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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