Select problems from Nilsson & Riedel 12ed
Forty-three worked examples from Electric Circuits, each described in Symbulator and checked against the answer the book prints. DC, AC, TR and FD, with Expert Mode and symbolic answers where they earn their place.
Last updated 2026-09-12
Nilsson and Riedel's Electric Circuits is the book a great many engineers learned this material from, and its worked examples are unusually well suited to showing what Symbulator is for: each one states a circuit, states a question, and then prints the answer. That last part is what makes this chapter checkable rather than merely illustrative — every answer below was compared against the number the book prints, and the page says so example by example.
The selection is deliberate. These are not the easiest problems in the book; they are the ones where the distance between describing a circuit and solving it by hand is widest — a delta that has to be transformed, a supermesh, a dependent source whose controlling current is three steps away, a transformer whose secondary floats, a switch that opens twice. The book meets each with a method. Symbulator meets all of them with the same six words: describe the circuit, choose an analysis.
How to read an entry
Each entry gives the book's question, the book's own figure, the Symbulator description, the analysis to choose, and the answers. Where an answer is a number it is quoted in prose, and the sentence says that it is the book's number too. Where an answer is an expression — a function of t, a transfer function in s, a formula in the circuit's own symbols — it is shown in a results panel exactly as the app prints it.
Four things recur, and they are the reason these particular examples were chosen:
- A dependent source is a value, not a device. Write
8*ir3in a source's value field and the controlling current is named; there is no constraint equation to write and none to get wrong. - The case is never chosen by anyone. Overdamped, critically damped and underdamped are the same three lines with different numbers, and the algebra decides which one comes out.
- A symbol left in the circuit stays in the answer. That is how a design problem gets checked, how a transfer function appears without being asked for, and how one run answers all three parts of a question.
- Expert Mode turns a question inside out. When the thing you know is an answer and the thing you want is a component value, state the answer as an equation and name the component as the unknown.
Every circuit is in the app already
Nothing here has to be typed. All forty-three circuits ship with Symbulator as a built-in example book — open Built-in Examples and pick Nilsson & Riedel 12ed from the list of books; the entries are named for the example each one comes from, and each arrives with its note, its picture, its settings, its Expert Mode fields and the analysis it wants already set.
Pick one, press Run Symbulator, and the answers below are what you get. Working with input files explains what an entry remembers and how to save your own.
Direct current — DC
Sixteen resistive problems, and the running theme is that the book's method — node voltages, mesh currents, source transformations, superposition, a delta-to-wye — is a way of getting an answer by hand, not a property of the answer. Symbulator is told the circuit and never told the method. Two of these run in Expert Mode, and two are two-port problems.
NR12's Example 3.7
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Using Voltage Division and Current Division to Solve a Circuit.
Use current division to find the current \(i_o\) and use voltage division to find the voltage \(v_o\) for the circuit in Fig. 3.22.
Nilsson & Riedel, 12th edition — the circuit for Example 3.7
Solution
A one-line answer to a problem the book solves with two division formulas: the whole circuit is solved at once, and the source's own r_j is the 6 Ω equivalent resistance the book works out by hand.
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
r7,1,0,24Set Analysis to DC — direct current.
Symbulator returns i_r7 = 2 A, v_r6 = 18 V, v_r7 = 48 V and r_j = 6 Ω — the same answers the book prints.
NR12's Example 3.10
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Using a Wheatstone Bridge to Measure Resistance.
For the Wheatstone bridge in Fig. 3.30, R3 can be varied from 10 Ω to 2 kΩ. What range of resistor values can this bridge measure?
Nilsson & Riedel, 12th edition — the circuit for Example 3.10
Solution
Balance is a statement about an answer — no current in the detector — so it is written as an equation and the unknown is a resistor. The answer comes back as the symbolic 4*r3, and the range 40 Ω to 8 kΩ is that one line read twice.
e,1,0,vs
r1,1,a,1'k
r2,1,b,4'k
rg,a,b,500
r3,a,0,r3
rx,b,0,rxi_rg = 0rxSet Analysis to DC — direct current. This one needs Enable Expert Mode ticked in the Expert Mode box; the equations and unknowns go in the fields it reveals.
NR12's Example 3.11
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Applying a Delta-to-Wye Transform.
Find the current and power supplied by the 40 V source in the circuit shown in Fig. 3.35.
Nilsson & Riedel, 12th edition — the circuit for Example 3.11
Solution
The book needs a delta-to-wye transform to reduce this bridge. Symbulator needs nothing: the bridge is six lines, and r_e reports the 80 Ω the transform was for.
e,1,0,40
r1,1,2,5
r2,2,3,100
r3,2,4,125
r4,3,4,25
r5,3,0,40
r6,4,0,37.5Set Analysis to DC — direct current.
