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physics test 2

# physics test 2 - PHYSICS 1442-001_f TEST 2>{Efficacy...

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Unformatted text preview: PHYSICS 1442-001 __f TEST 2- . . - - > {Efficacy 22)1oo-; _,/ Bob O’Donnell ' A 1. Electrical current results from the slow “drift” of electrons through an electrical circuit. If it is measured that 5.0);10'3 coulombs are passing through the battery of a cell phone every 0.10 seconds, the actual current must be ‘ a). 5.0xlO4A :3... gm”)? b). 20A r: W @oosa , 5 d). 50A 2. If the total current (I) through the battery of a ﬂashlight is measured by an ammeter to - be 0.20A and the battery voltage is 6.0V, the total resistance(R) of the circuit is "Tap 3 0.0th ‘J 'f “Qt b). 12 ohms Q: # Lt) _ c). 0.333 ohms ' x; if} Q d). 0.24 ohms - l - r 1 3 -IL 3. In problem 2 above, how much power (P) is being supplied to the ﬂashlight circuit by the battery? . ' V - V a). 0.24%! Y} inlay] it? 1.2w ‘ ’ I j c). 7.2w 9 ” l-L d). 30.0w 4. Because gold is inert and resistant to corrosion, it is commonly used by NASA in its space vehicles. If the resistance of a gold resistor at 20°C is accurately measured to be 10 ohms and the temperature efficient of resistivity (0c) of gold at 20 PC is 0.0034(3'1, what will be its resistance at 220°C? . a). 10.0 ohm _ 'r,’ ' i t .. .' r _ n t. -_ b). 6.3 ohm n) Ll ._ "J ’. .4 his“! j 16.8 Ohm __ m d). 17.48 ohm 5. Current is due to electron movement in the wire. Such movement can be described _ brieﬂy as a). All electrons moving parallel to the wire at very high speed (drift speed). b). All electrons moving exactly parallel to the wire at very low speed (drift speed) @Electrons moving randomly very fast, bouncing off wire atoms but having an \“ average direction parallel to the wire due to the fame exerted on the electrons by the ' . electric ﬁeld in the wire. . (1). Studies involving electron movements in a wire have never been done. 6. My cordless telephone has a battery rated at 0.35 All. What is the numerical value of such a rating in terms of coulombs? , ski, .x’. :4.- ”“i if Q}. 1260.c C) ” ‘ ' J ' b). 0.35c 7 " 7.31;?) c). 21.0c d). 3600.c 7. What is the resistance of the iron circuit component shown below at 20°C? The resistivity (f) of iron at 20°C is 9.71x10'8 ohm-m. ‘ r a). 9.71am“S ohms I {2,—— p ,,= 1;). 9.71;:10'10 ohms f: [00”? {63.0.0971 ohms , -- v - r__ _ W ﬁg» 1 RREP: : l-O mm 8. As is known, household alternating voltage as supplied by the power company varies with time in a sinusoidal manner. The power requirement of say a toaster will therefore vary with time. Since, however, only average(rms) values are used in alternating current calculations, what is the average poll/g; requirement of a toaster if the rms value of the voltage dr0p(Vms) is 50.Vlwh'ile the rms value of the current ~ (Inns) is LSA? . 37.5w . i}, c). 53.0w .- " d}. 7S.w.-/" 9. Although rms values are used in alternating current calculations, the peak values can be substantially different. Thus, for the toaster in problem 8 in which the rms voltage drop was 50.v, the actual peak (maximum) voltage drop is v a). .77” \I' 10'. Electric power originating from power stations is transmitted over long distances at very high voltages. The reason for such high transmitted voltages is a). Keeps birds off the wires b). Voltages at the power station are of the same magnitude and thus can be transmitted directly with no additional equipment. . transmitted voltages corresPond to low currents and thus 12R losses are kept to a minimum d). the cost of the wiring is much lower. 