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phys fall 2008 key exam2

phys fall 2008 key exam2 - "‘.m/5?4 Keg Instructions...

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Unformatted text preview: "‘ ......m. .../5:??4? Keg/ .. Instructions: Please read each problem carefully. Circle the correct answer choice, or fill in the blanks as asked for in the pro em , A—iH D 4 g .1. 4. .L a, J.— Usethefigureatthe rightforproblems 1-3. c1 c2 c, '16 “I" 3‘3"? “'9‘ Forthccireuit consisting of air filled, parallel plate 3 - capuitm. c. = 1.50 W, c,= 3.80 pr, c,=- 5.20 pr, and C. = 7.60 21F, connected as shown at the right, C. .1 \[8] #1. Determinetheequivalent capacitance between pointsAandB. ’ Answer: (a) 1.03x10'pF. (b) 1.50x10‘pr. (c) 1.39:: 10' pF. (d) 1.76x 10' 14F. (e) 1.55x10‘pr. @ I K/OF \ [8] #2. A potential of21.6 volts is now applied across points D and B. What charge ' owbe on C4? Answer: (a) 3.70 x 10* c. (b) 5.50 x 10' c. (c) 9.05 x 10*5 c. (a) 1.37 x 10“ c. 1.64 x 10* c. (t) [5].#3. A piece of glass (dielectric constant 12.!) is placed so as to completely fill between the plates of capacitor C.. What will he the new value of the capacitance of C4? Answer: (a)3.55 x10'5 F. (b) 4.26 x 10" r. (c) 7.02 x10" F @20 x 10'5 F. (e) 2.33 x 104 r. (1) \ [8] #4. A 12.0 m long section of a certain wire (resistivity =5 8.52 x 10" Qm at room temperature), has a circular cross section of radius=5.0x10"rn. Determine the value ofits ‘ atroomtempenture. Answer: (a) 1.15 x 10'2 o. (b) 2.77 x 10‘2 o. .30 x 10‘2 n. (a) 2.58 x 10'2 a. (e) 1.30 x 10'2 o. (1) \ [8] #5. An electric current of l5.0 mA exists when an electric field of 50.0 V/m is setup in a wire. If the wire has a cross section area of 9. 10 x 10“ 111’, what is its conductivity? wer: (a) 2.98 x 10“ (9.111)". (b) 1.81 x 10“ (0.111)". (c) 2.77 x 10‘l (fl-m)". (a) 1.31 x 10" (ti-m)". .30 x 10" (Q~m)". (t) [I #6. Circle each of the true statements a) When a capacitor is fully charged the electric current that moves from one plate to the other through the capacitor is a maximum. If a point charge Q is placed in an infinitely large material of dielectric constant K, the magnitude of the electric field at a 1 ’te distance r fiom the point charge is E = (l/41t1ccoXQ/rz). (G, e . Law tells us that the current is directly proportional to the potential difference that is applied across the device. ((93 n a resistance the kinetic energy of the charge carrier is tramformed to heat energy. “1 terrain? a. dielectric wuh 16> Kw on a charged air-filled capacitnr'has the affect of reducing the electric field between the -. 3 acitor plates. \ [8] #7. For the circuit shown , R. = 2.00 9, R1 = 3.00 0, R3 = 4.009, and R. = 5.00 I) . Determine the equivalent resistance between a and b. Answer: (a) 8.92 o. @722 o. (c) 6.95 x 102 o. (d) 6.24 n. (e) 1.1 19. (t) [20] #8. For the circuit shown: Write three loop rule equations. and a useful junction rule equation inVOIVing 11, i3 , and i]. 32.0 0 1/ PHYS 222 EXAM #2 Spring 2000 Covering Chapter 25 - 27 Name S For #9 & #10, use the Circuit shown at the right. E = 4.50 KV, C == 6.80 uF, + l R andR=970xlo’n E _ C ‘5)”. Whatisflieminimumtimethatshouldpassafiertheswitchisclosed, to consider that the capacitor“ is fully ed? Answer. (3)2. as x10"s. (b) 2.95 x 10- s. (c)3.39x 10“ [email protected] 10“ s. (e) 5.21 x rod 5. \ [5] #10. Consider that the capacitor is initially uncharged before the switch S is closed. What is the current in the resistance R immediatelyafierswitc- ' closed? Answer: (a) 455 mA. @ ' v mA. (c) 555 mA. (d) 632 mA. (e) 645 mA. (0 \[22] #11. The figure attherightisoftwoconcentric cylindricalooudmtingshellsofradiusa and b, withb>a. The ~ ~ between the shells is filled with air (dielectric constant = 1.000). The length of each cylinders is L. Using the steps below, which are based on those given in Chapter 25, determine an expresion for the capacitance. [4] Step #1: Using Q to represent the m of the charge write on the figure to show the charges that you assume on the shells. (Don't forget the signs to indicate the polarity of the charges.) [ 6] Step #2: Choose the correct expression for the electric field in the space between the shells: (Note: Win and N/C are the same.) pa (a). E(a<r<b)=2£l’r (V/m) (b)E(a<r<b)=-2w ”(V/m) @Ewmmzfio (V/m) 0 — 9": = VI 2““, . (e) E (a < r < b) 27:502-1} 0( m) . (f) [6] Step #3: Determine the electric potential difference between the two cylinders. in) v =(Q/21teoa)[1n(b/L)] (V). = (Q/2neoL)[|n(b/a)] (V). (c) v =(Q2/21woab)[lfl(L)] (V). (d) V = (Qz/ZnoLlllnw/afl (V)- (e) V = (Q/4eob)[ln(Ua)] (V)- (f) V = [6] Step #4: Complete the problem by using the above information to determine the capacitance. (a) C = 41r$oL[1n(b/a)]'l (1")- = ZMUHWIIH" (F) (C) C = 4Sob[|n(L/a)] (F). (d) C = 21:30a[ln(b/L)]" (F)- (e) C = ZNanbUMLH" (F). (D C = \[18] #12 The circuit shows meters to measure currents in and voltages Va. Match the letter in the meter with what it is measuring. (For example, Va; means the potential difference across R2, and in means the current in R2) Meter # Measure Meter # Measure Meter # Measure in f in 2 Via in ( le Vs: g0 ...
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