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Many circuits are too complex to be solved using the rules for series or parallel circuits or the techniques for conversion to simpler circuits described in previous chapters. For these circuits we need more general solution methods. The most general method is given by Kirchhoff’s laws, which permit the calculation of all circuit voltages and currents of circuits by a solution of a system of linear equations.
There are two Kirchhoff laws, the voltage law И текущий law. These two laws can be used to determine all voltages and currents of circuits.
Kirchhoff’s voltage law (KVL) states that the algebraic sum of the voltage rises and voltage drops around a loop must be zero.
A loop in the above definition means a closed path in the circuit; that is, a path that leaves a node in one direction and returns to that same node from another direction.
In our examples, we will use clockwise direction for loops; however, the same results will be obtained if the counterclockwise direction is used.
In order to apply KVL without error, we have to define the so called reference direction. The reference direction of the unknown voltages points from the + to the – sign of the assumed voltages. Imagine using a voltmeter. You would place the voltmeter positive probe (usually red) at the component’s reference + terminal. If the real voltage is positive, it is in the same direction as we assumed, and both our solution and the voltmeter will show a positive value.
When deriving the algebraic sum of the voltages, we must assign a plus sign to those voltages where the reference direction agrees with the direction of the loop, and negative signs in the opposite case.
Another way to state Kirchhoff’s voltage law is: the applied voltage of a series circuit equals the sum of the voltage drops across the series elements.
The following short example shows the use of Kirchhoff’s voltage law.
Find the voltage across resistor R2, учитывая, что напряжение источника, ВS = 100 V and that the voltage across resistor R1 это V1 = 40 V.
The figure below can be created with TINA Pro Version 6 and above, in which drawing tools are available in the schematic editor.
The solution using Kirchhoff’s voltage law: -VS + V1 + V2 =0, or VS V =1 + V2
следовательно: V2 V =S - V1 = 100-40 = 60V
Note that normally we don’t know the voltages of the resistors (unless we measure them), and we need to use both Kirchhoff’s laws for the solution.
Kirchhoff’s current law (KCL) states that the algebraic sum of all the currents entering and leaving any node in a circuit is zero.
In the following, we give a + sign to currents leaving a node and a – sign to currents entering a node.
Here’s a basic example demonstrating Kirchhoff’s current law.
Найти тока я2 если источник тока IS = 12 A, and
I1 = 8 A.
Используя текущий закон Кирхгофа в обведенном узле: -IS + Я1 + Я2 = 0, следовательно:
I2= ЯS - Я1 = 12 - 8 = 4 A,
как вы можете проверить с помощью TINA (next
figure).
In the next example, we will use both Kirchhoff’s laws plus Ohm’s law to calculate the current and the voltage across the resistors.
In the figure below, you will note the Voltage Arrow выше резисторы. Это новый компонент, доступный в Version 6 of TINA and works like a voltmeter. If you connect it across a component, the arrow determines the reference direction (to compare to a voltmeter, imagine placing the red probe at the tail of the arrow and the black probe at the tip). When you run DC analysis, the actual voltage on the component will be displayed on the arrow.
To begin using Kirchhoff’s current law, we see that the currents through all the components are the same, so let’s denote that current by I.
According to Kirchhoff’s voltage law: VS V =1+V2+V3
Now using Ohm’s law: VS= I * R1+ I * R2+ I * R3
And from here the current of the circuit:
I = VS /(Р1+R2+R3) = 120 / (10 + 20 + 30) = 2 A
Finally the voltages of the resistors:
V1= I * R1 = 2 * 10 = 20 V; V2 = I * R2 = 2 * 20 = 40 V; V3 = I * R3 = 2 * 30 = 60 V
The same results will be seen on the Voltage Arrows by simply running TINA’s interactive DC analysis.
In this next, more complex circuit, we also use both Kirchhoff’s laws and Ohm’s law, but we find that we most solve a linear system of equations.
The total number of independent applications of Kirchhoff’s laws in a circuit is the number of circuit branches, while the total number of unknowns (the current and voltage of each branch) is twice that . However, by also using Ohm’s law at each resistor and the simple equations defining the applied voltages and currents, we get a system of equation where the number of unknowns is the same as the number of equations.
Find the branch currents I1, I2, I3 в схеме ниже.
