26_DC Currents
35 Pages

26_DC Currents

Course Number: PHYSICS 123, Fall 2008

College/University: Rutgers

Word Count: 15394

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DIRECT-CURRENT CIRCUITS 26 26.1. 26.2. 26.3. 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,...

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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.

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Rutgers - PHYSICS - 123
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Rutgers - PHYSICS - 123
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Rutgers - PHYSICS - 123
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Rutgers - PHYSICS - 123
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Rutgers - PHYSICS - 123
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Rutgers - PHYSICS - 123
THE SECOND LAW OF THERMODYNAMICS2020.1.IDENTIFY: SET UP: EXECUTE: (b) e =For a heat engine, W = QH - QC . e =W . QH &gt; 0, QC &lt; 0. QHW = 2200 J. QC = 4300 J.(a) QH = W + QC = 6500 J.2200 J = 0.34 = 34%. 6500 J EVALUATE: Since the engine
Rutgers - PHYSICS - 123
THE FIRST LAW OF THERMODYNAMICS19The pV-diagram is sketched in Figure 19.119.1.(a) IDENTIFY and SET UP:The pressure is constant and the volume increases.Figure 19.1 (b) W = V2 V1p dVV2 V1Since p is constant, W = p dV = p (V2 - V1 )
Rutgers - PHYSICS - 123
THERMAL PROPERTIES OF MATTER1818.1.(a) IDENTIFY: We are asked about a single state of the system. SET UP: Use Eq.(18.2) to calculate the number of moles and then apply the ideal-gas equation. m 0.225 kg = 56.2 mol EXECUTE: n = tot = M 4.00 10-
Rutgers - PHYSICS - 123
TEMPERATURE AND HEAT1717.1.IDENTIFY and SET UP: EXECUTE:TF = 9 TC + 32. 5(a) TF = (9/5)(-62.8) + 32 = -81.0F(b) TF = (9/5)(56.7) + 32 = 134.1F (c) TF = (9/5)(31.1) + 32 = 88.0F EVALUATE: Fahrenheit degrees are smaller than Celsius degrees
Rutgers - PHYSICS - 123
SOUND AND HEARING1616.1.IDENTIFY and SET UP: Eq.(15.1) gives the wavelength in terms of the frequency. Use Eq.(16.5) to relate the pressure and displacement amplitudes. EXECUTE: (a) = v / f = (344 m/s)/1000 Hz = 0.344 m (b) pmax = BkA and Bk i
Rutgers - PHYSICS - 123
MECHANICAL WAVES1515.1.IDENTIFY: v = f . T = 1/ f is the time for one complete vibration. SET UP: The frequency of the note one octave higher is 1568 Hz. v 344 m/s 1 EXECUTE: (a) = = = 0.439 m . T = = 1.28 ms . f 784 Hz f15.2.v 344 m/s =
Rutgers - PHYSICS - 123
FLUID MECHANICS1414.1.IDENTIFY: SET UP: EXECUTE:Use Eq.(14.1) to calculate the mass and then use w = mg to calculate the weight. = m / V so m = V From Table 14.1, = 7.8 103 kg/m3.For a cylinder of length L and radius R, V = ( R 2 ) L =
Rutgers - PHYSICS - 123
PERIODIC MOTION1313.1.IDENTIFY and SET UP: The target variables are the period T and angular frequency . We are given the frequency f, so we can find these using Eqs.(13.1) and (13.2) EXECUTE: (a) f = 220 HzT = 1/f = 1/220 Hz = 4.54 10-3 s =
Rutgers - PHYSICS - 123
GRAVITATION12Use the law of gravitation, Eq.(12.1), to determine Fg .12.1.IDENTIFY and SET UP: EXECUTE:FS on MGmm mSmM (S = sun, M = moon); FE on M = G E2 M (E = earth) 2 r SM r EM22 FS on M mSmM r EM mS rEM = G 2 = FE on M r
Rutgers - PHYSICS - 123
EQUILIBRIUM AND ELASTICITY1111.1.IDENTIFY: Use Eq.(11.3) to calculate xcm . The center of gravity of the bar is at its center and it can be treated as a point mass at that point. SET UP: Use coordinates with the origin at the left end of the ba
Rutgers - PHYSICS - 123
DYNAMICS OF ROTATIONAL MOTION10EXECUTE: = Fl l = r sin = (4.00 m)sin 90 l = 4.00 m = (10.0 N)(4.00 m) = 40.0 N m10.1.IDENTIFY: Use Eq.(10.2) to calculate the magnitude of the torque and use the right-hand rule illustrated in Fig.(10.4) to
Rutgers - PHYSICS - 123
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Rutgers - PHYSICS - 123
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Rutgers - PHYSICS - 123
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Rutgers - PHYSICS - 123
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Rutgers - PHYSICS - 123
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Arizona - MATH - 125
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Cal Poly - ME - 212
