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  • 平行电路

    Close-up of a circuit board showcasing integrated circuits and various electronic components.

    Electrical are everywhere: skyscrapers, jumbo jets, arcade games, lights, heating, security... Very few complex things work without electrical circuits. Since the late 1970s, electrical circuits have primarily looked like this. The circuits are formed by a thin layer of conducting material deposited on the surface of an insulating board. Individual components are soldered to the interconnecting circuits. boards are vastly more complicated than the previously discussed, but operate on many similar principles.
    ::电路无处不在:摩天大楼、大型喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式喷气式

    Parallel Circuits
    ::平行电路

    Parallel circuits are circuits in which the charges leaving the potential source have different paths they can follow to get back to the source. In the sketch below, the current leaves the battery, passes through the orange switch, and then has three different paths available to complete the circuit. Each individual electron in this circuit passes through only one of the light bulbs. After the current passes through the switch, it divides into three pieces and each piece passes through one of the bulbs. The three pieces of current rejoin after the light bulbs and continue in the circuit to the potential source.
    ::平行电路是电路, 离开潜在电源的电路有不同的路由, 以便返回源。 在下面的草图中, 电流留下电池, 穿过橙色开关, 然后有三种不同的路由来完成电路 。 电路中的每个电子通过只有一个灯泡。 在电路通过开关后, 将电路分割成三块, 每块通过一个灯泡。 电路中的三个电路在灯泡之后重新连接, 并在电路中继续连接到潜在电源 。

    Diagram of a parallel circuit with three light bulbs and a battery, illustrating current paths.

    In the design of this parallel circuit , each resistor (light bulb) is connected across the battery as if the other two resistors were not present. Remember that the current going through each resistor goes through only the one resistor. Therefore, the drop across each resistor must be equal to the total voltage drop though the circuit.
    ::在此平行电路的设计中, 每个阻力器( 灯泡) 都像另外两个阻力器不在场一样在电池上连接。 记住, 每个阻力器穿透的电流只穿过一个阻力器。 因此, 每个阻力器的下降必须等于电路通电压下降的总电压。

    @$\begin{align*}V_T = V_1 = V_2 = V_3\end{align*}@$
    ::@$\ begin{ align@V_ T = V_ 1 = V_ 2 = V_ 3\ end{ end{ align} $

    The total current passing through the circuit will be the sum of the individual currents passing through each resistor.
    ::通过电路的电流总流速将是通过每个阻力器的单个电流的总和。

    @$\begin{align*}I_T = I_1 + I_2 + I_3\end{align*}@$
    ::@ $\ begin{ aliign@ I_ 1 + I_ 2 + I_ 3\ end{ aliign} $

    If we return to the analogy of a river, a parallel circuit is the same as the river breaking into three streams, which later rejoin to one river again. The amount of water flowing in the river is equal to the sum of the amounts of water flowing in the individual streams.
    ::如果我们回到河流的类比,平行电路与河流破入三条溪流相同,三条溪流后来又重新回到一条河。 河中的水流量相当于各河流水量的总和。

    Ohm's Law applies to resistors in parallel, just as it did to resistors in a series. The current flowing through each resistor is equal to the total voltage drop divided by the in that resistor.
    ::《 Ohm法》平行地适用于抵抗者,正如它同时适用于抵抗者一样,它同样适用于一系列抵抗者。流经每个抵抗者的电流相当于总电压下降,由抵抗者除以。

    @$$\begin{align*}I_1=\frac{V_T}{R_1} \quad \text{and} \quad I_2=\frac{V_T}{R_2} \quad \text{and} \quad I_3=\frac{V_T}{R_3}\end{align*}@$$
    ::@ $\ begin{ align} I_ 1@ 1 @ frac{ V_ TR_ 1} \ quad\ text{ 和}\ quad I_ 2\ frac{ V_ TR_ 2} \ quad\ text{ 和} \ quad I_ 3\ frac{ V_ TR_ 3} end{ ALign${ $ </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-MjQyNmU1MDdkZjc0MTNkY2VlN2UwZGE3NjdiMGU1Y2U.-n4t"> Since <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="I_T%3DI_1%2BI_2%2BI_3"> @$\begin{align*}I_T=I_1+I_2+I_3\end{align*}@$ ,
    ::@$\ begin{ align} I_ I_ T=I_ 1+I_ 2+I_ 3\end{ end{ leign} $, @ $\ begin{ aliign} I_ T=I_ 1+I_ 2+I_ 2+I_ 3\end{ end{ leign} $,