Система уравнений выглядит следующим образом:
Узловое уравнение для обведенного узла:
–
I1 – I2
- Я3 = 0
или умножение на -1
I1 + I2 + Я3 = 0
The loop equations (using the clockwise direction) for the loop L1, containing V1, R1 и R3
-V1+I1*R1-I3*R3 = 0
и для цикла L2, содержащего V2, R2 и R3
I3*R3 - Я2*R2 +V2 = 0
Подставляя значения компонентов:
I1+ Я2+ Я3 = 0 -8 + 40 * I1 - 40 * I3 = 0 40 * I3 -20 * I2 + 16 = 0
Экспресс я1 используя узловое уравнение: I1
= -Я2 - Я3
затем подставьте его во второе уравнение:
-V1 – (I2 + Я3)*Р1 -Я3*R3 = 0 or –8- (я2 + Я3) * 40 - я3* 40 = 0
Экспресс я2 and substitute it into the third equation, from which you can already calculate I3:
I2 = - (V1 + Я3*(Р1+R3))/Р1 or I2 = - (8 + I3* 80) / 40
I3*R3 + R2* (V1 + Я3*(Р1+R3))/Р1 +V2 = 0 or I3* 40 + 20 * (8 + I3* 80) / 40 + 16 = 0
А также: I3 = - (V2 + V1*R2/R1)/(Р3+ (R1+R3)*Р2/R1) or I3 = -(16+8*20/40)/(40 + 80*20/40)
Поэтому I3 = - 0.25 А; I2 = - (8-0.25 * 80) / 40 = 0.3 A и I1 = - (0.3-0.25) = - 0.05 А
Или: I1 = -50 мА; I2 = 300 мА; I3 = -250 мА.
Now let’s solve the same equations with TINA’s
interpreter:
Sys I1, I2, I3
I1 + I2 + I3 = 0
-V1+I1*R1-I3*R3=0
I3*R3-I2*R2+V2=0
конец;
I1 = [- 50m]
I2 = [300m]
I3 = [- 250m]
импортировать numpy как np,sympy как s
#У нас есть линейная система
#уравнения, которые мы хотим решить:
#I1+I2+I3=0
#-V1+I1*R1-I3*R3=0
#I3*R3-I2*R2+V2=0
I1,I2,I3=s.symbols([‘I1′,’I2′,’I3’])
соль = s.solve([
И1+И2+И3,
-V1+I1*R1-I3*R3,
I3*R3-I2*R2+V2], [I1, I2, I3])
печать (соль)
A= np.array([[1,1,1],[R1,0,-R3],[0,-R2,R3]])
б = np.array([0,V1,-V2])
x=np.linalg.solve(A,b)
#I1=х[0]
#I2=х[1]
#I3=х[2]
# I1
print("I1= %.3f"%x[0])
# I2
print("I2= %.3f"%x[1])
# I3
print("I3= %.3f"%x[2])
Наконец, давайте проверим Результаты с использованием TINA:
Next, let’s analyze the following even more complex circuit and determine its
branch currents and voltages.
Let’s denote the unknown voltages and currents by adding voltage and current arrows to components, and also show the loops (L1,L2, L3) and the nodes (N1,N2) where we will use the Kirchhoff’s equations.
 |
Вот набор Kirchhoff equations for the loops (using the clockwise direction) and the nodes.
-IL + ЯR1 - Яs = 0 (для N1)
- ЯR1 + ЯR2 + Яs3 = 0 (для N2)
-Vs1 - VR3 + VIs + VL = 0 (для L1)
-VIs + Vs2 +VR2 +VR1 = 0 (для L2)
-VR2 - Vs2 + Vs3 = 0 (для L3)
Applying Ohm’s law:
VL = ЯL*RL
VR1 =IR1*R1
VR2 = ЯR2*R2
VR3 = - ЯL*R3
This is 9 unknowns and 9 equations. The easiest way to solve this is to use TINA’s
interpreter. However, if we are pressed to use hand calculations, we note that this set of equations can be easily reduced to a system of 5 unknowns by substituting the last 4 equations into the L1, L2, L3 loop equations. Also, by adding equations (L1) and (L2), мы можем устранить VIs , сводя задачу к системе уравнений 4 для неизвестных 4 (IL, IR1 IR2, Is3). When we have found these currents, we can easily determine VL, VR1, VR2, and VR3 используя последние четыре уравнения (закон Ома).
Substituting VL ,VR1,VR2 ,VR3 :
-IL + ЯR1 - Яs = 0 (для N1)
- ЯR1 + ЯR2 + Яs3 = 0 (для N2)
-Vs1 + ЯL*R3 + VIs + ЯL*RL = 0 (для L1)
-VIs + Vs2 + ЯR2*R2 + ЯR1*R1 = 0 (для L2)
- ЯR2*R2 - Vs2 + Vs3 = 0 (для L3)
Добавляя (L1) и (L2) мы получаем
-IL + ЯR1 - Яs = 0 (для N1)
- ЯR1 + ЯR2 + Яs3 = 0 (для N2)
-Vs1 + ЯL*R3 + ЯL*RL + Vs2 + ЯR2*R2 + ЯR1*R1 = 0 (L1) + (L2)
- ЯR2*R2 - Vs2 + Vs3 = 0 (для L3)
After substituting the component values, the solution to these equations comes readily.
