Current sources, conductances and dependent sources

B11's Example 8.2

Determine the values of V_S, I₁ and I₂.

In the line below we join two commands with a colon. The first runs a DC simulation of the circuit described in the string; the second asks the calculator for three variables that answer the questions.

s\dc("j,0,1,7:e,1,0,12:r1,1,0,4"):
{v1,ie,ir1}

The calculator returns {12,4,3}, meaning V_S is 12V, I₁ is 4A and I₂ is 3A. These are the correct answers. We will keep using the single-line instruction.

B11's Example 8.15

Determine the current through each resistor. My solution, with the direction marked in blue:

s\dc("j6,0,1,6:r2,1,0,2:r6,1,2,6:r8,0,2,8:j8,2,0,8"):
approx({ir2,ir6,ir8})

We get these answers: {1.25,4.75,3.25}. So I_R2 is 1.25A, I_R6 is 4.75A, and I_R8 is 3.25A.

B11's Example 8.21

Determine the voltage in each node and the current through each resistor.

My solution:

s\dc("j1,0,1,4:r1,1,0,2:r3,1,2,12:r2,0,2,6:j2,2,0,2"):
{v1,v2,ir1,ir2,ir3}

We get the following answers: {6,-6,3,1,1}. So V₁=6V, V₂=-6V, I_R1=3A, and I_R2=I_R3=1A.

Bo2's Drill Exercise 2.2 (Conductances)

Given the circuit, determine the voltages in the nodes.

Below my solution:

s\dc("r10,1,0,1/3:r12,1,2,1/2:r13,1,3,1/2:r23,2,3,1/6:
r20,2,0,1/8:j12,1,2,17:j03,0,3,2"):approx({v1,v2,v3})

The answer, {-2.,1.,.5}, indicates v₁=-2V, v₂=1V, v₃=0.5V. This is correct.

HK5's Figure 1-24b (Expert)

Determine i_x and v_x in the following circuit.

My solution:

s\ex("j6,0,1,6:r5,1,2,5:r2,2,0,2:r1,1,3,1:
r3,0,4,3:j10,2,3,10:rx,3,4,rx"):{ir1,vrx}

Select DC. Add equation ir2=4. Add unknown rx. Run the simulation. The answer, {-8,80}, means that I_X is -8A and that V_X is 80V.

HK5's Example 2.2 (Conductances)

Determine the voltages in the nodes. My solution:

s\dc("j01,0,1,–8:j30,3,0,–25:j21,2,1,–3:r12,1,2,1/3:r23,2,3,1/2:r13,1,3,1/4:r20,2,0,1:r30,3,0,1/5"):{v1,v2,v3}

The answer, {1,2,3}, is correct: v₁=1V, v₂=2V, v₃=3V.

B11's Example 8.5

Determine the current I₂ in the circuit shown. My solution:

s\dc("j1,1,0,4:r1,1,0,3:e2,1,2,5:r2,0,2,2"):approx(ir2)

This gives us a value for I₂ of 3.4 A. This is correct.

Bo2's Example 2.5 (Conductances)

Determine the voltages in the nodes, and the current through the voltage source.

s\dc("j,0,1,3:e,3,2,3:r12,1,2,1/7:r20,2,0,1/3:r30,3,0,1/5:r13,1,3,1/2"):
approx({v1,v2,v3,–ie})

The answer, {-.5,-1.5,1.5,11.5}, is correct: v₁=-.5, v₂=-1.5, v₃=1.5, i=11.5

B11's Example 8.22

Determine V₁ and V₂. My solution:

s\dc("j1,0,1,6:r1,1,0,4:e,1,2,12:r3,1,2,10:r2,2,0,2:j2,2,0,4"):
approx({v1,v2})

The answer, {10.67,-1.33}, tells us that V₁ is 10.67V and V₂ is -1.33V.

B11's Example 8.19

Determine V₁, I₁ and I₂, in the circuit:

My solution:

s\dc("e,2,0,24:r1,1,2,6:r2,1,0,12:j,0,1,1"):approx({v1,ir1,ir2})

The answer, {20.,-.667,1.67}, tells us that V₁ is 20V, I₁ is -.667A and I₂ is 1.67 A.

B11's Example 8.14

Determine I₂ and I₃.

s\dc("e1,1,0,20:r1,1,2,6:r2,2,a,4:j,a,0,4:r3,a,3,2:e2,0,3,12"):
approx({ir2,ir3})

Answer: {3.33,-.667}. This is correct.