    then @$\begin{align*}I_T=\frac{V_T}{R_1} + \frac{V_T}{R_2} + \frac{V_T}{R_3}\end{align*}@$ ,
    ::+\frac{V_TR_1}+\frac{V_TR_2}+\frac{V_TR_3{end{align}$,

    and @$\begin{align*}\frac{V_T}{R_T}=\frac{V_T}{R_1} + \frac{V_T}{R_2} + \frac{V_T}{R_3}\end{align*}@$ .
    ::@ $\ begin{ legin{ legin} {V_ TR_ Tfrac{ V_ TR_ 1} +\ frac{ V_ TR_ 2} +\ frac{ V_ TR_ 3{ end{ align} $。

    If we divide both sides of the final equation by @$\begin{align*}V_T\end{align*}@$ , we get the relationship between the total resistance of the circuit and the individual parallel resistances in the circuit. The total resistance is sometimes called the equivalent resistance .
    ::如果我们将最后方程的两边除以 @$\ begin{ align} @ V_ T\end{ align} $, 我们就会得到电路完全抗力与电路中个体平行抗力之间的关系。 总抗力有时被称为等效抗力 。

    @$$\begin{align*}\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\end{align*}@$$
    ::@ $\ begin{ leign@ frac{ 1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\$\$$$$$$$$$$$$\\\\\\\\\\\\\\$$$$$$$$$\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\$$$$$$\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\$$$$$\\\\\\\\\\\\\\\\\\\$$$$$$$$$$$$$\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-MTBhY2ExMzczMzJjYjQzNjI2N2RkZGNhYzExYjczYjg.-2nc"> Consider the parallel circuit sketched below. <br/> <span style="color: green; "> ::考虑下面的平行电路图 </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-ZWU1NDQ0MDllOTBlNWExYzk3NmMwMDUwMTVjZjYyMmE.-pra"> The voltage drop for the entire circuit is 90. V. Therefore, the voltage drop in each of the resistors is also 90. V. <br/> <span style="color: green; "> ::整个电路的电压下降为90伏。 因此,每个阻力器的电压下降也是90伏。 </span> </p> <p id="x-ck12-MTQ2NmIzNTQ2MWJlZmRhNTEzNTE1NzNiYTQ1NjIyYjY.-zko"> <span class="x-ck12-img-inline"> <!-- @@author="Samantha Bacic" --> <!-- @@url="CK-12 Foundation" --> <img alt="Schematic of a parallel circuit illustrating multiple resistors connected to a single voltage source." data-flx-url="/flx/show/THUMB_POSTCARD/image/user%3AY2sxMnNjaWVuY2VAY2sxMi5vcmc./98045-1370320248-83-6-IntPhysC-ParallelCircuit_NEW.png" data-imageresourceid="7234792" data-onerror="null" loading="lazy" onerror="null" src="https://dr282zn36sxxg.cloudfront.net/datastreams/f-d%3A29e91b0f4436361bc95d9d4e57c9c9533915677412f808a1653b2264%2BIMAGE_THUMB_POSTCARD_TINY%2BIMAGE_THUMB_POSTCARD_TINY.1" style="" width="250"/> </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-NWNkYjJlOWE2ZWE5MDM3MjlmNmI3NDczNzc1NWI5MGQ.-ype"> The current through each resistor can be found using the voltage drop and the resistance of that resistor: <br/> <span style="color: green; "> ::通过电压下降和抵抗力的抵抗力 可以发现每个抵抗器的电流: </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-rgt"> <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-block-math" data-mathmethod="block" data-tex="I_1%3D%5Cfrac%7BV_T%7D%7BR_1%7D%3D%5Cfrac%7B90.%20%5C%20V%7D%7B60.%20%5C%20%5COmega%7D%3D1.5%20%5C%20A%20%5Cqquad%20I_2%3D%5Cfrac%7BV_T%7D%7BR_2%7D%3D%5Cfrac%7B90.%20%5C%20V%7D%7B30.%20%5C%20%5COmega%7D%3D3.0%20%5C%20A%20%5Cqquad%20I_3%3D%5Cfrac%7BV_T%7D%7BR_3%7D%3D%5Cfrac%7B90.%20%5C%20V%7D%7B30.%20%5C%20%5COmega%7D%20%3D%203.0%20%5C%20A"> @$$\begin{align*}I_1=\frac{V_T}{R_1}=\frac{90. \ V}{60. \ \Omega}=1.5 \ A \qquad I_2=\frac{V_T}{R_2}=\frac{90. \ V}{30. \ \Omega}=3.0 \ A \qquad I_3=\frac{V_T}{R_3}=\frac{90. \ V}{30. \ \Omega} = 3.0 \ A\end{align*}@$$
    ::@ $\ begin{ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ 等/ </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-NWYwZjlmMjQxZjA1MDZiYjQ3YjQ1ZmNjMmMwMzhlZTU.-oep"> The total current through the circuit would be the sum of the three currents in the individual resistors. <br/> <span style="color: green; "> ::通过电路的电流总量将是各抗体中三股电流的总和。 </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-hwj"> <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="I_T%3DI_1%2BI_2%3DI_3%3D1.5%20%5C%20A%20%2B%203.0%20%5C%20A%20%2B%203.0%20%5C%20A%20%3D%207.5%20%5C%20A"> @$\begin{align*}I_T=I_1+I_2=I_3=1.5 \ A + 3.0 \ A + 3.0 \ A = 7.5 \ A\end{align*}@$
    ::@$\ begin{ legin@I_ 1+I_ 2=I_ 3=1.5 \ A + 3. 0\ A + 3. 0\ A = 7. 5\ A\ end{ end{ align$}

