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We have already shown how the elementary methods of DC circuit analysis can be extended and used in AC circuits to solve for the complex peak or effective values of voltage and current and for complex impedance or admittance. In this chapter, we’ll solve some examples of voltage and current division in AC circuits.

例1

Find the voltages v₁（t）とv₂（t） v_s（T）= 110cos（2p50t）

Let’s first obtain this result by hand calculation using the voltage division formula.

The problem can be considered as two complex impedances in series: the impedance of the resistor R1, Z₁=R₁ オーム（実数）とRの等価インピーダンス₂ and L₂ シリーズで、 Z₂ = R₂ + j w L_2.

Substituting the equivalent impedances, the circuit can be redrawn in TINA as follows:

Note that we have used a new component, a complex impedance, now available in TINA v6. You can define the frequency dependence of Z by means of a table that you can reach by double clicking the impedance component. In the first row of the table you can define either the DC impedance or a frequency independent complex impedance (we have done the latter here, for the inductor and resistor in series, at the given frequency).

Using the formula for voltage division:

V₁ = V_s*Z₁ /（Z₁ + Z₂)

V₂ = V_s*Z₂ /（Z₁ + Z₂)

数値的に：

Z₁ = R₁ = 10オーム

Z₂ = R₂ + j w L = 15 + j 2*p* 50*0.04 =15 + j 12.56 Ω

V₁= 110*10/ (25+j12.56）= 35.13-j17.65 V = 39.31 e ^{–j26.7 °} V

V₂= 110*(15+j12.56）/（25 +）j12.56）= 74.86 +j17.65 V = 76.92 e ^{j 13.3°} V

The time function of the voltages:

v₁（t）= 39.31 cos（wt – 26.7°) V

v₂（t）= 76.9 cos（wt + 13.3°) V

V1

V2

Next let’s check these results with TINA’s Interpreter:

Note that when using the Interpreter we did not have to declare the values of the passive components. This is because we are using the Interpreter in a work session with TINA in which the schematic is in the schematic editor. TINA’s Interpreter looks in this schematic for the definition of the passive component symbols entered into the Interpreter program.

The diagram shows that V_s is the sum of the phasors V₁ & V₂, V_s = V₁ + V₂.

By moving the phasors we can also demonstrate that V₂ はどう違いますか V_s & V_1, V₂ = V_s – V_1.

This figure also demonstrates the subtraction of vectors. The resultant vector should start from the tip of the second vector, V₁.

In a similar way we can demonstrate that V₁ = V_s – V_2. 繰り返しますが、結果のベクトルは2番目のベクトルの先端から開始します。 V₁.

Of course, both phasor diagrams can be considered as a simple triangle rule diagram for V_s = V₁ + V₂ .

The phasor diagrams above also demonstrate Kirchhoff’s voltage law (KVL).

As we have learned in our study of DC circuits, the applied voltage of a series circuit equals the sum of the voltage drops across the series elements. The phasor diagrams demonstrate that KVL is also true for AC circuits, but only if we use complex phasors!

例2

In this circuit, R₁ represents the DC resistance of the coil L; together they model a real world inductor with its loss component. Find the voltage across the capacitor and the voltage across the real world coil.

L = 1.32 h, R₁ = 2 kohms、R₂ = 4 kohms、C = 0.1 mF, v_S（t）= 20 cos（wt) V, ｆ ＝ ＸＮＵＭＸＨｚ。

V2

Solving by hand using voltage division:

= 13.91 e ^{j 44.1°} V

&

v₁（t）= 13.9 cos（w×t + 44°) V

= 13.93 e ^{–j 44.1°} V

&

v₂(t) = 13.9 cos(w×t – 44.1°) V

Notice that at this frequency, with these component values, the magnitudes of the two voltages are nearly the same, but the phases are of opposite sign.

Once again, let’s have TINA do the tedious work by solving for V1 and V2 通訳付き

And finally, take a look at this result using TINA’s Phasor Diagram. Connecting a voltmeter to the voltage generator, invoking the Analysis/AC Analysis/Phasor Diagram command, setting the axes, and adding the labels will yield the following diagram (note that we have set 表示/ベクトルラベルスタイル 〜へ Real + j * Imag この図では）

例3

IR

IL

Using the formula for current division:

i_R（t）= 4 cos（w×t + 37.2°） NS

同様に：

i_L(t) = 3 cos(w×t – 53.1°)

And using the Interpreter in TINA:

We can also demonstrate this solution with a phasor diagram:

フェーザ図は、発電機電流ISが複素電流ILとIRの結果のベクトルであることを示しています。 また、キルヒホッフの電流法則（KCL）も示しており、回路の上位ノードに入る電流ISは、ノードから出る複素電流であるILとIRの合計に等しいことを示しています。

例4

Determine i₀（t）、 i₁(t) and i₂(t). The component values and the source voltage, frequency, and phase are given on the schematic below.

i₀

i₁

i₂

In our solution, we will use the principle of current division. First we find the expression for the total current i₀:

I_0M = 0.315 e ^{j 83.2°} A & i₀（t）= 0.315 cos（w×t + 83.2°） NS

次に、電流分割を使用して、コンデンサCの電流を求めます。

I_1M = 0.524 e ^{j 91.4°} A & i₁（t）= 0.524 cos（w×t + 91.4°） NS

そして、インダクタを流れる電流：

I_2M = 0.216 e^{–j 76.6°} A & i₂(t) = 0.216 cos(w×t – 76.6°） NS

With anticipation, we seek confirmation of our hand calculations using TINA’s Interpreter.

Another way of solving this would be to first find the voltage across the parallel complex impedance of Z_LR とZ_C. Knowing this voltage, we could find the currents i₁ そして私₂ by then dividing this voltage first by Z_LR そしてZによって_C. We will show next the solution for voltage across the parallel complex impedance of Z_LR and Z_C。 私達は方法に沿って分圧プリンシパルを使用しなければならないでしょう：

V_RLCM = 8.34 e ^{j 1.42°} V

&

I_C = I₁= V_RLCM*jwC = 0.524 e ^{j 91.42°} A

それゆえ

i_C （t）= 0.524 cos（w×t + 91.4°） NS。