26_DC Currents

# 26_DC Currents - DIRECT-CURRENT CIRCUITS 26 26.1 26.2 26.3...

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26-1 D IRECT -C URRENT C IRCUITS 26.1. 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 R equiv = R /3 + R /3 + R /12 = 3 R /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 X directly. But note that X is 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: X is greater than the equivalent parallel resistance of 2.00 . 26.3. (a) IDENTIFY: Suppose we have two resistors in parallel, with 12 R R < . SET UP: The equivalent resistance is eq 1 2 111 R RR =+ EXECUTE: It is always true that 121 R +> . Therefore eq 1 11 R R > and eq 1 R R < . EVALUATE: The equivalent resistance is always less than that of the smallest resistor. (b) IDENTIFY: Suppose we have N resistors in parallel, with N R <<< " . SET UP: The equivalent resistance is eq 1 2 1 N R R =+++ " EXECUTE: It is always true that 1 1 1 N R R +++ > " . Therefore eq 1 R R > and eq 1 R R < . 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: VI R = . eq 1 2 R . EXECUTE: (a) 1 eq 12.3 . 32 20 R ⎛⎞ = += Ω ⎜⎟ ΩΩ ⎝⎠ (b) eq 240 V 19.5 A. 12.3 V I R == = Ω (c) 32 20 240 V 240 V 7.5 A; 12 A. 32 20 VV II = = 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 ) + 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. 26

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26-2 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. 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- resistor.
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## This note was uploaded on 04/02/2008 for the course PHYSICS 123 taught by Professor Madey during the Fall '08 term at Rutgers.

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26_DC Currents - DIRECT-CURRENT CIRCUITS 26 26.1 26.2 26.3...

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