Mutual inductance and ideal transformers

Calculate the phasor currents I₁ and I₂ in the circuit.

Solution

Three things to notice in the description. The bottom node of both halves is ground. The coils are given in jΩ, so they are r elements with imaginary values, and the mutual is imaginary too, to match. And the dotted node of each coupled element is named first.

Circuit Description

e,1,0,12
r1,1,2,-4j
r2,2,0,5j
m,r2,r3,3j
r3,3,0,6j
r4,3,0,12

AC, and leave RMS phasors unticked. Then read the two currents with Mini-Tools set to aa:

aa(ir2) reads 13.02∠-49.4° A and aa(ir4) reads 2.910∠14.04° A.

Both are correct.

AS7's Practice Problem 13.1

Determine the voltage V_o in the circuit.

Solution

Again the dotted node of the coupled inductor comes first.

Circuit Description

e,1,0,(200∠45°)
r1,1,2,4
r2,2,0,8j
m,r2,r3,1j
r3,0,o,5j
r4,o,0,10

aa(vo) reads 20.00∠-134.43° V.

Correct.

AS7's Example 13.2

Calculate the mesh currents in the circuit.

Solution

Notice the node order again, matching the dots.

Circuit Description

e,1,0,100
r1,1,2,4-3j
r2,2,0,6j
m,r2,r3,2j
r3,3,2,8j
r4,3,0,5

aa(ir1) reads 20.30∠3.5° A and aa(ir4) reads 8.693∠19.03° A.

Correct.

NR9's Assessment Problem 9.14

A linear transformer couples a load of a 360 Ω resistor in series with a 0.25 H inductor to a sinusoidal source of 245.20 V maximum at 800 rad/s, whose internal impedance is 184 Ω. The transformer has R₁ = 100 Ω, L₁ = 0.5 H, R₂ = 40 Ω, L₂ = 0.125 H and a coefficient of coupling k = 0.4, with the dots at the top of both coils. Calculate (a) the reflected impedance, (b) the primary current and (c) the secondary current.

Solution

The book gives the coupling as a coefficient, so the m line takes it as it is, m,l1,l2,k=0.4, and there is no mutual inductance to work out first. Both dots are at the top, so both coils are written top node first. We call the source's internal impedance rs, and the load's resistor and inductor rl and ll. The book draws the two sides on separate bottom rails. We join them at ground, which changes no current, because nothing conducts from one winding to the other.

Circuit Description

e,1,0,245.2
rs,1,a,184
r1,a,p,100
l1,p,0,0.5
m,l1,l2,k=0.4
l2,q,0,0.125
r2,q,c,40
rl,c,d,360
ll,d,0,0.25

Set Analysis to AC — alternating current and put 800 in the ω — angular frequency box. In Settings, set Rounding to approx to n digits with n = 4 and tick Show AC answers as polar phasors.

Parts (b) and (c) are read straight off the cards. The primary current is the current through r1, and the secondary current the current through r2, each in the direction of the book's arrow:

current through r1

\[i_{r1} = 0.5000∠-53.13°\]

current through r2

\[i_{r2} = 0.08000∠0°\]

So (b) the primary current is 0.5 cos(800t − 53.13°) A, and (c) the secondary current is 0.08 cos 800t A.

(a) The reflected impedance is what the secondary adds to the impedance the source sees. The source's card reports that impedance as ze. Without the coupling it would be the source's own 184 Ω, the primary winding's 100 Ω and the primary coil's jωL₁ = j800 × 0.5 = j400 Ω, so we type the difference into Evaluate:

Evaluate

ze - (184 + 100 + 400j)

It gives 10.24 - 7.68j Ω. All three agree with the answers the book prints. The coupling written as the inductance it stands for, m,l1,l2,0.1, gives the same, since M = 0.4 × √(0.5 × 0.125) = 0.1 H.

10.3How to describe ideal transformers

An ideal transformer has four terminals — two a side. Symbulator simulates the case where one terminal on each side is grounded, which covers the textbook problems; you name only the two live nodes.

A t element takes five fields: a name starting with t, the live node on the first side, the live node on the second side, and the turns on each side. For example: t1,1,2,80,800

That is 80 turns on the side connected to node 1 and 800 on the side connected to node 2, wound the same way. Only the ratio matters, so 1,10 would describe the same transformer.