Symbulator returns -i_e = 0.5 A, -p_e = 20 W and r_e = 80 Ω — the same answers the book prints.
NR12's Example 4.4
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Using the Node-Voltage Method with Dependent Sources.
Use the node-voltage method to find the power dissipated in the 5 Ω resistor in the circuit shown in Fig. 4.10.
Nilsson & Riedel, 12th edition — the circuit for Example 4.4
Solution
A dependent source is written by naming another answer in its value — 8*ir3 — so no constraint equation has to be written by hand.
e,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*ir3Set Analysis to DC — direct current.
Symbulator returns v_2 = 16 V, v_3 = 10 V, i_r3 = 1.2 A and p_r3 = 7.2 W — the same answers the book prints.
NR12's Example 4.7
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Using the Mesh-Current Method with Dependent Sources.
Use the mesh-current method to find the power dissipated in the 4 Ω resistor in the circuit shown in Fig. 4.23.
Nilsson & Riedel, 12th edition — the circuit for Example 4.7
Solution
The same circuit the book solves with three mesh equations and a constraint. Symbulator is told the circuit, not the method.
e,1,0,50
r1,1,3,1
r2,1,2,5
r3,2,3,4
r4,2,0,20
e2,3,0,15*ir4Set Analysis to DC — direct current.
Symbulator returns i_r4 = 1.6 A, i_r3 = 2 A and p_r3 = 16 W — the same answers the book prints.
NR12's Example 4.8
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A Special Case in the Mesh-Current Method.
Use the mesh-current method to find branch currents \(i_a\), \(i_b\) and \(i_c\) in the circuit for Example 4.3, repeated here as Fig. 4.25.
Nilsson & Riedel, 12th edition — the circuit for Example 4.8
Solution
The book's 'special case' is a current source shared by no other mesh, which needs a rule of its own. Symbulator has no meshes, so there is no special case.
e,1,0,50
r1,1,2,5
r2,2,0,10
r3,2,0,40
j,0,2,3Set Analysis to DC — direct current.
Symbulator returns i_r1 = 2 A, i_r2 = 4 A, i_r3 = 1 A and v_2 = 40 V — the same answers the book prints.
NR12's Example 4.13
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Using Special Source Transformation Techniques.
a) Use source transformations to find the voltage \(v_o\) in the circuit shown in Fig. 4.42. b) Find the power developed by the 250 V voltage source. c) Find the power developed by the 8 A current source.
Nilsson & Riedel, 12th edition — the circuit for Example 4.13
Solution
Four source transformations in the book; one description here. The resistors the book has to put back before it can find the powers were never taken out.
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
r6,3,0,15Set Analysis to DC — direct current.
Symbulator returns v_r4 = 20 V, -i_e = 11.2 A, -p_e = 2800 W and -p_j = 480 W — the same answers the book prints.
NR12's Example 4.21
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Calculating the Condition for Maximum Power Transfer.
a) For the circuit shown in Fig. 4.65, find the value of \(R_L\) that results in maximum power being transferred to \(R_L\). b) Calculate the maximum power that can be delivered to \(R_L\).
Nilsson & Riedel, 12th edition — the circuit for Example 4.21
Solution
The Thévenin tool answers all three parts at once: z is the load for maximum transfer and pmax is the power it takes.
e,1,0,360
r1,1,2,30
r2,2,0,150Open Find equivalent, choose Thévenin / Norton, and give the two terminals 2 and 0.
Symbulator returns vth = 300 V, z = 25 Ω and pmax = 900 W — the same answers the book prints.
NR12's Example 4.23
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Using Superposition to Solve a Circuit with Dependent Sources.
Use the principle of superposition to find \(v_o\) in the circuit shown in Fig. 4.71.
Nilsson & Riedel, 12th edition — the circuit for Example 4.23
Solution
Superposition is a method for getting an answer by hand, not a property of the answer. Two dependent sources and two independent ones, solved once.
e,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*ir1Set Analysis to DC — direct current.
Symbulator returns v_r2 = 24 V, v_r3 = 10 V and i_r1 = -2.8 A — the same answers the book prints.
NR12's Example 5.1
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Analyzing an Op Amp Circuit.
The op amp in the circuit shown in Fig. 5.7 is ideal. a) Calculate \(v_o\) if \(v_a\) = 1 V and \(v_b\) = 0 V. b) Repeat for \(v_a\) = 1 V and \(v_b\) = 2 V. c) If \(v_a\) = 1.5 V, specify the range of \(v_b\) that avoids amplifier saturation.
