# barcode printing using vb.net Direct-current circuit analysis in Software Maker QR Code 2d barcode in Software Direct-current circuit analysis

88 Direct-current circuit analysis
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Power distribution in series circuits
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Let s switch back now to series circuits. This is a good exercise: getting used to thinking in different ways and to changing over quickly and often. When calculating the power in a circuit containing resistors in series, all you need to do is find out the current, I, that the circuit is carrying. Then it s easy to calculate the power Pn, based on the formula Pn I 2Rn.
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Problem 5-7
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Suppose we have a series circuit with a supply of 150 V and three resistors: R1 330 , R2 680 , and R3 910 . What is the power dissipated by R2 You must find the current in the circuit. To do this, calculate the total resistance first. Because the resistors are in series, the total is R 330 680 910 1920 . Then the current is I 150/1920 0. 07813 A 78.1 mA. The power in R2 is P2 I 2R2 0.07813 0.07813 680 4.151 W. Round this off to two significant digits, because that s all we have in the data, to obtain 4.2 W. The total power dissipated in a series circuit is equal to the sum of the wattages dissipated in each resistor. In this way, the distribution of power in a series circuit is like the distribution of the voltage.
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Problem 5-8
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Calculate the total power in the circuit of Problem 5-7 by two different methods. The first method is to figure out the power dissipated by each of the three resistors separately, and then add the figures up. The power P2 is already known. Let s bring it back to the four significant digits while we calculate: P2 4.151 W. Recall that the current in the circuit is I 0.07813 A. Then P1 0.07813 0. 07813 330 2.014 W, and P3 0.07813 0.07813 910 5.555 W. Adding these gives P 2.014 4.151 5.555 11.720 W. Round this off to 12 W. The second method is to find the series resistance of all three resistors. This is R 1920 , as found in Problem 5-7. Then P I 2R 0.07813 0.07813 1920 11.72 W, again rounded to 12 W. You might recognize this as an electrical analog of the distributive law you learned in junior-high-school algebra.
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Power distribution in parallel circuits
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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 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.
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Problem 5-9
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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-
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.
Problem 5-10
Find the total consumed power of the resistor circuit in Problem 5-9 using two different methods. The first method involves adding P1, P2, and P3. Let s use the four-significant-digit values for error reduction insurance. The sum is P 0.4091 0.1915 0.13240 7330 W. This can be rounded to 0.73 W or 730 mW. The second method involves finding the resistance R of the parallel combination. You can do this calculation yourself, keeping track for four digits for insurance reasons, getting R 12.28 . Then P E 2/R 9.0/12.28 0.7329 W. This can be rounded to 0. 73 W or 730 mW. In pure mathematics and logic, the results are all deduced from a few simple, intuitively appealing principles called axioms. You might already know some of these, such as Euclid s geometry postulates. In electricity and electronics, complex circuit analysis can be made easier if you are acquainted with certain axioms, or laws. You ve already seen some of these in this chapter. They are: The current in a series circuit is the same at every point along the way. The voltage across any component in a parallel circuit is the same as the voltage across any other, or across the whole set. The voltages across elements in a series circuit always add up to the supply voltage. The currents through elements in a parallel circuit always add up to the total current drawn from the supply. The total power consumed in a series or parallel circuit is always equal to the sum of the wattages dissipated in each of the elements. Now you will learn two of the most famous laws in electricity and electronics. These make it possible to analyze extremely complicated series-parallel networks. That s not what you ll be doing in this course, but given the previous axioms and Kirchhoff s Laws that follow, you could if you had to.
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