11. What is the equivalent resistance for the circuit below? f_ If)??? 14. ohms ‘e . 0.0714 ohms . a. c). 1.143 ohms d). 0.875 ohms 12. Calculate the equivalent resistance for the circuit below: a). 6.0 ohms g? + Z 13 . 1.50 Ohms ﬂ Le). 1.33 ohms Ell? i d). 0.667 ohms W ‘l‘ .j a}: . 13. Shown below is a simple series circuit that also includes the internal resistance of the battery. What is the current in the circuit? - \Jr 1. (L 2.0A - r v '1’) b). 2.5a ' 5” c). 3.33A E _ r d). 10.0A _ ‘ 4-1 14. Shown below is a battery having an internal resistance of 0.2 ohms, a current of 5A and an emf(8) of 12.v. What is the terminal voltage (V+-V_) of the battery? I '— ', '1‘ a). 12.v ‘J': I ”‘ b). 13mr \/ -'- ’QITHLV rd). Not enough information '0 I 15. Use Kirchhoffs Rules to determine the current in the circuit below. Va 1 3'12} ‘39— a. 3.0A b)..1.A _ c). 0.333A I _ V -d). Zero ‘ 7 E, 'T+3.+Lo:."’3~L *'-‘ ‘1 C9. Electron beams are sometimes used in hypervelocity wind tunnels to measure the static gas temperature at speciﬁc locations. If the current density( J) of such a beam is 5x10"I ° Alni‘," and the impressed electric ﬁeld is 1000.0 Vim. the electrical resistivity (lo) of the gas must be f. J“ F ,1... ._:: E J-. bis".er TIL“ M v} :1. ._a_. Neeci Tkﬁdisfﬁnee, i); l’i-Q . ,xto‘lJL-wi / a _ .. A meal, ..\‘ I —' '3, wr‘ .1 __, W \‘I’; lL—J nlz .7 H W“. _ x he s s l— ' .1.— ‘Hicx-J - ‘—. ‘ ‘H. try-7 l rd. .Wmmi aﬁa-.#_-.-- 17. In the circuit shown below1 the value of the potential energy of a positive charge at the positive terminal is qv+. What is the potential energy of the charge when it reaches points A and B? . - _ ~ '_ - ,..a. qv- _b qV+ 33)? ci(V+' V-) d). q(v.~v+) 18. Using conservation of energy, calculate the potential difference 7(v1,; v3) for the circuit element below. (I = LA). —-—-—-> 1: :1, A /'_ _ ﬂ '2: ,r J (II. F. '/\ _ a ,j i I U tel-312w Vb . " " I t.._.-.——/W——-+" 'i‘ + |——-a_ h b). 60v 'i‘J-Vrﬁk 2 J 42 a is) 10.11.. - 2.3. C). 24.1} Jib » \f3- ; li'f/ d). need a complete circuit 19. Shown below are two capacitors connected-in parallel, their equivalent capacitance is __l * - - ' ‘- = ’2: Duo F _ - l \ , i L! 1.87x107‘7F -- I _ i ,b). 8.0xlO‘F I i 1 I ‘_ I I q 1:). 0.533x10'71: . I V MO F (1). Need the battery voltage 20. Use Kirchhoffs function Rule {or chair? in = charge out) to determine an: expression describing the currents at junction A of the circuit below. 2 11: 12+ 13 b). I2: 11+ I; c‘) IgtII‘I‘Il a.) I‘+1,_+13=o 1‘ l...) 4%“? ‘ PHYSICS 1442—001 TEST 2 M ARC. 2004 Bob O’Donnell ‘ SECC STAFF ’i *2— The drift velocity is the average velocity that electrons move along a wire. Such movement provides the electrical current. Ifan experiment is perfom'iecl such that 1.0){1020 electrons are counted in 0.001 seconds as they pass the counting station, —_at is the currentfli in the wire? a}! 16,000A l LOXIU'fL‘) jierLEV-i4j b» LL \ .. ~ c). 1.6X10'16A L7 [titre] x d). 0.016A ' _ SiiPpose NASA requires the use of a gold resistor for a special mission in which seﬁere environmental conditions are expected. It is known that the resistivity (p0) of gold at 200 C is 2.443(10'8 ohm—m and its temperature coefﬁcient of resistivity (a) is +0.0034/0 c, What will be the resistivity (p) at 200° C? 3'”— g‘4-4X1e”g [l 4“ ‘OCM Q50”ﬂ°;>/:l 1* 5 a). 