Nilsson & Riedel, 12th edition — the circuit for Example 5.1
Solution
An ideal op amp does not know its supplies exist — it will report an output of 200 V as readily as 2 V — so saturation is a question you ask of the answer rather than something the solve enforces. Leave both inputs symbolic and one run gives the formula every part is then read off.
ea,1,0,va
r1,1,2,25'k
r2,2,3,100'k
eb,4,0,vb
o,4,2,3Set Analysis to DC — direct current.
a) With va = 1 V and vb = 0 V the formula gives vo = 5(0) − 4(1) = -4 V. That is inside the supplies, so the op amp is in its linear region and -4 V is the answer.
b) With va = 1 V and vb = 2 V, vo = 5(2) − 4(1) = 6 V. Inside the supplies again, so the op amp is still linear.
c) With va = 1.5 V the formula becomes vo = 5vb − 6. The op amp stays linear while that lies between the rails, so Symbulator is asked the question directly — put v_3 = 10 in Expert Mode with vb as the unknown, then again with v_3 = -10. The rails are reached at vb = 3.2 V and vb = -0.8 V, so the range is -0.8 V ≤ vb ≤ 3.2 V.
NR12's Example 5.3
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Designing a Summing Amplifier.
a) Design a summing amplifier whose output voltage is \(v_o = -4v_a — v_b — 5v_c\), using an ideal op amp with ±12 V power supplies and a 20 kΩ feedback resistor. b) Suppose \(v_a\) = 2 V and \(v_c\) = \(-\)1 V. What range of input voltages for \(v_b\) allows the op amp to remain linear?
Nilsson & Riedel, 12th edition — the circuit for Example 5.3
Solution
The design is checked in one run: the answer comes back as the very formula the problem asked the designer to hit.
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,4Set Analysis to DC — direct current.
a) The summing-amplifier formula is vo = -(Rf/Ra)va — (Rf/Rb)vb — (Rf/Rc)vc, so with a 20 kΩ feedback resistor the three input resistors are Ra = 20k/4 = 5 kΩ, Rb = 20k/1 = 20 kΩ and Rc = 20k/5 = 4 kΩ. Running that circuit returns the very formula the design was asked to hit, which is the check.
b) With va = 2 V and vc = -1 V the output collapses to vo = -vb − 3. Asking Expert Mode for the vb that puts v_4 on each rail gives 9 V at -12 V and -15 V at +12 V, so the op amp stays linear for -15 V ≤ vb ≤ 9 V.
NR12's Example 5.3 part c
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Designing a Summing Amplifier — part (c), in Expert Mode.
c) Suppose \(v_a\) = 2 V, \(v_b\) = 3 V and \(v_c\) = \(-\)1 V. Using the input resistor values found in part (a), how large can the feedback resistor be before the op amp saturates?
Nilsson & Riedel, 12th edition — the circuit for Example 5.3
Solution
An unknown that is a component value, not an answer: name the saturation voltage as an equation and ask for the resistor. This is what Expert Mode is for.
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,4v_4 = -12rfSet Analysis to DC — direct current. This one needs Enable Expert Mode ticked in the Expert Mode box; the equations and unknowns go in the fields it reveals.
Symbulator returns rf = 40000 — the same answers the book prints.
c) With va = 2 V, vb = 3 V and vc = -1 V the three input currents sum to a positive number, so the output swings negative and it is the -12 V rail that is reached first. Leave the feedback resistor as the symbol rf, put v_4 = -12 in Expert Mode and name rf the unknown: the answer is 40 kΩ. Any larger and the op amp saturates.
NR12's Example 5.5
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Designing a Difference Amplifier.
a) Design a difference amplifier that amplifies the difference between two input voltages by a gain of 8, using an ideal op amp and ±8 V power supplies. b) Suppose \(v_a\) = 1 V. What range of \(v_b\) keeps the op amp linear?
Nilsson & Riedel, 12th edition — the circuit for Example 5.5
Solution
The gain-of-8 design returns exactly 8(vb — va), and part (b)'s range follows from that one line.
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,3Set Analysis to DC — direct current.
a) The simplified difference-amplifier formula is vo = (Rb/Ra)(vb − va), so a gain of 8 wants two resistors in the ratio 8: Ra = Rc = 1.5 kΩ and Rb = Rd = 12 kΩ. The formula also requires Ra/Rb = Rc/Rd, which those four satisfy. The run returns exactly 8(vb − va).
b) With va = 1 V the output is vo = 8vb − 8, which reaches +8 V at vb = 2 V and -8 V at vb = 0 V. So the op amp remains in its linear region for 0 V ≤ vb ≤ 2 V.
NR12's Example 5.7
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Analyzing a Noninverting-Amplifier Circuit Using a Realistic Op Amp Model.
Analyze the noninverting amplifier of Example 5.4 using the realistic op amp model: open-loop gain \(A\) = 50,000, input resistance \(R_i\) = 100 kΩ, output resistance \(R_o\) = 7.5 kΩ. Find \(v_o/v_g\).