-IL+IR1 - 2 = 0 (для N1)
-IR1 + ЯR2 + ЯS3 = 0 (for N2)
-120 - + ЯL* 90 + IL* 20 + 60 + IR2* 40 + IR1*30 = 0 (L1) + (L2)
-IR2* 40 - 60 + 270 = 0 (для L3)
от L3 IR2 = 210 / 40 = 5.25 A (I)
из N2 IS3 - ЯR1 = - 5.25 (II),
от L1+L2 110 IL + 30 IR1 = -150 (III),
и для N1 IR1 - ЯL = 2 (IV)
Умножьте (IV) на –30 и добавьте к (III) 140 IL = -210 следовательно IL = - 1.5 А
Заменить яL в (IV) IR1 = 2 + (-1.5) = 0.5 A
и яR1 в (II), IS3 = -5.25 + IR1 = -4,75 A
И напряжения: VR1 = ЯR1*R1 = 15 V; VR2 = ЯR2*R2 = 210 V;
VR3 = - ЯL*R3= 135 V; VL = ЯL*RL = - 30 В; VIs V =S1+VR3-VL = 285 V
Sys IL,IR1,IR2,Is3,VIs,VL,VR1,VR3,VR2
-IL-Is + IR1 = 0
-IR1 + IR2 + Is3 = 0
-Vs1 + VR3 + Вис-VL = 0
-Vis + VR1 + VR2 + Vs2 = 0
-Vs3 + VR2 + Vs2 = 0
VR1 = IR1 * R1
VR2 = IR2 * R2
VR3 = -IL * R3
VL = IL * RL
конец;
IL = [- 1.5]
IR1 = [500m]
IR2 = [5.25]
Is3 = [- 4.75]
= ВП [285]
VL = [- 30]
VR1 = [15]
VR2 = [210]
VR3 = [135]
#Ax=b
импортировать numpy как np,sympy как s
#Символическое решение с использованием numpy.solve
#Уравнения:
#IL=-Is+IR1
#IR1=IR2+Is3
#Vs1+VR3-Вис-VL=0
#Вис=ВР1+ВР2+Вс2
#Vs3=VR2+Vs2
#VR1=IR1*R1
#VR2=IR2*R2
#VR3=-IL*R3
#ВЛ=ИЛ*РЛ
#Решить для:
#IL,IR1,IR2,
#Is3,Вис,ВЛ,
#ВР1,ВР3,ВР2
IL,IR1,IR2,Is3,Vis,VL,VR1,VR3,VR2=s.symbols([‘IL’,’IR1′,’IR2′,’Is3′,’Vis’,’VL’,’VR1′,’VR3′,’VR2′])
соль = s.solve([
-Is+IR1-IL,
ИР2+Ис3-ИР1,
Вс1+ВР3-Вис-ВЛ,
ВР1+ВР2+Вс2-Вис,
ВР2+Вс2-Вс3,
IR1*R1-VR1,IR2*R2-VR2,
-IL*R3-VR3,IL*RL-VL],[IL,IR1,IR2,Is3,Vis,VL,VR1,VR3,VR2])
печать (соль)
#Еще один метод решения с использованием numpy.linalg
A=np.array(
[[-1,1,0,0,0,0,0,0,0],
[0,-1,1,1,0,0,0,0,0],
[0,0,0,0,-1,-1,0,1,0],
[0,0,0,0,-1,0,1,0,1],
[0,0,0,0,0,0,0,0,1],
[0,R1,0,0,0,0,-1,0,0],
[0,0,R2,0,0,0,0,0,-1],
[-R3,0,0,0,0,0,0,-1,0],
[RL,0,0,0,0,-1,0,0,0]])
b=np.array([Is,0,-Vs1,-Vs2,Vs3-Vs2,0,0,0,0])
x=np.linalg.solve(A,b)
#IL=x[0] IR1=x[1] IR2=x[2]
#Is3=x[3] Вис=x[4] VL=x[5]
#VR1=x[6] VR2=x[8] VR3=x[7]
print("IL= %.3f"%x[0])
print("IR1= %.3f"%x[1])
print("IR2= %.3f"%x[2])
print("Is3= %.3f"%x[3])
print("Vis= %.3f"%x[4])
print("VL= %.3f"%x[5])
print("VR1= %.3f"%x[6])
print("VR2= %.3f"%x[8])
print("VR3= %.3f"%x[7])
Solution of the reduced set of equations using the interpreter:
|
{Решение сокращенной системы уравнений интерпретатором TINA}
Sys Il, Ir1, Ir2, Is3 -Il + Ir1-2 = 0 -Ir1 + Ir2 + Is3 = 0 -120+110*Il+60+40*Ir2+30*Ir1=0 -40 * Ir2 + 210 = 0 конец; Ил = [- 1.5] Ir1 = [500m] Ir2 = [5.25] Is3 = [- 4.75] |
We can also enter expressions for the voltages and have TINA’s Interpreter calculate them:
|
Il: = - 1.5;
Ir1: = 0.5; Ir2: = 5.25; Is3: = - 4.75; Вл: = Il * RL; Vr1: = Ir1 * R1 Vr2: = Ir2 * R2; Vr3: = - Il * R3; ВП: = Vs1-Vl + Vr3; Vl = [- 30] Vr1 = [15] Vr2 = [210] Vr3 = [135] = ВП [285] |
 |
We can check the result with TINA by simply turning on TINA’s DC interactive mode or using Analysis / DC Analysis / Nodal Voltages
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