    The equivalent resistance for this circuit is found using the equation above.
    ::此电路的等效抗力使用以上方程式找到。

    @$$\begin{align*}\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{60. \ \Omega} + \frac{1}{30. \ \Omega} + \frac{1}{30. \ \Omega} = \frac{1}{60. \ \Omega} + \frac{2}{60. \ \Omega} + \frac{2}{60. \ \Omega} = \frac{5}{60. \ \Omega}\end{align*}@$$
    ::@ $\ begin{ legn{ legn{ 1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-wuf"> <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="R_T%3D%5Cfrac%7B60.%20%5C%20%5COmega%7D%7B5%7D%3D12%20%5C%20%5COmega"> @$\begin{align*}R_T=\frac{60. \ \Omega}{5}=12 \ \Omega\end{align*}@$
    ::@ $\ begin{ legign}R_ Tfrac{ 60.\\\ \ \ \ \ \ \ \\ \ \\ \\\\\ \\\\\ \\ \\\ \\\ \\ \\\ end{ end{ \ legn\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \\ \ \ \ \\\\\\\ \\ \\ \\ \\\\\\\\\\\\\ \ \\\ \\\\\\\ \\\\\\\\\ \ \\\\\\\\\\\ \\\\\\\\\\\ \\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-ZDIyZjE2ZDY5OGVkNWU0NzFmZWYxNGQwNjdhNmZiYjI.-htd"> The equivalent resistance for the circuit could also be found by using the total voltage drop and the total current. <br/> <span style="color: green; "> ::使用电压总下降和电流总和,也可以发现电路的等效抗力。 </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-lji"> <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="R_T%3D%5Cfrac%7BV_T%7D%7BI_T%7D%3D%5Cfrac%7B90.%20%5C%20%5COmega%7D%7B7.5%20%5C%20A%7D%3D12%20%5C%20%5COmega"> @$\begin{align*}R_T=\frac{V_T}{I_T}=\frac{90. \ \Omega}{7.5 \ A}=12 \ \Omega\end{align*}@$
    ::@$\ begin{ legign}R_ Tfrac{ V_ TI_ Tfrac{ 90.\\\\ \ \ \ \ \ \\ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \\ \\ \\ \\\ end{ \ \ \ leg \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ </span> </p> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-ZDZkN2U5MmNjOWFmODk0NjhjNGMwZTlmNTU5MzQ5OWU.-7bh"> <span style="font-weight: 400;"> Use the PLIX below to observe the relationship between the individual voltage drops and the total voltage in a parallel circuit: </span> <br/> <span style="color: green; "> ::使用下面的PLIX观察单电压下降与平行电路总电压之间的关系: </span> </p> <div class="x-ck12-video-object plix-fs" id="x-ck12-0vq"> <div class="iframe-wrapper ck12-no-annotation"> <iframe allowfullscreen="allowfullscreen" data-artifactlang="en" data-embed="PGlmcmFtZSBzcmM9Imh0dHBzOi8vd3d3LmNrMTIub3JnL2Fzc2Vzc21lbnQvdG9vbHMvZ2VvbWV0cnktdG9vbC9mdWxsc2NyZWVuLmh0bWw/cUlEPTU3M2I2MTE3NWFhNDEzNTEzMWUyOTIxYSZhaWQ9MjYwMTI5NyIgc2Nyb2xsaW5nPSJubyIgZnJhbWVib3JkZXI9IjAiIHdpZHRoPSIzNTBweCIgaGVpZ2h0PSIzNTBweCIgYWxsb3dmdWxsc2NyZWVuLz4=" frameborder="0" height="370" id="x-ck12-MTU0MjY5NjUzNjI0Mg.." name="238730" src="https://flexbooks.ck12.org/flx/show/interactive/https%3A//www.ck12.org/assessment/tools/geometry-tool/fullscreen.html%3FqID%3D573b61175aa4135131e2921a%26aid%3D2601297%26hash%3D7d3fe4e08d5cfde986cdd7eb0da4a449?lang=en" style="max-width: 100%;" title="video" width="370"> </iframe> </div> </div> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <p id="x-ck12-OTljOWQ4ZGJlODIxM2FjN2Y5OTI4NmQ5MzEzMzYwMWE.