Nilsson & Riedel, 12th edition — the circuit for Example 5.7
Solution
No o element at all: the realistic op amp is a dependent source with its own input and output resistances, and the finite-gain answer 5.9988 falls out against the ideal 6.
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.
Symbulator returns v_3/vg = 5.9988 — the same answers the book prints.
NR12's Example 18.1
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Finding the z Parameters of a Two-Port Circuit.
Find the z parameters for the circuit shown in Fig. 18.3.
Nilsson & Riedel, 12th edition — the circuit for Example 18.1
Solution
Four separate open-circuit measurements in the book — two of them with a source moved to the other port. One run here, and z12 = z21 because the network is reciprocal, which is a result rather than an assumption.
r5,1,2,5
r20,1,0,20
r15,2,0,15Open Find equivalent, choose Two-port parameters, kind z, with the ports at 1 and 2.
Symbulator returns 11 = 10 Ω, 12 = 7.5 Ω, 21 = 7.5 Ω and 22 = 9.375 Ω — the same answers the book prints.
NR12's Example 18.6
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Analyzing Cascaded Two-Port Circuits.
Two identical amplifiers are connected in cascade. Each is described by its h parameters: h11 = 1000 Ω, h12 = 0.0015, h21 = 100, h22 = 100 µS. The source has 500 Ω of internal resistance and the load is 10 kΩ. Find the voltage gain V2/Vg.
Nilsson & Riedel, 12th edition — the circuit for Example 18.6
Solution
The book converts h to a, multiplies the two transmission matrices, then reads a gain formula out of Table 18.3. Here the two blocks are two lines of circuit wired end to end, and the gain is a division.
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'kSet Analysis to DC — direct current.
Symbulator returns v_c/vg = 33333.3 — the same answers the book prints.
Transients — TR
Fourteen transient problems. The pattern is always the same and always the one you would follow by hand: run the t < 0 circuit in DC to read the capacitor voltages and inductor currents, put those numbers in the fifth field of the c and l lines, and run the t ≥ 0 circuit in TR. No time constant is ever computed, no solution form is ever selected, and a sequential-switching problem is simply one more run.
NR12's Example 7.1
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Determining the Natural Response of an RL Circuit.
The switch in the circuit shown in Fig. 7.6 has been closed for a long time before it is opened at \(t\) = 0. Find a) \(i_L(t)\) for t ≥ 0, b) \(i_o(t)\) for t ≥ 0+, c) \(v_o(t)\) for t ≥ 0+.
Nilsson & Riedel, 12th edition — the circuit for Example 7.1
Solution
A switched circuit is two runs: a DC pass for the initial current, then TR with that number in the inductor's fifth field. No time constant is ever computed.
l,1,0,2,20
r1,1,2,2
r2,2,0,10
r3,2,0,40Set Analysis to TR — transient / time domain.
NR12's Example 7.3
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Determining the Natural Response of an RC Circuit.
The switch has been in position x for a long time. At t = 0 it moves instantaneously to position y. Find a) vC(t), b) vo(t) and c) io(t).
Nilsson & Riedel, 12th edition — the circuit for Example 7.3
Solution
The capacitor's starting voltage goes in its fifth field and the answers come back as functions of t. No time constant is computed anywhere.
c,1,0,0.5'u,100
r1,1,2,32'k
r2,2,0,240'k
r3,2,0,60'kSet Analysis to TR — transient / time domain.
NR12's Example 7.5
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Determining the Step Response of an RL Circuit.
The switch has been in position a for a long time. At t = 0 it moves from a to b. a) Find i(t) for t ≥ 0. b) What is the initial voltage across the inductor just after the switch has been moved?
Nilsson & Riedel, 12th edition — the circuit for Example 7.5
Solution
A step response with a non-zero, negative starting current — the inductor was carrying the 8 A source the other way. Part (b) is the answer at t = 0.
e,1,0,24
r1,1,2,2
l,2,0,0.2,-8Set Analysis to TR — transient / time domain.
NR12's Example 7.10
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Determining the Step Response of a Circuit with Magnetically Coupled Coils.
There is no energy stored in the circuit at the time the switch is closed. Find the solutions for io, vo, i1 and i2.
Nilsson & Riedel, 12th edition — the circuit for Example 7.10
Solution
The book replaces the coupled pair with one 1.5 H equivalent and then works back to i1 and i2 through KVL. The m element is one line, and both coil currents come back on their own.
e,1,0,120
r1,1,2,7.5
l1,2,0,3
l2,2,0,15
m,l1,l2,6Set Analysis to TR — transient / time domain.
NR12's Example 7.11 to 35 ms
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Analyzing an RL Circuit That Has Sequential Switching (0 to 35 ms).