B11's Example 8.20

Determine V₁, V₂, I₁, I₂ and I₃. My solution:

s\dc("e,3,0,64:r1,3,1,8:r2,1,2,4:j,1,2,2:r3,2,0,10"):
approx({v1,v2,ir1,ir2,ir3})

Answer: {37.82,32.73,3.27,1.27,3.27}. You should know how to read these by now, but here it is just in case: V₁ = 37.82V, V₂ = 32.73V, I₁ = 3.27A, I₂ = 1.27A, I₃ = 3.27A.

RM3's Example 9-12 (solve)

If R₃ is to be replaced with R₄ and I₄, determine the value and direction of the source.

So that we can keep the name of node b, we will use the top left node as reference.

First, make sure you understand what this problem is asking you to do. The idea is that, despite the change, we keep the same voltage drop and current flow between nodes a and b. We must first know what they are. So we simulate the original circuit:

s\dc("e,0,b,20:r1,0,a,16:r2,a,b,40:r3,a,b,60"):{va-vb,ir3}

We find that the voltage drop is 12V and the current is 0.2A. These are the currents and voltages that we have to keep once we do the replacement. Simulate the circuit now replacing R₃ with a resistor R₄ of 240Ω and a source j with value I₄. Run this:

s\dc("e,0,b,20:r1,0,a,16:r2,a,b,40:r4,a,b,240:j,a,b,i4")

Notice that both the voltage drop (given by va-vb) and the current (given by ir4+ij) are algebraic functions in terms of i4. Now you can find i4 solving by voltage drop:

solve(va-vb=12.,i4)       …or by current flow…       solve(ir4+ij=0.2,i4)

The result is the same: i4 = .15 A. The required current source is .15A from a to b.

RM3's Example 8-13

Solve for the currents through R₂ and R₃ in the circuit shown. My solution:

s\dc("r1,a,0,10'k:r2,1,0,5'k:r3,b,a,6'k:r4,0,2,16'k:
j,a,b,2'm:e1,1,b,10:e2,b,2,8"):approx({ir2,ir3})

The answer, {.00154,.00111} is correct: I_R2 = 1.54 mA and I_R3 = 1.11 mA.

B11's Example 6.3 (Hidden source)

Determine V_S and I₁.

The problem presents a seemingly ‘hanging' node with 20V. To simulate this, imagine a ‘hidden' 20V voltage source connected to the node.

s\dc("j,0,1,6:r1,1,2,2:r2,1,2,1:e,2,0,20"):
{v1,ir1}

We get the answer: {24,2}, which is correct. V_S is 24V and I₁ is 2A.

HK5's Figure 1-24c (Expert)

Determine i_x and v_x in the following circuit.

s\ex("e,1,0,60:r8,1,2,8:r10,2,0,10:r4,2,3,4:r2,3,0,2:j,0,3,ix"):
{ix,v3}

Select DC. Add equation ir8=5. Add unknown ix. Run the simulation, and you will get: {1,8}. This is correct: I_X is 1A and that V_X is 8V.

B11's Example 6.22 (Hidden source)

Determine I₁.

Although no source is shown, the circuit has a current. I use a ‘hidden' current source.

s\dc("jt,0,1,12'm:r1,1,0,1'k:r2,1,0,10'k:r3,1,0,22'k"):approx(ir1)

We get I1 = 10.48 mA, which is correct.

B11's Example 6.21 (Hidden source, Expert)

Determine I_S, I₁ and I₃.

This ‘hidden source' problem is perfect for Expert. My solution:

s\ex("js,0,1,is:r1,1,0,6:r2,1,0,3:r3,1,0,1"):approx({is,ir1,ir3})

Select DC, and press Enter. Add equation ir2=2'm. Add unknown is. Run the simulation.

The answer, {.009,.001,.006}, is correct, since the currents are as follows: I_S is 9mA, I₁ is 1mA and I₃ is 6mA.

Bo2's Example 1.9

Determine I₁, I₂ and v.

This is my solution.

s\dc("ji,0,1,2:r1,1,0,3:jd,0,1,4v1:r2,1,0,5"):{ir1,v1,ir2}

The answer, {-5/26,-15/26,-3/26}, is correct: I₁=-5/26 A, I₂=-3/26 A and v=-15/26 V.

AS2's Practice Problem 2.7

Find v_o and i_o in the circuit. My solution:

s\dc("ji,0,o,6:ro,o,0,2:jd,o,0,iro/4:r8,o,0,8"):{vo,iro}

The answer, {8,4}, is correct: v_o = 8 V and i_o = 4 A.

HK5's Example 1-3

Determine the power delivered by each source and consumed by both resistors.

s\dc("ei,1,0,120:r1,1,2,30:ed,2,3,2vra:ra,0,3,15"):{–pei,–ped,pr1+pra}

The answer, {960,1920,2880}, is right: the independent source delivers 960W, the dependent source delivers 1920W, and the resistors consume 2880W together.

