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In a network of water pipes that does not leak, and into which no water is added along the way, the total number of cubic feet going in has to be the same as the total volume going out. Water can t form from nothing, nor can it disappear, inside a closed system of pipes. Charge carriers, thought Kirchhoff, must act the same way in an electric circuit. This is Kirchhoff s First Law. An alternative name might be the law of conservation of current.

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That way, when you beg the computer to find the file after you ve done something stupid (this is where many people kill the chicken and light candles), the utility has a trivial task: simply restore the pointers As long as there hasn t been too.

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Resistances in series produce ratios of voltages, and these ratios can be tailored to meet certain needs by means of voltage divider networks When a voltage divider network is designed and assembled, the resistance values should be as small as possible without causing too much current drain on the battery or power supply (In practice, the optimum values depend on the nature of the circuit being designed This is a matter for engineers, and specific details are beyond the scope of this course) The reason for choosing the smallest possible resistances is that, when the divider is used with a circuit, you do not want that circuit to upset the operation of the divider The voltage divider fixes the intermediate voltages best when the resistance values are as small as the current-delivering capability of the power supply will allow Figure 5-7 illustrates the principle of voltage division.

Problem 5-11

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The individual resistances are R1, R2, R3, , Rn The total resistance is R = R1 + R2 + R3 + + Rn The supply voltage is E, and the current in the circuit is therefore I = E/R At the various points P1, P2, P3, , Pn, the potential differences relative to the negative battery terminal are E1, E2, E3, , En, respectively The last voltage, En, is the same as the battery voltage, E All the other voltages are less than E, and ascend in succession, so that E1 < E2 < E3 < < En (The mathematical symbol < means is less than ) The voltages at the various points increase according to the sum total of the resistances up to each point, in proportion to the total resistance, multiplied by the supply voltage Thus, the voltage E1 is equal to ER1/R.

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

Refer to Fig. 5-5A. Suppose all three resistors have values of 100 , and that I1 2.0 A while I2 1.0 A. What is the battery voltage First, find the current I drawn from the battery. It must be 3.0 A; I I1 I2 2.0 1.0 3.0 A. Next, find the resistance of the whole combination. The two 100- resistors in series give a value of 200 , and this is in parallel with 100 . You can do the

The voltage E2 is equal to E(R1 + R2)/R The voltage E3 is equal to E(R1 + R2 + R3)/R This process goes on for each of the voltages at points all the way up to En = E(R1 + R2 + R3 + + Rn)/R = ER/R = E..

5-5 Kirchhoffs First Law. At A, the current into either X or Y is the same as the current out of that point: I = Il + I2. At B, the current into Z equals the current out of Z: Il + I2 = I3 + I4+ I5. Also see quiz questions 13 and 14.

Kirchhoff s second law 91 calculations and find that the total resistance, R, across the battery, E, is 66.67 . Then E IR 66.67 3.0 200 volts. (Some battery.)

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