Polarity is a minus sign. If the dots are on opposite ends, make one of the two turn counts negative: t,a,b,1,-5

It does not matter which of the two you make negative, as long as one of them is.

The turns may be written as a pair in brackets, t1,1,2,[80,800], which means exactly the same as the five-field form.

All four terminals. When neither bottom terminal is ground, write each side as a bracketed pair, top node then bottom node, and the turns as a pair:

Circuit Description

t1,[1,3],[2,4],[80,800]

That is a primary between nodes 1 and 3, a secondary between nodes 2 and 4, 80 and 800 turns. The dots sit at the two top nodes, as before. An ideal transformer conducts nothing from one side to the other, so a side with no path of its own to ground is an island: Symbulator takes one of its nodes — the winding's lower terminal — as that side's own reference, holds it at 0, and says so in a note above the results. Every current and every voltage difference on that side is the same whichever node is chosen; only its absolute potentials depend on it, and they have no meaning anyway. Either terminal of a pair may be 0, so t1,[1,0],[2,0],[80,800] is the two-node form written out.

What answers do you get? The voltages at the live nodes, as for any node in the circuit, and the current entering the transformer at each of its live nodes, named with the transformer and the node — for a transformer t on nodes 1 and 2, that is it1 and it2. With all four terminals named, there are four such currents, one per node.

10.4Instructive t problems

AS7's Figure 13.33

Obtain V_Th and Z_Eq for the part of the circuit to the right of nodes a and b.

Solution

Everything left of a and b is ignored. Note that this one is entirely symbolic — the turns ratio, the impedance and the source are all names.

Circuit Description

t,2,3,1,n
r2,3,4,z2
e2,4,0,vs2

Find equivalent, Thévenin / Norton, nodes 2 and 0, in DC.

Results gives

Thévenin voltage

\[v_{th} = \dfrac{vs_{2}}{n}\]

equivalent resistance

\[R_{eq} = \dfrac{z_{2}}{n^{2}}\]

Correct — the transformer refers the source by the turns ratio and the impedance by its square, which is exactly what an ideal transformer does.

AS7's Example 13.8

For the ideal transformer circuit, find the source current I₁, the output voltage V_o, and the complex power supplied by the source.

Solution

One of the turn counts is negative here, because the dots are of opposite polarity.

Circuit Description

e,1,0,120
r1,1,2,4-6j
t,2,3,1,-2
ro,3,0,20

AC, with RMS phasors ticked this time — the question asks for complex power, and that is the convention it wants.

Three readings from Mini-Tools with aa: ir1 is 11.09∠33.69° A, vro is 110.9∠-146.31° V, and -se — the complex power supplied, so the opposite of the power consumed by the source — is 1331∠-33.69° VA.

Three conversions in a row is the point at which the setting is less work than the tool: tick Show AC answers as polar phasors in Settings and the first two are already in that form when the circuit solves. The third still wants aa, because -se is an expression rather than an answer.

All three are correct.

AS7's Example 13.11

Refer to the autotransformer circuit. Calculate I₁, I₂ and I_o if Z_L = 8 + j6 Ω, and the complex power supplied to the load.

Solution

Symbulator has no dedicated element for an autotransformer. The same t element does it: the input is node 1, the output is node 2, and the turns are the 80 of the lower winding and the 80 + 120 of the whole.

Circuit Description

e,1,0,(120∠30°)
t,1,2,80,80+120
rl,2,0,8+6j

AC, with RMS phasors ticked.

This is how the three currents come out of the two the transformer reports:

I₁ is the current entering the transformer at node 1, it1.
I₂ is the opposite of the current entering it at node 2, it2.
I_o is the opposite of the sum of the two, since what enters the transformer at its two live nodes leaves it through the common one.

Results gives, under t,

current into port at node 1

\[i_{t1} = 75∠-6.87°\]

current into port at node 2

\[i_{t2} = 30∠173.13°\]

so I₁ is 75∠-6.87° A, I₂ is the opposite of it2, 30∠-6.87° A, and I_o, read with aa on -it1-it2, is 45∠173.13° A.

All three are correct.

The autotransformer as one tapped winding

The same autotransformer, described as what it physically is: one winding with a tap. The lower section, 80 turns, runs from the tap to ground; the upper section, 120 turns, runs from the top of the winding down to the tap. Find I₁ and I₂ again.

