Power distribution in series circuits

Chapter Power distribution in series circuits

Teach Yourself Electricity and Electronics Third Edition Book
Pages 748
Views 14,558
Downloads : 46 times
PDF Size : 4.4 MiB

Summary of Contents

Teach Yourself Electricity and Electronics Third Edition Book

  • Power distribution in series circuitsLet’s switch back now to series circuits. This is a good exercise: getting used to thinkingin different ways and to changing over quickly and often.When calculating the power in a circuit containing resistors in series, all you needto do is find out the current, I, that the circuit is carrying. Then it’s easy to calculate thepower Pn, based on the formula PnI 2Rn.Problem 5-7Suppose we have a series circuit with a supply of 150 V and three resistors: R1330Ω, R2680Ω, and R3910Ω. What is the power dissipated by R2?You must find the current in the circuit. To do this, calculate the total resistancefirst. Because the resistors are in series, the total is R3306809101920Ω.Then the current is I150/19200. 07813 A 78.1 mA. The power in R2 is P2I 2R20.078130.078136804.151 W. Round this off to two significant digits, becausethat’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 dis-sipated in each resistor. In this way, the distribution of power in a series circuit is likethe distribution of the voltage.Problem 5-8Calculate 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 resistorsseparately, and then add the figures up. The power P2 is already known. Let’s bring itback to the four significant digits while we calculate: P24.151 W. Recall that the cur-rent in the circuit is I0.07813 A. Then P10.078130. 07813 3302.014 W,and P30.078130.078139105.555 W. Adding these gives P2.0144.1515.55511.720 W. Round this off to 12 W.The second method is to find the series resistance of all three resistors. This is R1920Ω, as found in Problem 5-7. Then PI2R0.078130.07813192011.72W, again rounded to 12 W.You might recognize this as an electrical analog of the distributive law you learnedin junior-high-school algebra.Power distribution in parallel circuitsWhen resistances are wired in parallel, they each consume power according to the sameformula, PI2R. But the current is not the same in each resistance. An easier methodto find the power Pn, dissipated by resistor Rn, is by using the formula PnE2/Rnwhere E is the voltage of the supply. Recall that this voltage is the same across every re-sistor in a parallel circuit.Problem 5-9A circuit contains three resistances R122Ω, R247Ω, and R368Ω across a volt-age E3.0 V. Find the power dissipated by each resistor.88 Direct-current circuit analysis