Both switches have been closed for a long time. At t = 0 switch 1 is opened; 35 ms later switch 2 is opened. a) Find iL(t) for 0 ≤ t ≤ 35 ms.
Nilsson & Riedel, 12th edition — the circuit for Example 7.11
Solution
Sequential switching is just more runs. A DC pass on the t < 0 circuit gives iL(0) = 6 A, which is the number that goes in the inductor's fifth field here.
r6,2,0,6
r3,2,3,3
l,3,0,0.15,6
r18,3,0,18Set Analysis to TR — transient / time domain.
NR12's Example 7.11 after 35 ms
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Analyzing an RL Circuit That Has Sequential Switching (after 35 ms).
b) Find iL for t ≥ 35 ms. (Time is measured from the second switching.)
Nilsson & Riedel, 12th edition — the circuit for Example 7.11
Solution
The third run: switch 2 has dropped the 18 Ω, so the inductor now sees 9 Ω and the starting current is what the second run left at 35 ms.
r6,2,0,6
r3,2,3,3
l,3,0,0.15,1.47961Set Analysis to TR — transient / time domain.
NR12's Example 7.13
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Finding the Unbounded Response in an RC Circuit.
a) When the switch is closed at t = 0, find vo(t). The Thévenin resistance seen by the capacitor is negative, so the response grows without bound.
Nilsson & Riedel, 12th edition — the circuit for Example 7.13
Solution
A dependent source makes the Thévenin resistance -5 kΩ, and the exponent comes back positive. Nothing had to be told that this case was different.
c,1,0,5'u,10
r1,1,0,10'k
r2,1,0,20'k
j,0,1,7*ir2Set Analysis to TR — transient / time domain.
NR12's Example 8.2
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Finding the Overdamped Natural Response of a Parallel RLC Circuit.
For the circuit in Fig. 8.6, v(0+) = 12 V and iL(0+) = 30 mA. Find the expression for v(t). (Example 8.3 then asks for the three branch currents.)
Nilsson & Riedel, 12th edition — the circuit for Example 8.2
Solution
Overdamped, and nothing had to say so: the book compares alpha with omega-nought and picks a solution form. Example 8.3's branch currents are in the same run.
c,1,0,0.2'u,12
l,1,0,50'm,0.03
r,1,0,200Set Analysis to TR — transient / time domain.
NR12's Example 8.4
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Finding the Underdamped Natural Response of a Parallel RLC Circuit.
In the circuit shown, V0 = 0 and I0 = -12.25 mA. Calculate the voltage response for t ≥ 0.
Nilsson & Riedel, 12th edition — the circuit for Example 8.4
Solution
The same three lines give the underdamped case, damped sine and all. The book needs a different table row; the description does not change.
c,1,0,125'n,0
l,1,0,8,-0.01225
r,1,0,20'kSet Analysis to TR — transient / time domain.
NR12's Example 8.11
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Finding the Natural Response of a Series RLC Circuit.
The 0.1 µF capacitor is charged to 100 V. At t = 0 it is discharged through a series combination of a 100 mH inductor and a 560 Ω resistor. a) Find i(t). b) Find vC(t).
Nilsson & Riedel, 12th edition — the circuit for Example 8.11
Solution
Series rather than parallel, and again the case is not chosen by anyone. The inductor's node order is written to match the book's mesh arrow.
c,1,0,0.1'u,100
l,2,1,0.1
r,2,0,560Set Analysis to TR — transient / time domain.
NR12's Example 8.12
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Finding the Step Response of a Series RLC Circuit.
No energy is stored in the 100 mH inductor or the 0.4 µF capacitor when the switch is closed. Find vC(t) for t ≥ 0.
Nilsson & Riedel, 12th edition — the circuit for Example 8.12
Solution
A step response on a series RLC: the final value, the two roots and both coefficients arrive together in one expression.
e,1,0,48
l,1,2,0.1
r,2,3,1250
c,3,0,0.4'uSet Analysis to TR — transient / time domain.
NR12's Example 13.5
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Analyzing a Circuit with Multiple Meshes.
The circuit has no initial stored energy. At t = 0 the switch closes. Use Laplace methods to find i1(t) and i2(t) for t ≥ 0.
Nilsson & Riedel, 12th edition — the circuit for Example 13.5
Solution
Two coupled mesh equations, a partial-fraction expansion and two inverse transforms in the book. Choosing TR does all of it and prints i1 and i2.
e,1,0,336
l1,1,2,8.4
r1,2,0,42
l2,2,3,10
r2,3,0,48Set Analysis to TR — transient / time domain.
NR12's Example 13.7
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Analyzing a Circuit with Mutual Inductance.
The make-before-break switch has been in position a for a long time. At t = 0 it moves instantaneously to position b. Use Laplace methods to find i2(t) for t ≥ 0.
