联合系列-拉拉电路

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Electrical can become immensely complicated. This is a polynomial plotter, which allows users to plot polynomials and evaluate functions at various $\begin{align*}x\end{align*}$ values.
::电气可能变得非常复杂。 这是一个多式绘图器, 用户可以用不同的 x 值绘制多式绘图并评估函数 。

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Combined Series-Parallel Circuits
::联合系列-拉拉电路

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Most circuits are not just a series or parallel circuit ; most have resistors in parallel and in series. These circuits are called combination circuits . When solving problems with such circuits, use this series of steps.
::大多数电路不仅仅是一个系列或平行电路; 多数有平行和系列的阻力器。 这些电路被称为组合电路。 当解决这些电路的问题时, 请使用这一系列步骤 。

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1. For resistors connected in parallel, calculate the single equivalent resistance that can replace them.
   ::对于平行连接的抗体,计算能够取代它们的单等效抗药性。
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2. For resistors in series, calculate the single equivalent resistance that can replace them.
   ::对于一系列的抗体,计算单等效抗药性,以取代它们。
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3. By repeating steps 1 and 2, you can continually reduce the circuit until only a single equivalent resistor remains. Then you can determine the total circuit current. The drops and currents though individual resistors can then be calculated.
   ::通过重复步骤1和步骤2,您可以不断减少电路,直到只有一个等效的阻力器留下为止。然后您可以确定总电路电流。下降和电流,尽管单个阻力器可以计算出来。

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Example  
::示例示例示例示例

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In the combination circuit sketched below, find the equivalent resistance for the circuit, find the total current through the circuit, and find the current through each individual resistor.
::在下面绘制的组合电路图中,找到电路的同等抗力,通过电路找到总电流,通过每个单独的阻力器找到电流。

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We start by simplifying the parallel resistors $\begin{align*}R_2\end{align*}$ and $\begin{align*}R_3\end{align*}$ .
::我们首先简化平行抵抗者R2和R3。

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$$\begin{align*}\frac{1}{R_{23}}&=\frac{1}{180 \ \Omega}+\frac{1}{220 \ \Omega}=\frac1{99\ \Omega }\\ R_{23}&=99 \ \Omega\end{align*}$$
::1R23=1180 1220 199 R23=99

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We then simplify $\begin{align*}R_1\end{align*}$ and $\begin{align*}R_{23}\end{align*}$ which are series resistors.
::然后我们简化R1和R23, 它们是系列抵抗者。

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$\begin{align*}R_T=R_1+R_{23}=110 \ \Omega + 99 \ \Omega = 209 \ \Omega\end{align*}$
::RT=R1+R23=110 $99 $209 $______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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We can then find the total current, $\begin{align*}I_T=\frac{V_T}{R_T}=\frac{24 \ V}{209 \ \Omega}=0.11 \ A\end{align*}$ .
::然后我们就可以找到总电流,IT=VTRT=24 V209+0.11 A。

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All the current must pass through $\begin{align*}R_1\end{align*}$ , so $\begin{align*}I_1=0.11 \ A\end{align*}$ .
::所有电流必须经过R1, 所以I1=0. 11 A.

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The voltage drop through $\begin{align*}R_1\end{align*}$ is $\begin{align*}(110 \ \Omega) (0.11 \ A)=12.6 \text{ volts}\end{align*}$ .
::通过R1的电压下降为(110 }(0.11 A)= 12.6伏。

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Therefore, the voltage drop through $\begin{align*}R_2\end{align*}$ and $\begin{align*}R_3\end{align*}$ is 11.4 volts.
::因此,通过R2和R3的电压下降为11.4伏。

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$\begin{align*}I_2=\frac{V_2}{R_2}=\frac{11.4 \ V}{180 \ \Omega}=0.063 \ A\end{align*}$ and $\begin{align*}I_3=\frac{V_3}{R_3}=\frac{11.4 \ V}{220 \ \Omega}=0.052 \ A\end{align*}$
::I2=V2R2=11.4 V180=0.063 A和I3=V3R3=11.4 V220=0.052 A

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Use the Marquee Lights simulation below to arrange  many identical light bulbs in different configurations. Try to add several bulbs in series and observe the circuit diagram to see what happens to the current, resistance, and brightness of the bulbs. Then, do the same with several bulbs in parallel and compare. Lastly, set the Configuration slider to Mixed and observe what happens in a combined series-parallel circuit:
::使用下面的 Marquee Lights 模拟来安排不同配置中许多相同的灯泡。 尝试在序列中添加多个灯泡, 并观察电路图以观察灯泡的当前、 阻力和亮度。 然后对多个灯泡进行平行和比较。 最后, 将配置滑动器设置为混合灯泡, 并观察在组合系列平行电路中发生的情况 :

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Further Reading
::继续阅读

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Summary
::摘要

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• Combined circuit problems should be solved in steps.
  ::联合电路问题应分阶段解决。

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Review
::回顾

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1. Two 60.0 Ω resistors are connected in parallel and this parallel arrangement is then connected in series with a 30.0 Ω resistor. The combination is placed across a 120. V potential difference.
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   1. Draw a diagram of the circuit.
      ::绘制电路图 。
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   2. What is the equivalent resistance of the parallel portion of the circuit?
      ::电路平行部分的抵抗力是多少?
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   3. What is the equivalent resistance for the entire circuit?
      ::整个电路的抵抗力是多少?
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   4. What is the total current in the circuit?
      ::电路的总电流是多少?
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   5. What is the voltage drop across the 30.0 Ω resistor?
      ::电压在30.0的电阻器上 下降多少?
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   6. What is the voltage drop across the parallel portion of the circuit?
      ::电路平行部分的电压下降是多少?
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   7. What is the current through each resistor?
      ::每个抵抗者身上的流水是什么?
   
   ::2 60.0 × 阻力器同时连接,而这一平行安排随后又与 30.0 × 阻力器连成序列。 组合将放置在120 V 的潜在差异上。 绘制一个电路图。 电路平行部分的对应抗力是多少? 整个电路的对应抗力是多少? 电路的总抗力是多少? 电压在 30.0 × 阻力器之间是多少? 电压在电路平行部分的下降是多少? 每个阻力器之间的电流是多少?
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2. Three 15.0 Ω resistors are connected in parallel and the combination is then connected in series with a 10.0 Ω resistor. The entire combination is then placed across a 45.0 V potential difference. Find the equivalent resistance for the entire circuit.
   ::3个15.0 × 阻力器平行连接,然后组合与一个10.0 × 阻力器连成一系列。然后将整个组合置于45.0 V 潜在差异之间。 找出整个电路的同等阻力。

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Explore More
::探索更多

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Use this resource to answer the questions that follow.
::使用此资源回答下面的问题 。

 

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1. In a circuit that contains both series and parallel parts, which parts of the circuit are simplified first?
   ::在含有系列和平行部件的电路中,电路的哪些部分首先简化?
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2. In the circuit drawn below, which resistors should be simplified first?
   ::在下面绘制的电路中,应首先简化哪些阻力器?