1. 电气工程类专业英语翻译问题2
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The battery polarities, of course, remain as they were according to their symbology (short end negative, long and positive). 当然，电池的极性根据它们的符号（短端为负，长端为正）而保持其原样。It is okay if the polarity of a resistor’s voltage polarity is correctly based on the assumed direction of current through it.如果一个电阻器的电压的极性是正确地依据通过电阻器的假设的电流方向，那就行了。 In some cases we may discover that current will be forced backwards through a battery, causing this very effect. 在有些情况下，我们可以发现，电流将被迫通过电池返回，从而导致这一效应。The important thing to remember here is to base all your resistor polarities and subsequent calculations on the directions of current(s) initially assumed. 重要的是要记住，这里是把你所有的电池极性和其后的计算建立在一开始假设的电流方向的基础上的。As stated earlier, if your assumption happens to be incorrect, it will be apparent once the equations have been solved (by means of a negative solution). 正如前面说明的那样，如果你的假设偶然是错误的，那一旦这些方程式被求解（用一个负解），问题就会很明显。The magnitude of the solution, however, will still be correct.不过，解的大小（值）将仍是正确的
Kirchhoff’s Voltage Law (KVL) tells us that the algebraic sum of all voltages in loop must be equal zero, so we can create more equations with current terms (I1, I2, and I3) for our simultaneous equations. 基尔霍夫的电压定律（KVL）告诉我们，在环路中的所有电压的代数和必定等于零，因此我们可以为四个同时的方程式列出更多带有电流项（I1, I2, 和 I3）的方程式。（To obtain a KVL equation, we must tally voltage drops in a loop of the circuit, as though we were measuring with a real voltmeter. 要获得一个KVL方程式，我们必须计算该回路中某一环路中的电压降，好像我们是在用一台真正的电压表进行测量一样。I’ll choose to trace the left loop of this circuit first, starting from the upper-left corner and moving counterclockwise (the choice of starting points and directions is arbitrary). 我将首先选择跟踪此回路左面的环路，从左上角开始，并逆时针移动（起点和方向的选择是任意的）。
Notice how current is being pushed backwards through battery w (electrons flowing “up”) e to the higher voltage of battery 1 (whose current is pointed “down” as it normally would)! 注意，电流是如何因电池1较高的电压（其电流像通常那样指向“下”）被推回电池W的（电子流向“上”）！Despite the fact that battery E2’s polarity is trying to push electrons down in that branch of the circuit, electrons are being forced backwards through it e to the superior voltage of battery E1.尽管电池E2的极性试图在该回路的该分支中将电子向下推，但由于电池1的优势电压，电子还是被迫通过E2电池而向后移动。 Dose this mean that the stronger battery will always “win” and the weaker battery always get current forced through it backwards? 这是否意味着较强的电池将始终“赢”，而较弱的电池始终让电流被迫向回流过自己呢？No! It actually depends on both the batteries’ relative voltage and the resistor values in the circuit.不！它实际上取决于回路中两个电池的相对电压和电阻器的值。 The only sure way to determine what’s going on is to take the time to mathematically analyze the network.决定如何继续进行的唯一确定方法是花时间对网络进行数学分析。
Now that we know the magnitude of all currents in this circuit, we can calculate voltage drops across all resistor with Ohm’s Law (E=I*R).既然我们知道在此回路中的所有电流的大小，我们就可借助欧姆定律（E=I*R)来计算跨所有电阻器的电压降了。