The autotransformer as one tapped winding, drawn by Symbulator

Solution

The second winding's bottom terminal is the first winding's top, node 1, so this needs the four-terminal form: the first winding is the pair [1,0], the second the pair [2,1].

Circuit Description

e,1,0,(120∠30°)
t,[1,0],[2,1],[80,120]
rl,2,0,8+6j

AC, with RMS phasors ticked. Results gives, under t,

current into port at node 1

\[i_{t1} = 75∠-6.87°\]

current into port at node 2

\[i_{t2} = 30∠173.13°\]

Node 1 is a terminal of both windings, so it1 is the total entering the transformer there, which is the source current I₁, 75∠-6.87° A; and I₂ is the opposite of it2, 30∠-6.87° A. The same answers as with the two-node description above, from a description that looks like the circuit.

A transformer whose primary is not grounded

A 10 V source feeds a 1 Ω resistor and then the primary of a 2:1 ideal transformer; the far end of the primary returns to ground through a 3 Ω resistor. The secondary is grounded and drives an 8 Ω load. Find the load current and the current into each terminal of the transformer.

A transformer whose primary is not grounded, drawn by Symbulator

Solution

The primary sits between nodes 2 and 3, neither of them ground, so it is written as the pair [2,3]; the secondary's bottom is ground, and the pair [4,0] says so.

Circuit Description

e,1,0,10
r1,1,2,1
t,[2,3],[4,0],[2,1]
r3,3,0,3
rl,4,0,8

DC. Results gives, under t,

current into port at node 2

\[i_{t2} = \dfrac{5}{18}\]

current into port at node 3

\[i_{t3} = -\dfrac{5}{18}\]

current into port at node 4

\[i_{t4} = -\dfrac{5}{9}\]

and irl is 5/9 A. Check: the 8 Ω load reflects to the primary as 8 × (2/1)² = 32 Ω, in series with 1 + 3 Ω, so the primary current is 10/36 = 5/18 A, entering at node 2 and leaving at node 3; the 2:1 ratio doubles it on the secondary, where it4 is negative because that current is leaving the transformer into the load.

Both windings between live nodes, and why a side needs a ground

The same 2:1 transformer, now with its secondary between two live nodes as well: the 100 Ω load sits across the secondary, and node 5 is tied to ground through a 7 Ω resistor. Find the currents.

Both windings between live nodes, drawn by Symbulator

Solution

Circuit Description

e,1,0,10
r0,1,2,1
t,[2,4],[3,5],[2,1]
r4,4,0,3
rl,3,5,100
r5,5,0,7

DC. Results gives, under t,

current into port at node 2

\[i_{t2} = \dfrac{5}{202}\]

current into port at node 4

\[i_{t4} = -\dfrac{5}{202}\]

current into port at node 3

\[i_{t3} = -\dfrac{5}{101}\]

current into port at node 5

\[i_{t5} = \dfrac{5}{101}\]

and ir5 is 0: the secondary's current circulates through the load alone, and the resistor to ground carries nothing. What it does is give the secondary side a reference. An ideal transformer conducts nothing from one winding to the other, so without r5 that side would have no path to node 0 at all — delete it and Symbulator still solves, taking node 5 as the side's own reference and saying so in a note: Node(s) 5, 3 have no path to node 0 (they lie behind a port or a coupling), so their voltages are measured against 5, taken as 0. The currents and v3 − v5 come out the same either way; only the two absolute potentials differ, and on a real bench they would be whatever the wiring made them.

10.5Switching with coupled coils

Seven problems from Nilsson and Riedel's Electric Circuits: four from the 11th edition, each with a switch that moves at t = 0 and a pair of coupled coils on the far side of it, and three from the 12th, in which a switch closes onto a source with nothing stored. The method is the one Lesson 6 uses for every switch: a DC solve before the switch gives the coil currents, and those become the initial conditions of a transient solve after it — or simply zeros, when the book says no energy is stored. In the first four the coupled side has no ground of its own, and none is needed: a coupling conducts nothing across, so Symbulator takes one node of that side as its reference and says so. The variable asked for is the one the book marks in blue.

NR11's 60 V source and coupled 2 H and 8 H coils

A 60 V source feeds a 9 Ω resistor and, through a switch at a, a 3 Ω resistor in series with a 2 H coil. The coil is coupled, M = 2 H, to an 8 H coil closed on 2 Ω and 10 Ω. The switch has been at a a long time and moves to b at t = 0, shorting the 3 Ω and the 2 H together. Find i₁ in the 2 H coil and i₂ in the 8 H coil for t > 0. Both dots are at the coils' upper ends.

