u,100
r1,1,2,32'k
r2,2,0,240'k
r3,2,0,60'k

FDOpen in app ↗Open in split view

Set Analysis to FD — complex frequency domain.

v_2 is the book's \(V_o(s)\).

The question asks for \(v_o(t)\), and the run has given its transform. Inverting it is the third step, and ours to take: s2t turns a function of \(s\) back into a function of \(t\), and we give it the answer just read.

s2t(v_2)

It gives:

That is what Example 7.3 reached in a single TR run, which transforms, solves and inverts out of sight. Here the halves are separate and visible. FD does what the book does down to its \(V_o(s)\), and s2t does the inversion the book performs with a table of transforms.

Solution

The circuit is a parallel RLC circuit holding an initial current and an initial voltage, with a constant current source switched on at \(t\) = 0. The question wants the voltage across it, and the book works it by the Laplace method. A constant source switched on at \(t\) = 0 is a step, \(24u(t)\) mA, so we type it as a function of time inside curly brackets, {24'm*u(t)}, and FD converts it to the \(s\) domain. We put the two initial conditions the question gives in the fifth fields of the capacitor and the inductor, 50 V and 29 mA written 0.029, and name the four elements j, c, l and r. We set the analysis to FD, which returns the transform \(V(s)\) of the voltage asked for, the voltage at node 1.

j,0,1,{24'm*u(t)}
c,1,0,25'n,50
l,1,0,25'm,0.029
r,1,0,500

Set Analysis to FD — complex frequency domain.

v_1 is the book's \(V(s)\).

The question asks for \(v(t)\). s2t inverts a function of \(s\) back into a function of \(t\), so we give it the answer just read.

s2t(v_1)

It gives:

The book arrives at the same expression by a partial fraction expansion and a table of transforms, its last two steps. The same description run in TR returns it in one go, the solver transforming and inverting out of sight.

Solution

The circuit is a source switched at \(t\) = 0 onto two resistors and an inductor, and the question wants the Thévenin equivalent of that part of the circuit, seen from two terminals, as functions of \(s\). The 480 V source switched on at \(t\) = 0 is a step, \(480u(t)\) V, so we type it as a function of time inside curly brackets, {480u(t)}, and FD converts it to the \(s\) domain. We call the terminals a and b node a and ground, and name them to the Find equivalent card with Thévenin / Norton chosen. The card works in FD as it does in DC and AC, and returns the equivalent's voltage and impedance as functions of \(s\). We leave out the capacitor to the right of the terminals, since the question asks for the equivalent seen from them.

e,1,0,{480u(t)}
r1,1,2,20
l,2,0,0.002
r2,2,a,60

Open Find equivalent, choose Thévenin / Norton, and give the two terminals a and 0. Set Analysis to FD — complex frequency domain.

zeq is the book's \(Z_{Th}\).

Solution

The circuit consists of a source driving a resistor, a capacitor and an inductive branch, with the capacitor's voltage as the output. The question wants the transfer function from source to output, and its poles and zeros. A transfer function is the ratio of an output to the input that produced it, as functions of \(s\), so we leave the source as the symbol vg, so that the output comes back as a multiple of it. We number the nodes from the source, so the output, the capacitor's voltage, is the voltage at node 2, and the transfer function is that voltage divided by vg, which we will read in the Evaluate card.

e,1,0,vg
r1,1,2,1000
r2,2,3,250
l,3,0,50'm
c,2,0,1'u

Set Analysis to FD — complex frequency domain.

a) is the transfer function above.

b) The poles of a transfer function are the values of \(s\) that make its denominator zero, and the zeros are the values that make its numerator zero. The Mini-Tools card finds both at once: we choose pz — poles and zeros in its Tool menu and give it the transfer function.

v2/vg

Press Run.

The card returns the poles -3000 - 4000j and -3000 + 4000j, and the zero -5000.

So the transfer function has poles at \(s = -3000 \pm j4000\) and a zero at \(s = -5000\). The card cancels any factor the numerator and the denominator share before it reads the roots, so a pole and a zero at the same place are not reported.

Solution

The circuit consists of a series resistor feeding a parallel inductor and capacitor, with the voltage across the pair as the output. The question wants its transfer function shown to be a bandpass, its centre frequency, bandwidth and \(Q\) in general, and then the resistor and inductor for a given centre frequency and bandwidth. Nothing in the question is a number until part (d), so we leave the three elements as the symbols R, L and C and the source as vi, naming the resistor rr to keep its name apart from its value. We set the analysis to FD. The transfer function is the output voltage, at node 2, divided by the input, which we will read in the Evaluate card, and it comes back in terms of the three symbols.

e,1,0,vi
rr,1,2,R
c,2,0,C
l,2,0,L

Set Analysis to FD — complex frequency domain. Set Rounding in Settings to approx to n digits with n = 4.

a) Dividing numerator and denominator by \(RLC\) puts the transfer function in the standard form \(H(s) = \dfrac{s/RC}{s^2 + s/RC + 1/LC}\): a first-order numerator in \(s\) over a second-order denominator, which is the shape of a bandpass filter.

b) The centre frequency is the \(\omega_0\) of that denominator, \(\omega_0 = 1/\sqrt{LC}\).

c) The bandwidth is the coefficient of \(s\) in the denominator, \(\beta = 1/RC\). The cutoff frequencies are \(\omega_{c} = \mp\beta/2 + \sqrt{(\beta/2)^2 + \omega_0^2}\), and \(Q = \omega_0/\beta\).

d) The question gives the centre frequency, the bandwidth and the capacitor and asks for \(R\) and \(L\). Those are two equations in two unknowns, which the Solve card takes as they stand: \(\omega_0 = 1/\sqrt{LC}\) at 2π × 5000 rad/s and \(\beta = 1/RC\) at 2π × 200 rad/s, with the capacitor as a condition:

1/sqrt(L*C)=2*pi*5000
1/(R*C)=2*pi*200

L, R

C=5'u

Tick real solutions only and press Solve equations.

The card returns L = 0.0002026 H and R = 159.2 Ω.

That is \(L\) = 202.6 µH and \(R\) = 159.2 Ω.