KIRCHHOFF의 법

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예제를 편집하거나 자체 회로를 생성하려면 TINACloud에 저가의 액세스 권한을 얻으십시오.

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 및 전류 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 R_2, 소스 전압 V_S = 100 V and that the voltage across resistor R₁ V이다.₁ = 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: -V_S + V₁ + V₂ =0, or V_S = V₁ + V₂

금후: 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.

현재 I 찾기₂ 소스 전류 I_S = 12 A, and I₁ = 8 A.

회람 된 노드에서 Kirchhoff의 현재 법칙 사용하기 : -I_S + I₁ + I₂ = 0, 따라서 : I₂= I_S - 나₁ = 12 - 8 = 4 A, TINA를 사용하여 확인할 수 있습니다. (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 위의 저항. 이것은에서 사용할 수있는 새로운 구성 요소입니다. 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.

According to Kirchhoff’s voltage law: V_S = V₁+V₂+V₃

Now using Ohm’s law: V_S= I * R₁+ I * R₂+ I * R₃

And from here the current of the circuit:

I = V_S /(아르 자형₁+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.

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 아래 회로에서.

원이있는 노드의 노드 방정식

- I₁ - I₂ - 나₃ = 0

또는 -1

I₁ + I₂ + I₃ = 0

The loop equations (using the clockwise direction) for the loop L1, containing V₁, R₁ 및 R₃

-V₁+I₁*R₁-I₃*R₃ = 0

V를 포함하는 루프 L2의 경우₂, R₂ 및 R₃

I₃*R₃ - 나₂*R₂ +V₂ = 0

구성 요소 값 대체 :

I₁+ I₂+ I₃ = 0 -8 + 40 * I₁ - 40 * I₃ = 0 40 * I₃ -20 * I₂ + 16 = 0

익스프레스 I₁ 절점 방정식을 사용하여 : I₁ = -I₂ - 나₃

두 번째 방정식으로 대체하십시오.

-V₁ – (I₂ + I₃)*아르 자형₁ -나는₃*R₃ = 0 or -8- (I₂ + I₃) * 40 - 나₃* 40 = 0

익스프레스 I₂ and substitute it into the third equation, from which you can already calculate I₃:

I₂ = – (V₁ + I₃*(아르 자형₁+R₃))/아르 자형₁ or I₂ = - (8 + I₃* 80) / 40

I₃*R₃ + R₂*(V₁ + I₃*(아르 자형₁+R₃))/아르 자형₁ +V₂ = 0 or I₃* 40 + 20 * (8 + I₃* 80) / 40 + 16 = 0

과 : I₃ = – (V₂ + V₁*R₂/R₁)/(아르 자형₃+ (R₁+R₃)*아르 자형₂/R₁) or I₃ = -(16+8*20/40)/(40 + 80*20/40)

따라서 I₃ = – 0.25A; I₂ = - (8-0.25 * 80) / 40 = 0.3 A 과 I₁ = – (0.3-0.25) = – 0.05A

또는 : I₁ = -50 mA; I₂ = 300 mA; I₃ = -250 mA.

Now let’s solve the same equations with TINA’s interpreter:

마지막으로 TINA를 사용한 결과 :

Next, let’s analyze the following even more complex circuit and determine its branch currents and voltages.

위의 회로를 클릭 / 탭하여 온라인으로 분석하거나 Windows에서 저장하려면이 링크를 클릭하십시오.

-I_L + I_R1 - 나_s = 0 (N1 용)

- 나_R1 + I_R2 + I_s3 = 0 (N2 용)

-V_s1 - V_R3 + V_Is + V_L = 0 (L1 용)

-V_Is + V_s2 +V_R2 +V_R1 = 0 (L2 용)

-V_R2 - V_s2 + V_s3 = 0 (L3 용)

Applying Ohm’s law:

V_L = I_L*R_L

V_R1 =I_R1*R₁

V_R2 = I_R2*R₂

V_R3 = – 나_L*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), 우리는 V_Is , 문제를 4 미지수에 대한 4 방정식 시스템으로 줄인다._L, I_R1 I_R2, I_s3). When we have found these currents, we can easily determine V_L, V_R1, V_R2, and V_R3 마지막 4 개의 방정식 (옴의 법칙)을 사용합니다.

Substituting V_L ,V_R1,V_R2 ,V_R3 :

-I_L + I_R1 - 나_s = 0 (N1 용)

- 나_R1 + I_R2 + I_s3 = 0 (N2 용)

-V_s1 + I_L*R₃ + V_Is + I_L*R_L = 0 (L1 용)

-V_Is + V_s2 + I_R2*R₂ + I_R1*R₁ = 0 (에 대한 L2)

- 나_R2*R₂ - V_s2 + V_s3 = 0 (L3 용)

우리가 얻는 (L1)과 (L2) 추가

-I_L + I_R1 - 나_s = 0 (N1 용)

- 나_R1 + I_R2 + I_s3 = 0 (N2 용)

-V_s1 + I_L*R₃ + I_L*R_L + V_s2 + I_R2*R₂ + I_R1*R₁ = 0 (L1) + (L2)

- 나_R2*R₂ - V_s2 + V_s3 = 0 (L3 용)

After substituting the component values, the solution to these equations comes readily.

-I_L+I_R1 - 2 = 0 (N1 용)

-I_R1 + I_R2 + I_S3 = 0 (for N2)

-120 – + 나_L* 90 + I_L* 20 + 60 + I_R2* 40 + I_R1*30 = 0 (L₁) + (L₂)

-I_R2* 40 - 60 + 270 = 0 (L₃)

L에서₃ I_R2 = 210 / 40 = 5.25 A (나는)

N에서₂ I_S3 - 나_R1 =-5.25 (II)

L에서₁+L₂ 110 I_L + 30 I_R1 = -150 (III)

및 N₁ I_R1 - 나_L = 2 (IV)

-30만큼 곱하고 (III) 140 I_L = -210 금후 I_L = – 1.5A

대체 내가_L (IV) I_R1 = 2 + (-1.5) = 0.5 A

및 _R1 으로 (II) I_S3 = -5.25 + I_R1 = -4,75 A

그리고 전압 : V_R1 = I_R1*R₁ = 15 V; V_R2 = I_R2*R₂ = 210 V;

V_R3 = – 나_L*R₃= 135 V; V_L = I_L*R_L = – 30V; V_Is = V_S1+V_R3-V_L = 285 V

Solution of the reduced set of equations using the interpreter:

We can also enter expressions for the voltages and have TINA’s Interpreter calculate them: