1-22Figure 1-5.—Wheatstone bridge.The operation of the bridge is explained in a few logical steps. When the battery switch S1 is closed,electrons flow from the negative terminal of the battery to point a. Here the current divides as it would inany parallel circuit. Part of it passes through R1 and R2; the remainder passes through R3 and R_{x}. Thetwo currents, I_{1} and I_{2}, unite at point c and return to the positive terminal of the battery. The value of I_{1}depends on the sum of resistance R1 and R2, and the value of I_{2} depends on the sum of resistances R3 andR_{x}. In each case, according to Ohm’s law, the current is inversely proportional to the resistance.R1, R2, and R3 are adjusted so that when S1 is closed, no current flows through G. When thegalvanometer shows no deflection, there is no difference of potential between points b and d. All of I1follows the a b c path and all I_{2} follows the a b c path. This means that a voltage drop E_{1} (across R1between points a and b) is the same as voltage drop E_{3} (across R3 between points a and d). Similarly, thevoltage drops across R2 and R_{x} (E_{2} and E_{x}) are also equal. Expressed algebraically,andWith this information, we can figure the value of the unknown resistor R_{x}. Divide the voltage dropsacross R1 and R3 by their respective voltage drops across R2 and R_{x} as follows:We can simplify this equation:

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