Nilsson & Riedel, 12th edition — the circuit for Example 13.7
Solution
The book replaces the coupled coils with a T-equivalent and adds two voltage sources for the initial currents. Here the coupling is one m line and the initial currents are fifth fields. The secondary is an island, and Symbulator says so in a note rather than refusing the circuit.
r3,0,p,3
l1,p,0,2,5
m,l1,l2,2
l2,q,d,8,0
r2b,q,c,2
r10,c,d,10Set Analysis to TR — transient / time domain.
NR12's Example 13.13
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A Series Inductor Circuit with an Impulsive Response.
The switch has been closed for a long time and opens at t = 0. Find vo(t).
Nilsson & Riedel, 12th edition — the circuit for Example 13.13
Solution
Opening the switch forces two inductors carrying different currents into series, so the voltage has to contain an impulse. The answer says DiracDelta(t) — the current jumps from 10 A to 6 A, and the algebra is what noticed.
e,1,0,100
r1,1,2,10
l1,2,3,3,10
r2,3,4,15
l2,4,0,2,0Set Analysis to TR — transient / time domain.
Sinusoidal steady state — AC
Eight problems in the sinusoidal steady state. Impedances given in ohms go in as they are written, complex ones included, and then the frequency never enters — which is why several of these leave omega as a symbol. Where the book gives henries and farads instead, the frequency goes in the ω — angular frequency box and the conversion is the solver's.
NR12's Example 9.9
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Combining Impedances in Series and in Parallel.
The sinusoidal current source produces is = 8 cos 200,000t A. b) Find the equivalent admittance to the right of the source. c) Find the phasor voltage V. d) Find the phasor current I. e) Find the steady-state expressions for v and i.
Nilsson & Riedel, 12th edition — the circuit for Example 9.9
Solution
Henries and farads go in as they are given: the solver turns them into j8 and -j5 at the stated frequency. V comes back 40 at -36.87 degrees and I 4 at -90.
j,0,1,8
r1,1,0,10
r2,1,2,6
l,2,0,40'u
c,1,0,1'uSet Analysis to AC — alternating current. Put 200000 in the ω — angular frequency box.
Symbulator returns v_1 = 32 - 24j V (40.00∠-36.87°) and i_r2 = -4j A (4.000∠-90.00°) — the same answers the book prints.
NR12's Example 9.10
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Using a Delta-to-Wye Transform in the Frequency Domain.
Use a delta-to-wye impedance transformation to find I0, I1, I2, I3, I4, I5, V1 and V2 in the circuit in Fig. 9.23.
Nilsson & Riedel, 12th edition — the circuit for Example 9.10
Solution
Impedances go in as Ω, complex ones included, so the frequency never has to be known. Eight answers the book gets by transforming and working back.
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,-20jSet Analysis to AC — alternating current. Every impedance is given in ohms, so the frequency never enters: leave omega in the ω — angular frequency box.
Symbulator returns -i_e = 2.4 + 3.2j A (4.000∠53.13°), i_r1 = 2 + 2.66667j A (3.333∠53.13°), i_r3 = 1.33333 + 4.26667j A (4.470∠72.65°), i_r4 = 0.666667 - 1.6j A (1.733∠-67.38°), i_r5 = 1.73333 + 4.8j A (5.103∠70.14°), v_b = 109.333 + 8j V (109.6∠4.185°) and v_c = 96 - 34.6667j V (102.1∠-19.86°) — the same answers the book prints.
NR12's Example 9.12
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Finding a Thévenin Equivalent in the Frequency Domain.
Find the Thévenin equivalent circuit with respect to terminals a,b for the circuit shown in Fig. 9.32.
Nilsson & Riedel, 12th edition — the circuit for Example 9.12
Solution
A dependent source means the Thévenin resistance cannot be found by inspection; the book needs a test source. The tool returns both numbers.
e,1,0,120
r1,1,2,12
r2,2,0,60
r3,2,9,-40j
e2,3,0,10*v2
r4,3,9,120Open Find equivalent, choose Thévenin / Norton, and give the two terminals 9 and 0. Set Analysis to AC — alternating current. Every impedance is given in ohms, so the frequency never enters: leave omega in the ω — angular frequency box.
Symbulator returns vth = 784 - 288j V (835.2∠-20.17°) and z = 91.2 - 38.4j Ω (98.95∠-22.83°) — the same answers the book prints.
NR12's Example 9.14
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Using the Mesh-Current Method in the Frequency Domain.
Use the mesh-current method to find the voltages V1, V2 and V3 in the circuit shown in Fig. 9.39.