AS2's Practice Problem 2.6

Find v_x and v_o in the circuit. My solution:

s\dc("ei,x,1,35:rx,x,0,10:ed,0,2,2vx:ro,1,2,5"):{vx,vro}

The answer, {10,-5}, is correct: v_x =10 and v_o =-5.

Bo2's Example 1.10

Determine v₁, v₂ and i. My solution: The simulation took 14 seconds.

s\dc("ei,1,0,2:r1,1,2,1/3:ed,3,2,4*ir1:r2,3,0,1/5"):{ir1,vr1,vr2}

The answers, {-15/26,-5/26,-3/26}, is correct: v₁=-5/26 V, v₂=-3/26 V and i=-15/26 A.

AS2's Example 2.6

Determine v_o and i in the circuit. My solution is shown below:

s\dc("e12,1,o,12:ri,1,2,4:ed,2,3,2vo:e4,0,3,4:ro,o,0,6"):{vo,iri}

The answer, {48,-8}, is correct: v_o =48 and i = -8.

HK5's Drill Problem 1.11

Find the power absorbed by each element in the circuit. My solution:

s\dc("r1,x,0,30:ei,1,x,12:r2,1,2,8:r3,2,3,7:ed,3,0,4vx"):
approx({pr1,pei,pr2,pr3,ped})

The answer, {.768,1.92,.2048,.1792,-3.072}, is correct.

AS2's Example 3.6

Determine the value of I_o in the circuit. My solution:

s\dc("ei,a,0,24:ro,a,b,10:r12,b,0,12:r4,b,c,4:r24,a,c,24:ed,c,0,4iro"):
approx(iro)

The answer, 1.5 A, is correct.

Bo2's Drill Exercise 1.12

Determine i, v and i_d.

We are given an unnecessary piece of information: the 4V drop in the 2Ω resistor.

My solution:

s\dc("ei,1,0,10:r1,1,2,1:r2,2,3,2:r3,2,0,3:
r4,3,0,2:ed,2,3,ir1/2"):{ir1,vr3,ied}

The answer, {4,6,1}, is correct: i=4, v=6 and i_d=1.

AS2's Example 3.2

Determine the voltages at the nodes.

My solution:

s\dc("ji,0,1,3:jd,3,0,2ir2:r2,1,2,2:r4a,1,3,4:r8,2,3,8:r4b,2,0,4"):
approx({v1,v2,v3})

The answer, {4.8,2.4,-2.4}, is correct.

AS2's Example 3.4

Find the node voltages in the circuit. My solution:

s\dc("r2,1,0,2:e,1,2,20:j,0,2,10:r6,2,3,6:rx,1,4,3:r4,3,0,4:
ed,3,4,3vrx:r1,4,0,1"):approx({v1,v2,v3,v4})

The answer, {26.67,6.67,173.33,-46.67}, is correct.

Bo2's Drill Exercise 2.6

Determine the voltage in each node.

Notice that in the schematic, the resistors' values are given in siemens. Symbulator has to be fed the resistors with values in ohms. This means that, when we simulate them, we have to convert them from siemens to ohms by dividing 1 over the siemens value.

My solution:

s\dc("j,0,1,6:ei,3,1,6:ed,2,3,3v1:r5,1,0,1/5:r2,1,2,1/2:
r3,2,0,1/3:r1,2,3,1:r4,3,0,1"):{v1,v2,v3}

The answer, {-1,2,5}, is correct.

Bo2's Example 2.7

Determine the voltages in all nodes. My solution:

s\dc("j,0,1,1:r3,0,1,3:r4,2,1,4:r1,2,0,1:r2,2,3,2:
r5,3,0,5:ei,3,4,1.5:ed,4,0,2vr4"):{v1,v2,v3,v4}

The answer, {1.5,-.5,-2.5,-4.}, is correct.

Bo2's Example 2.6

Determine the voltages in all nodes.

My solution:

s\dc("e1,0,1,1:e2,3,4,.5:ed,3,2,3vr4:j,0,4,2:r4,1,2,1/4:
r1,2,0,1:r8,3,0,1/8:r2,2,4,1/2"):{v1,v2,v3,v4}

The answer, {-1,-2.,1.,.5}, is correct.

HK5's Drill Problem 1-12

Find i_A, i_B and i_C.

My solution:

s\dc("jl,x,0,5.6:ra,0,x,18:jb,0,x,.1vx:r9,0,x,9:jr,0,x,2"):
approx({ira,ijb,ir9})

The answer, {3.,-5.4,6.}, is correct.

Bo2's Drill Exercise 1.10

Determine i, v, i_s and v_s.

With one unknown value and one known solution, this problem is a job for Expert.

