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Kirchhoff reasoned that dc ought to behave something like water in a network of pipes, and that the current going into any point ought to be the same as the current going out of that point. This, Kirchhoff thought, must be true for any point in a circuit, no matter how many branches lead into or out of the point. Two examples of this principles are shown in Fig. 5-5. Examine illustration A. At point X, I, the current going in, equals I1 + I2, the current going out. At point Y, I2 + I1, the current going in, equals I, the current going out. Now look at illustration B. In this case, at point Z, the current I1 + I2 going in is equal to the current I3 + I4 + I5 going out. These are examples of Kirchhoff s First Law. We can also call it Kirchhoff s Current Law or the principle of conservation of current.

When resistances are wired in parallel, they each consume power according to the same formula, P I 2R. But the current is not the same in each resistance. An easier method to find the power Pn, dissipated by resistor Rn, is by using the formula Pn E 2/Rn where E is the voltage of the supply. Recall that this voltage is the same across every resistor in a parallel circuit.

4

Problem 5-9

5-5 Kirchhoff s First Law. At A, the current into point X or point Y is the same as the current out of that point. That is, I = I1 + I2.

A circuit contains three resistances R1 22 , R2 47 , and R3 age E 3.0 V. Find the power dissipated by each resistor. 68 across a volt-

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and aviation, for example, enhance safety through the use of these technologies, while chemistry and architecture use them to enhance the understanding of highly complex structures and interactions. In education, the virtual classroom delivers teachers and schoolrooms to far-flung locations. Today the cost of these technologies is somewhat prohibitive because they are new and are provided by a relatively small number of pioneer manufacturers. As applications and the market drive them toward commodity, the cost will come down, and digital video and virtual reality will become mainstream technologies. Ellen Beth Van Buskirk, former Director of Corporate Communications for the SEGA Channel, agrees. Video games have long been the bastion of teenage boys, she says. These new technologies will offer entire families the opportunity to interact together in virtual worlds. It s going to be a very exciting future.

At B, the current into point Z equals the current flowing out of point Z. That is, I1 + I2 = I3 + I4 + I5. Illustration for Quiz Questions 13 and 14.

Kirchhoff s first law 89 First find E 2, because you ll be needing that number often: E 2 3.0 3.0 9.0. Then P1 9.0/22 0.4091 W, P2 9.0/47 0.1915 W, P3 9.0/68 0.1324 W. These can be rounded off to P1 0.41 W, P2 0.19 W, and P3 0.13 W. But remember the values to four places for the next problem. In a parallel circuit, the total power consumed is equal to the sum of the wattages dissipated by the individual resistances. In this respect, the parallel circuit acts like the series circuit. Power cannot come from nowhere, nor can it vanish. It must all be accounted for.

Summary

Problem 5-11 Refer to Fig. 5-5A. Suppose all three resistors have values of 100 , and that I1 = 2.0 A and I2 = 1.0 A. What is the battery voltage First, find the current I drawn from the battery: I = I1 + I2 = 2.0 + 1.0 = 3.0 A. Next, find the resistance of the entire network. The two 100- resistances in series give a value of 200 , and this is in parallel with 100 . You can do the calculations and find that the total resistance, R, connected across the battery is 66.67 . Then E = IR = 66.67 3.0 = 200 V. Problem 5-12 In Fig. 5-5B, suppose each of the two resistors below point Z has a value of 100 , and all three resistors above point Z have values of 10.0 . Suppose the current through each 100- resistor is 500 mA. What is the current through any one of the 10.0- resistors, assuming that the current through all three 10.0- resistors is the same What is the voltage across any one of the three 10.0- resistors The total current into point Z is 500 mA + 500 mA = 1.00 A. This is divided equally among the three 10- resistors. Therefore, the current through any one of them is 1.00/3 A = 0.333 A. The voltage across any one of the 10.0- resistors can thus found by Ohm s Law: E = IR = 0.333 10.0 = 3.33 V.

Problem 5-10

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