INSTRUCTOR'S MANUAL To Accompany ENGINEERING MECHANICS - DYNAMICS Volume 2 Fifth Edition, 2002 J. L. Meriam and L. G. Kraige Copyright 2002 by John Wiley &amp; Sons, Inc.USE OF THE INSTRUCTOR'S MANUAL The problem solution portion of this manual has be
WVU - MAE - 244
1Tension, Compression, and ShearNormal Stress and StrainProblem 1.2-1 A solid circular post ABC (see figure) supports a load P1 2500 lb acting at the top. A second load P2 is uniformly distributed around the shelf at B. The diameters of the upper
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32CHAPTER 1Tension, Compression, and ShearProblem 1.6-10 A flexible connection consisting of rubber pads (thickness t 9 mm) bonded to steel plates is shown in the figure. The pads are 160 mm long and 80 mm wide. (a) Find the average shear strai
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106CHAPTER 2Axially Loaded MembersProblem 2.5-3 A rigid bar of weight W 750 lb hangs from three equally spaced wires, two of steel and one of aluminum (see figure). The diameter of the wires is 1/8 in. Before they were loaded, all three wires h
WVU - MAE - 244
122CHAPTER 2Axially Loaded MembersStresses on Inclined SectionsProblem 2.6-1 A steel bar of rectangular cross section (1.5 in. 2.0 in.) carries a tensile load P (see figure). The allowable stresses in tension and shear are 15,000 psi and 7,000
WVU - MAE - 244
134CHAPTER 2Axially Loaded MembersProblem 2.6-16 A prismatic bar is subjected to an axial force that produces a tensile stress 63 MPa and a shear stress 21 MPa on a certain inclined plane (see figure). Determine the stresses acting on all faces
WVU - MAE - 244
144CHAPTER 2Axially Loaded MembersProblem 2.7-9 A slightly tapered bar AB of rectangular cross section and length L is acted upon by a force P (see figure). The width of the bar varies uniformly from b2 at end A to b1 at end B. The thickness t
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CHAPTER TWO SOLUTIONS 1. (a) 12 s (b) 750 mJ (c) 1.13 k (d) 3.5 Gbits (e) 6.5 nm (f) 13.56 MHz (g) 39 pA (h) 49 k (i) 11.73 pAEngineering Circuit Analysis, 6th EditionCopyright 2002 McGraw-Hill, Inc. All Rights Reserved.CHAPTER TWO SOLUTIONS2
WVU - MAE - 320
CHAPTER THREE SOLUTIONS 1.Engineering Circuit Analysis, 6th EditionCopyright 2002 McGraw-Hill, Inc. All Rights Reserved.CHAPTER THREE SOLUTIONS 2. (a) six nodes; (b) nine branches.Engineering Circuit Analysis, 6th EditionCopyright 2002 McGr
WVU - EE - 221
CHAPTER FOUR SOLUTIONS 1. (a) 0.1 -0.5 -0.2 -0.3 0.1 -0.3 -0.4 0 0.4 v1 v2 v3 0 4 6=Solving this matrix equation using a scientific calculator, v2 = -8.387 V (b) Using a scientific calculator, the determinant is equal to 32.Engineering Circuit
WVU - MAE - 320
CHAPTER FIVE SOLUTIONS 1. Define percent error as 100 [ex (1 + x)]/ ex x 0.001 0.005 0.01 0.05 0.10 0.50 1.00 5.00 1+x 1.001 1.005 1.01 1.05 1.10 1.50 2.00 6.00 ex 1.001 1.005 1.010 1.051 1.105 1.649 2.718 148.4 % error 510-5 110-3 510-3 0.1 0.5 9 2
WVU - MAE - 320
CHAPTER SIX SOLUTIONS 1. The first step is to perform a simple source transformation, so that a 0.15-V source in series with a 150- resistor is connected to the inverting pin of the ideal op amp. Then, vout = - 2200 (0.15) = - 2.2 V 150Engineering
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CHAPTER SEVEN SOLUTIONS1.(a) C =A 8.854 10 -12 (78.54 10 -6 ) = = 6.954 pF d 100 10 - 61 2E 2( 10 - 3 ) 1 CV 2 V = = = 16.959kV 2 C 6.954 10 -12(b) Energy , E =(c) E =1 2 E 2(2.5 10 -6 ) CV 2 C = 2 = = 500 pF 2 (100 2 ) VC=A Cd
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CHAPTER EIGHT SOLUTIONS 1. (a)iL (0) =100 = 2A iL (t ) = 2e -80t / 0.2 50 -400 t = 2e A, t &gt; 0(b) (c)iL (0.01) = 2e-4 = 36.63mA 2e -400t1 = 1, e400t1 = 2, t1 = 1.7329msEngineering Circuit Analysis, 6th EditionCopyright 2002 McGraw-Hill,
WVU - MAE - 320
CHAPTER NINE SOLUTIONS 1. o L = 10, s1 = -6s -1 , s2 = -8s -12 2 -6 = + 2 - o , - 8 = - - 2 - o adding,-14 = -2 = 7 s -12 2 -6 = -7 + 49 - o o = 481 , o = 6.928 LCrad/s 6.928 L = 10, L = 1.4434H, 1 1 C= = 14.434mF, = 7 R = 4.949 4
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CHAPTER TEN (Phasor Analysis) SOLUTIONS 1. (a) 2103 = 290.9t rad/s 21.6 f (t ) = 8.5sin (290.9t + ) 0 = 8.5sin (290.9 2.110 -3 + ) T = 4 (7.5 - 2.1)10-3 = 21.6 10-3 , = = -0.6109rad + 2 = 5.672rad or 325.0 f (t ) = 8.5sin (290.9t + 325.0) (b