2.44X10_8 ohms—m -. 1.4-9X10‘l3 ohms—m c). 3.93.:{10'3 ohms—m i. 4.103(10'3 ohms—m 1 Shown below is the gold resistor, its dimensions and the direcrion of the cmrent. As noted above, the resistivity (pg) at 200 C is 2.443(10‘3 ohms-m, the resistance (R) at 20° C will therefore be . A3; a e c: e. K 'm 2 afoul 0'00 QC q" ' O. S .3. 44K 10‘5 74% -k o L y 0). “Ll-4.0111115 imam i {)an m \ c). ExilieRZlO'a ohms H (i). 2.44x10‘1“ ohms " 3‘44 Current is a circuit is due to electron movement in the wire. Such movement can be u described brieﬂy as average direction parallel to the Wire due to The force exerted on die electrons by the eiectric ﬁeld in :he wire. ' Science has never reached any conciusrons regarding electron movement in conducmrs. w‘w;‘wz-M.v': ?,,.Wg-nw1'r . s -- -;.WW_ 7_ _‘ _. . ,_ a) 3.33 A - b) 120.0 A D 1.80A 7. My cordless phone has a battery rated at 1. coulombs‘? é ‘ 1.50 C 1:27;: 5,400. C 4/ t c . 2,400. c 1' :1). Need the number of hours. 8. If the alternating current power rating ff (ls-bar» on a light bulb is 60. W. and the average voltage of the household c1rcuit is 1 10V (Vm), the actual resistance of the bulbis P,_ {V V2110— 201331315 ?= (cow “D 7 (a. “a b). 1.330 s »; m3, --———-— c). 6,600. ohms 1 : P we I [A (a, 4 9 641:4 ohms JV; F "’ '34 291‘ —_ a \D Information given on packaging regarding alternating current is generally given in terms of ms values for current and voltage and average values for power. IF on a certain label the true value for voltage is given as 1 10V, the actual oeuk value (V0) is L/ c}. 2:0.V (‘9 1551-5 v {If}. ‘-N?:at is the equivalent resistance for the circuit below? 1 1 F—‘A/i—n—IKNL—m ' : :Li. ;0 our-is : ’,- Cedar?) 3'11'15 M n . c}. _7=J~1-onms F '9‘ A die 0.950 onrns. i __ s A S .3, __‘ i ll.The equivaleutresistance for the circuitbelow is 141 ['JL @06630111113 , I F Jmi 1 : g I j;1.50ohms 5+7 ” z a 2 1,41 , 'Q‘ C). 3.000th . l d. 03"?) ohms ___, __ - ) 3 —e’ “5 “"5" L5 ' “W? 12. Show below is a circuit that also incorporates the ' battery What is the current (I; in the circuit? a). OSOA - V“ "0“ C b.SO.A 1.: AL ID @0495A Q, 26, L d). 2.02A I .—_ 9 4% , 13. For the battery below, what is the terminal voltage (W. - VJ? Q. V V L1 ﬂ @0301] ' . 2.0 a ' ' 1 [Us ID V—i- o . 11.0 V :1). Need the complete circuit E 15. Shown below is part of a circuit. As is know, potential energy (q the potential (V) Varies around the circuit. Although the potential may not be known at every point of the circuit, the potential difference can be determined. For the A circuit element below, what is the potential difference (Va —- Vb) between Va and Vb? - Va— 51+) 120 -— 2{4)—5 Vb \ A ‘ V - \j 5. A Q‘H1:_' w: Lg¥éllgov V0“ % " g/C -— E5 “‘3: 3 Yb C). lT‘OV V/C\—-\[b: 215 -i- 2 i i ti). Need :1 complete circuit \fék— VD : | I 16. Power companies transfer power at very high voltages because a]. Power [s originally generated at the identical high voltages. . b‘i. It’s :1 :ederal law that power be generated at Very voltages q- . - . .‘ . — .. ,2 l”:ng TITLI‘lSl'I‘lIFEECI voltages cause very our current and tbererore :ovv L R ransmttteti power Losses. a) l . The i e ofwirjng used requires very high voltages. ...
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