Nilsson & Riedel, 12th edition — the circuit for Example 9.14
Solution
Two mesh equations and a constraint in the book. Here the controlling current is just the name of the answer it is, written into the dependent source's value.
e,1,0,150
r1,1,2,1
r2,2,a,2j
r3,a,c,12
r4,c,0,-16j
r5,a,4,1
r6,4,b,3j
e2,b,0,39*ir3Set Analysis to AC — alternating current. Every impedance is given in ohms, so the frequency never enters: leave omega in the ω — angular frequency box.
Symbulator returns v_1 - v_a = 78 - 104j V (130.0∠-53.13°), v_a = 72 + 104j V (126.5∠55.30°), v_a - v_b = 150 - 130j V (198.5∠-40.91°) and i_r3 = -2 + 6j A (6.325∠108.4°) — the same answers the book prints.
NR12's Example 9.15
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Analyzing a Linear Transformer in the Frequency Domain.
A linear transformer has R1 = 200 Ω, R2 = 100 Ω, L1 = 9 H, L2 = 4 H, k = 0.5, and couples an 800 Ω + 1 µF load to a 300 V (rms) source of internal impedance 500 + j100 at 400 rad/s. g) Calculate the Thévenin equivalent with respect to the terminals of the load.
Nilsson & Riedel, 12th edition — the circuit for Example 9.15
Solution
The primary is grounded and the secondary is not: its foot is node d, a name like any other. Nothing conducts from one winding to the other, so the secondary's absolute potentials are undefined — its currents and its voltage differences are not — and Symbulator says so in a note, measuring that side against d. Self-impedance, reflected impedance and the scaling factor are three of the book's seven parts; name the load's two terminals and the tool answers the last outright.
e,1,0,300
r1,1,2,500
r2,2,a,100j
r3,a,p,200
r4,p,0,3600j
m,r4,r5,1200j
r5,q,d,1600j
r6,q,c,100Open Find equivalent, choose Thévenin / Norton, and give the two terminals c and d. Set Analysis to AC — alternating current. Every impedance is given in ohms, so the frequency never enters: leave omega in the ω — angular frequency box. Tick RMS phasors in Settings, since the book's source is given in rms.
Symbulator returns vth = 93.9351 + 17.7715j V (95.60∠10.71°) and z = 171.086 + 1224.26j Ω (1236∠82.04°) — the same answers the book prints.
NR12's Example 10.8
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Balancing Power Delivered with Power Absorbed in an AC Circuit.
a) Calculate the total average and reactive power delivered to each impedance in the circuit shown in Fig. 10.18. b) Calculate the average and reactive powers associated with each source. c) Verify that the average power delivered equals the average power absorbed, and likewise for the reactive power.
Nilsson & Riedel, 12th edition — the circuit for Example 10.8
Solution
Every element reports its own complex power, so part (a) and part (b) are one run. Part (c) — the balance the book checks by hand — is the sum of the s answers, and it is exactly zero.
e,1,0,150
r1,1,2,1
r2,2,a,2j
r3,a,c,12
r4,c,0,-16j
r5,a,4,1
r6,4,b,3j
e2,b,0,39*ir3Set Analysis to AC — alternating current. Every impedance is given in ohms, so the frequency never enters: leave omega in the ω — angular frequency box.
Symbulator returns s_r1 + s_r2 = 1690 + 3380j VA (3779∠63.43°), s_r3 + s_r4 = 240 - 320j VA (400.0∠-53.13°), s_r5 + s_r6 = 1970 + 5910j VA (6230∠71.57°), s_e = 1950 - 3900j VA (4360∠-63.43°), s_e2 = -5850 - 5070j VA (7741∠-139.1°) and the sum of all eight = 0 VA — the same answers the book prints.
NR12's Example 10.12
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Finding Maximum Power Transfer in a Circuit with an Ideal Transformer.
The variable resistor is adjusted until maximum average power is delivered to RL. a) What is the value of RL in Ω? b) What is the maximum average power delivered to RL?
Nilsson & Riedel, 12th edition — the circuit for Example 10.12
Solution
An ideal transformer whose windings share a node, so both ports are written as bracketed terminal pairs. The book works the constraint equations twice, once open-circuit and once short-circuit; the tool returns -210 V, 35 Ω and 315 W.
e,1,0,840
r60,1,p,60
t,[p,x],[x,a],[4,1]
r20,x,0,20Open Find equivalent, choose Thévenin / Norton, and give the two terminals a and 0. Set Analysis to AC — alternating current. Every impedance is given in ohms, so the frequency never enters: leave omega in the ω — angular frequency box. Tick RMS phasors in Settings, since the book's source is given in rms.
Symbulator returns vth = -210 V, z = 35 Ω and pmax = 315 W — the same answers the book prints.
NR12's Example 11.1
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Analyzing a Wye-Wye Circuit.