The circuit after the switch, drawn by Symbulator

Solution

Before the switch, in DC, with the source in and the switch at a:

Circuit Description

e,1,0,60
r9,1,a,9
r3,a,2,3
l1,2,0,2
l2,3,4,8
r2,3,5,2
r10,5,4,10
m,l1,l2,2

Both coils are shorts in DC, so il1 = 5 A (60 V over 9 + 3 Ω) and il2 = 0. The secondary has no path to node 0, and the results say so in a note: its voltages are measured against node 4.

After the switch the source is gone and the 3 Ω and the 2 H are shorted through b; the 5 A is the coil's initial condition. In TR:

Circuit Description

r3,0,2,3
l1,2,0,2,5
l2,3,4,8,0
r2,3,5,2
r10,5,4,10
m,l1,l2,2

current through l1

\[i_{l1} = \dfrac{5}{2}\left(e^{-t} + e^{-3t}\right)\ \mathrm{A}\]

current through l2

\[i_{l2} = \dfrac{5}{4}\left(e^{-t} - e^{-3t}\right)\ \mathrm{A}\]

Two modes, at 1 and 3 per second: i₁ starts from 5 A and i₂ from 0, and both vanish. Both satisfy the pair of coil equations exactly, which is how they were checked.

NR11's two switches and coupled 3 H and 15 H coils

On the left, a 24 V source and 120 Ω feed a 3 H coil through switch 1, which has been open a long time and closes at t = 0. On the right, a 15 H coil, coupled to the first by M = 3 H, is connected through switch 2 to either a — a 10 Ω resistor and a 20 V source — or b, a 360 Ω resistor. Switch 2 has been at a a long time and moves to b at t = 0. Find i₁ for t > 0. The dot of the 3 H coil is at its upper end, the dot of the 15 H coil at its lower end.

The circuit after the switches, drawn by Symbulator

Solution

Before the switches there is no primary current at all, and the 20 V drives 2 A through the 10 Ω and down through the 15 H coil. In DC:

Circuit Description

e,a,0,20
r10,a,c,10
l2,d,c,15

The dot of the 15 H coil is at its bottom, so its first node is d, the bottom, and the current from d to c reads il2 = -2 A.

After the switches the 24 V drives the primary from rest and the secondary, starting at 2 A, decays through the 360 Ω. In TR, ticking Do you want to limit the results to save time? and asking for the one current:

What results are you after? List the variables here

il1

Circuit Description

e1,1,0,24
r120,1,2,120
l1,2,0,3,0
l2,d,c,15,-2
r360,c,d,360
m,l1,l2,3

current through l1

\[i_{l1} = \dfrac{1}{5} - \dfrac{31}{20}\,e^{-20t} + \dfrac{27}{20}\,e^{-60t}\ \mathrm{A}\]

It starts from 0, settles at 24/120 = 0.2 A, and its two modes are at 20 and 60 per second. The dots are at opposite ends, which is why the secondary is written bottom node first with a positive M; writing it top node first with M = −3 gives the same answer.

NR11's 90 V source and coupled 3 H and 2 H coils

A 90 V source and 5 Ω feed, through a switch at a, a 3 H coil whose far end joins a 2 H coil and a 10 Ω resistor to ground; the coils are coupled, M = 1 H, dots at both upper ends. The switch has been at a a long time and moves to b at t = 0, replacing the source by a 20 Ω resistor. Find i_o, the current into the 3 H coil, for t > 0.

Solution

Before the switch both coils are shorts in DC, and the 2 H shorts the 10 Ω:

Circuit Description

e,1,0,90
r5,1,t,5
l1,t,m,3
l2,m,0,2
r10,m,0,10
m,l1,l2,1

il1 = 18 A, il2 = 18 A and ir10 = 0: all of it goes through the coils.

After the switch, in TR, asking only for il1 in the same way:

Circuit Description

r20,t,0,20
l1,t,m,3,18
l2,m,0,2,18
r10,m,0,10
m,l1,l2,1

current through l1

\[i_{l1} = 12\,e^{-2t} + 6\,e^{-20t}\ \mathrm{A}\]

From 18 A to zero, with modes at 2 and 20 per second.

