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 circle’s 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 Ω) = 1/X+ 1/(15.0 Ω) + 1/(5.0 Ω) + 1/(10.0 Ω), so X= 7.5 Ω. EVALUATE:Xis greaterthan the equivalent parallel resistance of 2.00 Ω. 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 121R+>. 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 NR<<<". SET UP:The equivalent resistance is eq121NRR=+++"EXECUTE:It is always true that 11 1NRR+++ >". Therefore eq1RR>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=. eq12R. EXECUTE:(a) 1eq12.3 .32 20 R−⎛⎞=+=Ω⎜⎟ΩΩ⎝⎠(b) eq240 V19.5 A.12.3VIR===Ω(c) 3220240 V240 V7.5 A;12 A.3220VVII==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 Ohm’s law (V = RI) to find the current. EXECUTE:(a)1/R= 1/(15.0 Ω) + 1/(30.0 Ω) gives R= 10.0 Ω. I= V/R = (35.0 V)/(10.0 Ω) = 3.50 A. (b)1/R= 1/(10.0 Ω) + 1/(35.0 Ω) gives R= 7.78 Ω. I= (35.0 V)/(7.78 Ω) = 4.50 A (c)1/R= 1/(20.0 Ω) + 1/(25.0 Ω) gives R= 11.11 Ω, so I= (35.0 V)/(11.11 Ω) = 3.15 A. 26
26-2 Chapter 26 (d)From part (b), the resistance of the triangle alone is 7.78 Ω. Adding the 3.00-Ωinternal resistance of the battery gives an equivalent resistance for the circuit of 10.78 Ω. Therefore the current is I= (35.0 V)/(10.78 Ω) = 3.25 A EVALUATE:It makes a big difference how the triangle is connected to the battery. 26.6. IDENTIFY:The potential drop is the same across the resistors in parallel, and the current into the parallel combination is the same as the current through the 45.0-Ωresistor.
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