A balanced, positive-sequence Y-connected generator with internal impedance 0.2 + j0.5 Ω per phase and internal voltage 120 V per phase feeds a balanced Y-connected load of 39 + j28 Ω per phase over a line of 0.8 + j1.5 Ω per phase. b) Calculate the three line currents. c) Calculate the phase voltages at the load. d) Calculate the line voltages at the load. e) Calculate the phase voltages at the generator terminals.
Nilsson & Riedel, 12th edition — the circuit for Example 11.1
Solution
There is no three-phase mode and none is needed. The whole circuit goes in — three sources carrying their phase in the value — rather than the single-phase equivalent the book has to construct first. The neutral comes back at exactly zero, which is the balance, measured rather than assumed.
ea,ga,0,120
eb,gb,0,120*exp(-2j*pi/3)
ec,gc,0,120*exp(2j*pi/3)
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+28jSet Analysis to AC — alternating current. Every impedance is given in ohms, so the frequency never enters: leave omega in the ω — angular frequency box.
Symbulator returns i_rla = 1.92 - 1.44j A (2.400∠-36.87°), |v_pa - v_nn| = 115.225 V, |v_pa - v_pb| = 199.576 V, |v_a| = 118.898 V and v_nn = 0 V — the same answers the book prints.
The s domain — FD
Five problems in the s domain. FD returns every answer as a function of s, initial conditions included, which makes a transfer function nothing more than the answer with the source left as a symbol. Nothing on this page is labelled filter or transfer function, because nothing needs to be.
NR12's Example 13.2
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The Natural Response of an RC Circuit.
The switch has been in position x for a long time; at t = 0 it moves to y. Use the Laplace transform method to find vo(t). (The same circuit as Example 7.3, worked in the s domain.)
Nilsson & Riedel, 12th edition — the circuit for Example 13.2
Solution
The same description as Example 7.3 with FD chosen instead of TR: the answers come back as transforms rather than as functions of t. One circuit, two domains, no re-typing.
c,1,0,0.5'u,100
r1,1,2,32'k
r2,2,0,240'k
r3,2,0,60'kSet Analysis to FD — complex frequency domain.
NR12's Example 13.3
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The Step Response of an RLC Circuit.
The initial current in the inductor is 29 mA and the initial voltage across the capacitor is 50 V. Use the Laplace transform method to find v(t) for t ≥ 0.
Nilsson & Riedel, 12th edition — the circuit for Example 13.3
Solution
The book combines three parallel impedances and adds three current sources, two of them standing for the initial conditions. Here the initial conditions are the fifth field of c and l, and V(s) is the answer.
j,0,1,0.024/s
c,1,0,25'n,50
l,1,0,25'm,0.029
r,1,0,500Set Analysis to FD — complex frequency domain.
NR12's Example 13.6
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Creating a Thévenin Equivalent in the s Domain.
Find the Thévenin equivalent with respect to terminals a,b for the circuit shown in Fig. 13.20.
Nilsson & Riedel, 12th edition — the circuit for Example 13.6
Solution
The Thévenin tool works in the s domain too, so the equivalent comes back as a pair of rational functions rather than a pair of numbers.
e,1,0,480/s
r1,1,2,20
l,2,0,0.002
r2,2,a,60Open Find equivalent, choose Thévenin / Norton, and give the two terminals a and 0. Set Analysis to FD — complex frequency domain.
NR12's Example 13.9
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Deriving the Transfer Function of a Circuit.
Derive the transfer function H(s) = Vo/Vg for the circuit in Fig. 13.31.
Nilsson & Riedel, 12th edition — the circuit for Example 13.9
Solution
Nothing is labelled 'transfer function' because nothing needs to be: leave the source as a symbol, run FD, and divide. The poles and zeros are then the expression's own.
e,1,0,vg
r1,1,2,1000
r2,2,3,250
l,3,0,50'm
c,2,0,1'uSet Analysis to FD — complex frequency domain.
NR12's Example 14.6
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Designing a Parallel RLC Bandpass Filter.
a) Show that the RLC circuit in Fig. 14.22 is a bandpass filter by deriving an expression for the transfer function H(s). b) Compute the centre frequency. c) Calculate the cutoff frequencies, the bandwidth and Q. d) Compute R and L for a centre frequency of 5 kHz and a bandwidth of 200 Hz, using a 5 µF capacitor.
Nilsson & Riedel, 12th edition — the circuit for Example 14.6
Solution
Nothing in Symbulator is filter-shaped. Leave R, L and C as symbols, run FD, and the standard bandpass form appears — from which the centre frequency, the bandwidth and part (d)'s R = 159.2 Ω and L = 202.6 µH are ordinary algebra.
e,1,0,vi
rr,1,2,R
c,2,0,C
l,2,0,LSet Analysis to FD — complex frequency domain.