NR11's capacitor shorted by a switch, coupled 0.8 H and 1.6 H coils

A 48 V source feeds a 10 mF capacitor, a 4 Ω resistor and a 0.8 H coil in series. A switch across the capacitor has been closed a long time and opens at t = 0. The coil is coupled, M = 0.8 H, to a 1.6 H coil across a 20 Ω resistor; both dots are at the upper ends. Find v_o, the voltage across the 20 Ω, for t > 0.

The circuit after the switch opens, drawn by Symbulator

Solution

Before the switch the capacitor is shorted, so it is left out of the DC picture:

Circuit Description

e,1,0,48
r4,1,3,4
l1,3,0,0.8
l2,4,5,1.6
r20,4,5,20
m,l1,l2,0.8

il1 = 12 A and il2 = 0.

After the switch opens the capacitor enters the loop, uncharged, with the coil carrying its 12 A. In TR, and then in Evaluate:

Evaluate

v4-v5

Circuit Description

e,1,0,48
c,1,2,10'm,0
r4,2,3,4
l1,3,0,0.8,12
l2,4,5,1.6,0
r20,4,5,20
m,l1,l2,0.8

\[v_o = e^{-5t}\left(60\cos 10t - 120\sin 10t\right) - 60\,e^{-25t}\ \mathrm{V}\]

A damped oscillation at 10 rad/s, the primary's series RLC, plus the secondary's own mode at 25 per second; it starts from 0 and returns to 0. Since the secondary's reference is node 5, v4 alone reads the same thing.

NR12's Problem 7.68

There is no energy stored in the circuit when the switch closes at t = 0. A 200 V source and 50 Ω feed a 4 H coil in series with an 8 H coil, coupled by M = 5 H; the 4 H coil's dot is at its left end, the 8 H coil's at its bottom. Find i(t), v₁(t) across the 4 H and v₂(t) across the 8 H, and say whether the answers make sense.

Solution

The current enters the 4 H at its dot and the 8 H at its undotted end, so the 8 H is written bottom node first and the pair opposes:

Circuit Description

e,1,0,200
r50,1,2,50
l1,2,3,4,0
l2,0,3,8,0
m,l1,l2,5

In TR, with the fifth field of each coil, the initial current, at 0:

current through l1

\[i_{l1} = 4 - 4\,e^{-25t}\ \mathrm{A}\]

voltage at node 3

\[v_{3} = 300\,e^{-25t}\ \mathrm{V}\]

and Evaluate with v2-v3 gives v₁ = \(-100\,e^{-25t}\) V. The pair reduces to 4 + 8 − 2·5 = 2 H, whence the 25 per second (2 H over 50 Ω), and v₁ is negative because the 5 H coupling outweighs the coil's own 4 H; the two voltages still add to \(200\,e^{-25t}\), which is 200 − 50 i.

NR12's Problem 7.70

There is no energy stored when the switch closes. A 10 V source and 250 Ω feed a 0.5 H coil and a 0.25 H coil in parallel, coupled by M = 0.25 H; the 0.5 H coil's dot is at its top, the 0.25 H coil's at its bottom. Find i_o, v_o, i₁ and i₂, both branch currents taken downward.

Solution

The 0.25 H coil is written bottom node first, since that is its dotted end:

Circuit Description

e,1,0,10
r250,1,a,250
l1,a,0,0.5,0
l2,0,a,0.25,0
m,l1,l2,0.25

TR. Results gives ir250 = i_o = \(0.04(1 -\,e^{-5000t})\) A, va = v_o = \(10\,e^{-5000t}\) V and il1 = i₁ = \(0.016(1 -\,e^{-5000t})\) A; i₂ is downward, the opposite of il2, so Evaluate with -il2 reads \(0.024(1 -\,e^{-5000t})\) A. The parallel pair opposes and reduces to (0.5·0.25 − 0.25²)/(0.5 + 0.25 + 2·0.25) = 0.05 H, whence the 5000 per second with 250 Ω.

NR12's Problem 7.71

There is no energy stored when the switch closes. A 15 V source and 75 Ω feed an 8 mH coil and a 20 mH coil in parallel, coupled by M = 10 mH, both dots at the top. Find i_o, v_o, i₁ and i₂.

Solution

Both coils are written top node first:

Circuit Description

e,1,0,15
r75,1,a,75
l1,a,0,8'm,0
l2,a,0,20'm,0
m,l1,l2,10'm