Determine i, v, i_s and v_s.

s\ex("es,2,0,vs:jd,0,3,2ir1:r7,0,1,7:r1,3,1,1:r3,3,2,3:r4,1,2,4"):
{ir1,vjd,–ies,vs}

Select DC, Add vr4=4 to the equations and vs to the unknowns. Run the simulation. The answer, {2,-9,-3,3}, is correct: i=2, v=-9, i_s=-3 and v_s=3.

Bo2's Drill Exercise 1.11

Determine i, v and v_d. (Since all element values are known, the tip we are given by the book – namely, that the voltage drop in the 6Ω resistor is 1.5V – is totally superfluous.)

s\dc("e,1,0,12:r1,1,2,1:r4,2,0,4:r10,2,3,10:r6,3,0,6:
r2,3,4,2:j,4,0,vr10/15"):approx({vr10,ir2,vj})

The answer, {7.5,.5,.5} is correct: i=.5, v=7.5 and v_d=.5.

TR5's Exercise 4.2

Find v_O and i_O in terms of i_S.

This is my solution. The simulation took 14 seconds.

s\dc("ji,0,x,is:r1,x,0,1'k:r2,x,o,2'k:jd,0,o,vx/500:ro,o,0,500"):
{vo,iro}

The answer, {1000*is,2*is}, is correct: v_O=1000 i_S and i_O=2 i_S.

TR5's Example 4.4

Find v_O and the equivalent resistance R_IN, in terms of v_S, when R₁ is 50 Ω, R₂ is 1 kΩ, R₃ is 100 Ω, R₄ is 5 kΩ and g is 100 mS (i.e. 100'm).

This is my solution. The simulation took 18 seconds.

s\dc("e,1,0,vs:r1,1,2,50:r2,2,o,1'k:r3,o,0,100:
r4,o,0,5'k:j,0,o,100'm*vr2"):approx({vo,re})

The answer, {.904*vs,10952.}, is correct: v_O=.904 v_S and R_IN = 10.95 kΩ.

TR5's Figure 4-4

Find the voltage drop, current, and power consumed by the 500Ω resistor, and the ratio of that power to that delivered by the independent source, all in terms of i_S.

I have not labeled the nodes in the figure, so you can practice doing it. My solution:

s\dc("js,0,1,is:r50,1,0,50:rx,1,0,25:jd,o,0,48irx:
r3,o,0,300:ro,o,0,500"):{iro,vo,pro,pro/(–pjs)}

The answers we get are correct: i_O=-12is, v_O=-6000is, p_O=72000is², and p_O/p_S=4320.

TR5's Example 4.1 (Symbolic)

Find v_O.

I did not label the nodes, so you can practice. Remember not to use rc: it is reserved.

s\dc("ei,1,0,vs:rs,1,2,rs:rx,2,0,rp:ed,0,3,r*irx:rrc,3,o,rrc:rl,o,0,rl"):vo

The answer, shown first below, is correct. The textbook's answer follows it.

Bo2's Example 1.11 (Symbolic)

Determine v₂.

My solution: In my solution I named the value of the source v₁, to keep it similar to the book. This required avoiding naming any node as 1: if there was a node 1, Symbulator would store in v1 the voltage of the node, creating trouble. There is no problem with using r1 as a value, since nothing will be stored in that r1 value.

s\dc("e,a,0,v1:r1,a,3,r1:rg,3,0,rg:j,2,0,gm*vrg:rd,2,0,rd:rl,2,0,rl"):v2

The simulation took 25 seconds. The answer I got is shown first below (the textbook's follows it.)

TR5's Example 4-7 (Symbolic)

Find R_IN, i.e. the resistance as seen by the current source. My solution:

s\dc("ji,0,a,is:re,a,0,re:jd,b,a,β*is:rl,b,0,rl"):rji

The answer we get — re*(β+1) — is correct, as can be seen by comparing it to the textbook's answer.

TR5's Exercise 4.3 (Symbolic)

Find v_O, in terms of the value in the circuit. For resistors, use their conductance value.

This is my solution. We use µ as a constant in the dependent source. It is not confused with the SI prefix for micro because the prefix has an apostrophe.

s\dc("ei,1,0,vs:ed,2,0,μ*(vrx):r1,1,2,1/g1:
r2,2,o,1/g2:rx,1,o,1/gx:rl,o,0,1/gl"):vo

This is the answer we get. It is correct. Compare it with the textbook's answer below it.

TR5's Example 4.5 (Symbolic)

Find i_B.

My solution is shown below. Be patient. This simulation takes more than one minute.

s\dc("e1,1,0,vcc:rb,1,b,rb:e2,e,b,vγ:re,e,0,re:rrc,1,c,rrc:j,c,e,β*irb"):irb

Compare my answer, first below, to the book's answer beneath it.