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In the circuit of Fig. 5-4, what does the ammeter A say This involves finding the current in any given branch. The voltage is 12 V across every branch; Rn 120. Therefore In, the ammeter reading, is found by Ohm s Law: In E/Rn 12/120 0.10 A 100 mA. All of the In s should add to get the total current, I .There are 12 identical branches, each carrying 0.1 A; therefore the sum is 0.1 12 1.2 A. It checks out. And there aren t any problems this time with significant figures. Those two problems were designed to be easy. Here are two that are a little more involved.

Problem 5-8 Calculate the total dissipated power P in the circuit of Problem 5-7 by two different methods First, let s figure out the power dissipated by each of the three resistances separately, and then add the figures up The power P2 is already known Let s use all the significant digits we have while we calculate Thus, as found in Problem 5-7, P2 = 4151 W Recall that the current is I = 007813 A Then P1 = 007813 007813 330 = 2014 W, and P3 = 007813 007813 910 = 5555 W Adding the three power figures gives us P = P1 + P2 + P3 = 2014 + 4151 + 5555 = 11720 W We should round this off to 117 W.

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So, where is all this image-based technology going In Snow Crash, author Neal Stephenson describes what is perhaps the ultimate marriage of virtual reality and the Information Superhighway. In his world, users log on to the network through a low-power wireless terminal that sports a lightweight head-mounted display about the size of a pair of glasses. Instead of using a video display terminal (VDT) or LCD displays, the HMD uses low-power, multicolored lasers to paint threedimensional images on the retinas of both eyes. The lasers have tracking circuitry that enables them to follow the minute movements of the user s eyes, thus guaranteeing a perfect set of images from any position.

Problem 5-5

The second method is to find the total series resistance and then calculate the power The series resistance is R = 1920 , as found in Problem 5-7 Then P = I 2R = 007813 007813 1920 = 1172 W Again, we should round this to 117 W..

Three resistors are in parallel across a battery that supplies E 12 V. The resistances are R1 22 , R2 47 , and R3 68 . These resistors carry currents I1, I2, and I3 respectively. What is the current, I3, through R3 This is done by means of Ohm s Law, as if R3 were the only resistor in the circuit. There s no need to worry about the parallel combination. The other branches do not affect I3. Thus I3 E/R3 12/68 0.18 A 180 mA. That problem wasn t hard at all. But it would have seemed that way, had you needlessly calculated the total parallel resistance of R1, R2, and R3.

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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 each of the various resistances Rn is to use the formula Pn = E 2/Rn, where E is the voltage of the supply or battery. This voltage is the same across every branch resistance in a parallel circuit.

Problem 5-6

Once logged on (or jacked in, to use the appropriate terminology), the user finds him- or herself in a virtual world The world, known as a metaverse, consists of a main street with shops, bars, and businesses the databases of the more routine Web that the user can stop in and visit Furthermore, rather than talking to other users via the crude medium of typed text, users actually see other users in the virtual world These manifestations of the real users are called avatars, a word that means the physical manifestation of a god Obviously, we re quite a ways from this level of capability, but not all that far Still-image applications are firmly entrenched and offer a broad variety of capabilities that have in many ways revolutionized banking, medicine, insurance, and architecture, to name a few.

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