-5d8"> <span style="font-weight: 400;"> Use the PLIX Interactive below to observe the differences between parallel and series circuits. Adjust the number of resistors in the circuit and observe the difference in total resistance between a parallel and <span class="x-ck12-vocab-interlink" data-definition="a%20closed%20circuit%20in%20which%20all%20of%20the%20current%20must%20pass%20through%20every%20resistor%20in%20the%20circuit." data-id="5259" data-interlink-id="x-ck12-w8a3g9xzdvucz7c6" data-json="eyJkYXRhLXBsdXJhbCI6ICIiLCAiZGF0YS1kZWZpbml0aW9uIjogImElMjBjbG9zZWQlMjBjaXJjdWl0JTIwaW4lMjB3aGljaCUyMGFsbCUyMG9mJTIwdGhlJTIwY3VycmVudCUyMG11c3QlMjBwYXNzJTIwdGhyb3VnaCUyMGV2ZXJ5JTIwcmVzaXN0b3IlMjBpbiUyMHRoZSUyMGNpcmN1aXQuIiwgImRhdGEtaWQiOiA1MjU5LCAiZGF0YS1sYW5ndWFnZUlEIjogMSwgImRhdGEtdGVybSI6ICJzZXJpZXMgY2lyY3VpdCJ9" data-languageid="1" data-plural="" data-term="series circuit" role="term" tabindex="0"> series circuit </span> . Then, try to explain the different levels of bulb brightness observed in each circuit configuration: </span> <br/> <span style="color: green; "> ::使用下面的 PLIX 互动来观察平行电路和序列电路之间的差异。 调整电路中的阻力器数量, 并观察平行电路和序列电路之间的总耐力差异。 然后, 尝试解释每条电路配置中观察到的不同灯泡亮度 : </span> </p> <div class="x-ck12-video-object plix-fs" id="x-ck12-l2b"> <div class="iframe-wrapper ck12-no-annotation"> <iframe allowfullscreen="allowfullscreen" data-artifactlang="en" data-embed="PGlmcmFtZSBzcmM9Imh0dHBzOi8vd3d3LmNrMTIub3JnL2Fzc2Vzc21lbnQvdG9vbHMvZ2VvbWV0cnktdG9vbC9mdWxsc2NyZWVuLmh0bWw/cUlEPTU3NTA5N2NlZGEyY2ZlMjJkNTE0MDZjMCZhaWQ9MjYwMDg4NCIgc2Nyb2xsaW5nPSJubyIgZnJhbWVib3JkZXI9IjAiIHdpZHRoPSIzNTBweCIgaGVpZ2h0PSIzNTBweCIgYWxsb3dmdWxsc2NyZWVuLz4=" frameborder="0" height="370" id="x-ck12-MTU1NTQ0NjM5MDQ2Mg.." name="238729" src="https://flexbooks.ck12.org/flx/show/interactive/https%3A//www.ck12.org/assessment/tools/geometry-tool/fullscreen.html%3FqID%3D575097ceda2cfe22d51406c0%26aid%3D2600884%26hash%3D5c49d11eec6cf3ec72a8808a9ab6d8f4?lang=en" style="max-width: 100%;" title="video" width="370"> </iframe> </div> </div> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <h3> Further Reading <br/> <span style="color: green; "> ::继续阅读 </span> </h3> <ul id="x-ck12-MDY2MzJmNmIyZTYxMDBmMDQ3MmMyODMzNzczYjVkYmY.-lw1"> <li> </li> </ul> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <h3> Summary <br/> <span style="color: green; "> ::摘要 </span> </h3> <ul id="x-ck12-MjQ0Y2RjMDViOTNmOWQxNDk1OTkzZmFiYmI4MmVhYTk.-xpa"> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <li> Parallel electrical circuits have multiple paths the current may take. <br/> <span style="color: green; "> ::平行电路有多个路口 电流可能走多条路 </span> </li> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <li> <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="V_T%20%3D%20V_1%20%3D%20V_2%20%3D%20V_3"> @$\begin{align*}V_T = V_1 = V_2 = V_3\end{align*}@$ .
    ::@$\ begin{ align@V_ T = V_ 1 = V_ 2 = V_ 3\ end{ align_ $ 。 </span> </li> <button class="play-button btn btn-success" style="float: right;" value="@s"> 播放段落 </button> <li> <span class="x-ck12-mathEditor" data-contenteditable="false" data-edithtml="" data-math-class="x-ck12-math" data-mathmethod="inline" data-tex="I_T%20%3D%20I_1%20%2B%20I_2%20%2B%20I_3"> @$\begin{align*}I_T = I_1 + I_2 + I_3\end{align*}@$ .
    ::@$\ begin{ align}I_ T = I_ 1 + I_ 2 + I_ 3\ end{ aliign}$.

