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26-1DIRECT-CURRENT CIRCUITS26.1. IDENTIFY: The newly-formed wire is a combination of series and parallel resistors. SET UP:Each of the three linear segments has resistance R/3. The circle is two R/6 resistors in parallel. EXECUTE:The resistance of the circle is R/12 since it consists of two R/6 resistors in parallel.The equivalent resistance is two R/3 resistors in series with an R/6 resistor, giving Requiv= R/3 + R/3 + R/12 = 3R/4. EVALUATE:The equivalent resistance of the original wire has been reduced because the circles resistance is less than it was as a linear wire. 26.2. IDENTIFY:It may appear that the meter measures Xdirectly. But note that Xis in parallel with three other resistors, so the meter measures the equivalent parallel resistance between ab. SET UP:We use the formula for resistors in parallel. EXECUTE:1/(2.00 ) = 1/X+ 1/(15.0 ) + 1/(5.0 ) + 1/(10.0 ), so X= 7.5 . EVALUATE:Xis greaterthan the equivalent parallel resistance of 2.00 . 26.3. (a)IDENTIFY:Suppose we have two resistors in parallel, with 12RR<. SET UP:The equivalent resistance is eq12111RRR=+EXECUTE:It is always true that 121111RRR+>. Therefore eq111RR>and eq1RR<. EVALUATE:The equivalent resistance is always less than that of the smallest resistor. (b)IDENTIFY:Suppose we have Nresistors in parallel, with 12NRRR<<<". SET UP:The equivalent resistance is eq121111NRRRR=+++"EXECUTE:It is always true that 1211111NRRRR+++>". Therefore eq111RR>andeq1RR<. EVALUATE:The equivalent resistance is always less than that of the smallest resistor. 26.4. IDENTIFY:For resistors in parallel the voltages are the same and equal to the voltage across the equivalent resistance. SET UP:VIR=. eq12111RRR=+. EXECUTE:(a) 1eq1112.3 .32 20 R=+= (b) eq240 V19.5 A.12.3VIR===(c) 3220240 V240 V7.5 A;12 A.3220VVIIRR======EVALUATE:More current flows through the resistor that has the smaller R.26.5. IDENTIFY:The equivalent resistance will vary for the different connections because the series-parallel combinations vary, and hence the current will vary. SET UP:First calculate the equivalent resistance using the series-parallel formulas, then use Ohms law (V = RI) to find the current. EXECUTE:(a)1/R= 1/(15.0 ) + 1/(30.0 ) gives R= 10.0 . I= V/R = (35.0 V)/(10.0 ) = 3.50 A. (b)1/R= 1/(10.0 ) + 1/(35.0 ) gives R= 7.78 . I= (35.0 V)/(7.78 ) = 4.50 A (c)1/R= 1/(20.0 ) + 1/(25.0 ) gives R= 11.11 , so I= (35.0 V)/(11.11 ) = 3.15 A. 2626-2 Chapter 26 (d)From part (b), the resistance of the triangle alone is 7.78 . Adding the 3.00-internal resistance of the battery gives an equivalent resistance for the circuit of 10.78 . Therefore the current is I= (35.0 V)/(10.78 ) = 3.25 A EVALUATE:It makes a big difference how the triangle is connected to the battery. ... View Full Document

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