College Phyic Student Manual Chater CHAPT : CCUTS, BOLCTCTY, AND DC NSTUMNTS. SSTOS N SS AND PAALLL. (a) What i the reitance of ten 75 -Ω reitor connected in erie? (b) n arallel? (a) From the equation + +... we know that reitor in erie add: + 0 ( 75 Ω)( 0).75 Ω + + +... + k (b) From the equation + +..., we know that reitor in erie add like: + +... + 0 75 Ω ( 0).64 0 Ω So that.64 0 Ω 7.5 Ω 7. eferring to the eamle combining erie and arallel circuit and Figure.6, calculate in the following two different way: (a) from the known value of and ; (b) uing Ohm law for. n both art elicitly how how you follow the te in the Problem- Solving Strategie for Serie and Parallel eitor. Ste : The circuit diagram i drawn in Figure.6. Ste : Find. Ste : eitor and are in arallel. Then, reitor i in erie with the combination of and. Ste 4: 57
College Phyic Student Manual Chater (a) Looking at the oint where the wire come into the arallel combination of and, we ee that the current coming in i equal to the current going out and, o that + or.5 A.6A 0.74 A, (b) Uing Ohm law for, and voltage for the combination of and, found in amle., we can determine the current: 9.65.0 Ω 0.74 A Ste 5: The reult i reaonable becaue it i maller than the incoming current,, and both method roduce the ame anwer.. LCTOMOT FOC: TMNAL OLTAG 5. Carbon- zinc dry cell (ometime referred to a non- alkaline cell) have an emf of.54, and they are roduced a ingle cell or in variou combination to form other voltage. (a) How many.54- cell are needed to make the common 9- battery ued in many mall electronic device? (b) What i the actual emf of the aroimately 9- battery? (c) Dicu how internal reitance in the erie connection of cell will affect the terminal voltage of thi aroimately 9- battery. (a) To determine the number imly divide the 9- by the emf of each cell: 9.54 5.84 6 (b) f i dry cell are ut in erie, the actual emf i.54 6 9.4 (c) nternal reitance will decreae the terminal voltage becaue there will be voltage dro acro the internal reitance that will not be ueful in the oeration of the 9- battery. 0. Unreaonable eult (a) What i the internal reitance of a.54- dry cell that ulie.00 W of ower to a 5.0-Ω bulb? (b) What i unreaonable about thi reult? (c) Which aumtion are unreaonable or inconitent? 58
College Phyic Student Manual Chater (a) Uing the equation P Ohm Law and, we have r we have P.00 W 5.0 Ω 0.58 A. So uing.54 r, or : r 5.0 Ω 9.04 Ω 0.58 A (b) You cannot have negative reitance. (c) The voltage hould be le than the emf of the battery; otherwie the internal reitance come out negative. Therefore, the ower delivered i too large for the given reitance, leading to a current that i too large.. KCHHOFF S ULS. Aly the loo rule to loo abcdefgha in Figure.5. Uing the loo rule for loo abcdefgha in Figure.5 give: - + r + r + r 0 7. Aly the loo rule to loo akledcba in Figure.5. Uing the loo rule to loo akledcba in Figure.5 give: - r - + + r - + 0 5.4 DC OLTMTS AND AMMTS 44. Find the reitance that mut be laced in erie with a 5.0-Ω galvanometer having a 50.0-µA enitivity (the ame a the one dicued in the tet) to allow it to be ued a a voltmeter with a 0.00- full- cale reading. 59
College Phyic Student Manual Chater We are given r 5.0 Ω, 0.00, and 50.0 µ A. Since the reitor are in erie, the total reitance for the voltmeter i found by uing + +... So, uing Ohm law we can find the reitance : + 0.00 tot + r, o that r 5.0 Ω 975 Ω.98 kω - 5 5.00 0 A 50. Suoe you meaure the terminal voltage of a.585- alkaline cell having an internal reitance of 0.00 Ω by lacing a.00 - k Ω voltmeter acro it terminal. (Figure.54.) (a) What current flow? (b) Find the terminal voltage. (c) To ee how cloe the meaured terminal voltage i to the emf, calculate their ratio. (a) r Going counterclockwie around the loo uing the loo rule give: + r + 0, or + r (.00 0 Ω ).585.5848 0 A.58 0 + 0.00 Ω A (b) The terminal voltage i given by the equation r : r.585 (.5848 0 A)( 0.00 Ω).5848 Note: The anwer i reorted to 5 ignificant figure to ee the difference. (c) To calculate the ratio, divide the terminal voltage by the emf:.5848.585 0.99990 60
College Phyic Student Manual Chater.5 NULL MASUMNTS 58. Calculate the emf of a dry cell for which a otentiometer i balanced when.00 Ω, while an alkaline tandard cell with an emf of.600 require.47 Ω to balance the otentiometer. We know and, o that, or.00 Ω.47 Ω (.600 ).540.6 DC CCUTS CONTANNG SSTOS AND CAPACTOS 6. The timing device in an automobile intermittent wier ytem i baed on an C time contant and utilize a 0.500 - µf caacitor and a variable reitor. Over what range mut be made to vary to achieve time contant from.00 to 5.0? From the equation C, C.00 5.00 0 4.00 0 F we know that: 6 7-7 -7 Ω and C 5.0 5.00 0.00 0 F Ω Therefore, the range for i: 6 7 4.00 0 Ω.00 0 Ω 4.00 to 0.0 MΩ 69. A heart defibrillator being ued on a atient ha an C time contant of 0.0 m due to the reitance of the atient and the caacitance of the defibrillator. (a) f the defibrillator ha an 8.00 - µf caacitance, what i the reitance of the ath through the atient? (You may neglect the caacitance of the atient and the reitance of the defibrillator.) (b) f the initial voltage i.0 k, how long doe it take to decline to 6.00 0? (a) Uing the equation C we can calculate the reitance: 6
College Phyic Student Manual Chater C.00 0 8.00 0-6.5 0 F Ω.5 kω (b) Uing the equation C e, we can calculate the time it take for the 0 - voltage to dro from.0 k to 600 : -C ln 0-6 600 - (.5 0 Ω)( 8.00 0 F) ln.996 0 0.0 m.0 0 4 74. ntegrated Concet f you wih to take a icture of a bullet traveling at 500 m/, then a very brief flah of light roduced by an C dicharge through a flah tube can limit blurring. Auming.00 mm of motion during one C contant i accetable, and given that the flah i driven by a 600 -µf caacitor, what i the reitance in the flah tube? Uing c or t and the equation for the time contant, we can write the time t v contant a C, o getting thee two time equal give an ereion from which we can olve for the required reitance: v C,or vc.00 0 m (500 m/)(6.00 0 4. 0 F) Ω 6