  • @$\begin{align*}\frac{1}{R_T}=\frac{1}{R_1}=\frac{1}{R_2}+\frac{1}{R_3}\end{align*}@$ .
    ::@$\ begin{ align}\\ frac{ 1\\\\\\\\\R_ T\\frac{ 1\R_ 1\\\\\\\frac{ 1\R_ 2\\\\\\frac{ 1\R_3} end{ leign} $。

Review
::回顾

  1. Three 15.0 Ω resistors are connected in parallel and placed across a 30.0 V potential difference.
    1. What is the equivalent resistance of the parallel circuit?
      ::平行电路的抵抗力是多少?
    2. What is the total current through the circuit?
      ::电路的电流总量是多少?
    3. What is the current through a single branch of the circuit?
      ::电路的一个分支的电流是什么?

    ::3个15.0 × 阻力器平行连接, 并放置在30.0 V 的潜在差异上。 平行电路的同等抗力是多少? 电路的电流总量是多少? 电路的一个分支的电流是多少?
  2. A 12.0 Ω and a 15.0 Ω resistor are connected in parallel and placed across a 30.0 V potential.
    1. What is the equivalent resistance of the parallel circuit?
      ::平行电路的抵抗力是多少?
    2. What is the total current through the circuit?
      ::电路的电流总量是多少?
    3. What is the current through each branch of the circuit?
      ::电路每个分支的电流是什么?

    ::A 12.0 和 a 15.0 阻力器平行连接, 并放置在 30.0 V 潜能值上。 平行电路的同等阻力是什么? 电路的电流总量是多少? 电路的每个分支的电流是多少?
  3. A 120.0 Ω resistor, a 60.0 Ω resistor, and a 40.0 Ω resistor are connected in parallel and placed across a potential difference of 12.0 V.
    1. What is the equivalent resistance of the parallel circuit?
      ::平行电路的抵抗力是多少?
    2. What is the total current through the circuit?
      ::电路的电流总量是多少?
    3. What is the current through each branch of the circuit?
      ::电路每个分支的电流是什么?

    ::A 120.0 抗体, a 60.0 抗体, 和 a 40. 0 抗体平行连接, 并放置在12. 0 V 的潜在差异上。 平行电路的抗力是多少? 电路的电流总量是多少? 电路的每个分支的电流是多少?

Explore More
::探索更多

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

  

  1. Why do the light bulbs glow less brightly when connected across a 120 V source in a series circuit than when connected across the same 120 V source in a parallel circuit?
    ::为什么光灯灯泡在一系列电路中连接120V源,而在同一120V源的平行电路中连接时,光灯灯灯泡的亮度却不那么高呢?
  2. Why do the other bulbs go dark when one bulb is removed in the series circuit but the other bulbs do not go dark when one bulb is removed in the parallel circuit?
    ::为什么其他灯泡在序列电路中被除掉一个灯泡,而其他灯泡在平行电路中被除掉一个灯泡时却不变暗?