CHAPTER 21:CIRCUITS,BIOELECTRICITY, ANDDC ... ... College"Physics" Student"Solutions"Manual"...

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  • College  Physics   Student  Solutions  Manual   Chapter  21  

    157  

     

    CHAPTER  21:  CIRCUITS,  BIOELECTRICITY,   AND  DC  INSTRUMENTS   21.1  RESISTORS  IN  SERIES  AND  PARALLEL  

    1.   (a)  What  is  the  resistance  of  ten   Ω-275  resistors  connected  in  series?  (b)  In  parallel?    

    Solution   (a)  From  the  equation   ...RRRR 321s +++=  we  know  that  resistors  in  series  add:  

    ( )( ) Ω=Ω=++++= k 75.2 10 275 10321s R........ RRRR  

    (b)  From  the  equation   .... RRR

    ++= 21p

    111 ,  we  know  that  resistors  in  series  add  like:  

    ( ) Ω ×=⎟

    ⎠

    ⎞ ⎜ ⎝

    ⎛ Ω

    =+++= −2

    1021p

    1064.3 275

    1 101111 R

    ......... RRR

     

    So  that   Ω=Ω⎟ ⎠

    ⎞ ⎜ ⎝

    ⎛ ×

    = −

    5.27 1064.3

    1 2pR  

    7.   Referring  to  the  example  combining  series  and  parallel  circuits  and  Figure  21.6,   calculate   3I  in  the  following  two  different  ways:  (a)  from  the  known  values  of   I  and  

    2I ;  (b)  using  Ohm’s  law  for   3R .  In  both  parts  explicitly  show  how  you  follow  the  steps   in  the  Problem-­‐Solving  Strategies  for  Series  and  Parallel  Resistors.  

    Solution   Step  1:  The  circuit  diagram  is  drawn  in  Figure  21.6.  

    Step  2:  Find   3I .  

    Step  3:  Resistors   2R and   3R  are  in  parallel.  Then,  resistor   1R  is  in  series  with  the   combination  of   2R and   3R .  

    Step  4:  

  • College  Physics   Student  Solutions  Manual   Chapter  21  

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    (a)  Looking  at  the  point  where  the  wire  comes  into  the  parallel  combination  of   2R   and     3R  ,  we  see  that  the  current  coming  in   I  is  equal  to  the  current    going  out  

    2I  and   3I  ,  so  that   ,III 32 += or   A 74.0A 61.1A 35.223 =−=−= II  

    (b)  Using  Ohm’s  law  for   3R ,  and  voltage  for  the  combination  of   2R  and   3R  ,  found  in  

    Example  21.3,  we  can  determine  the  current:   A 742.0 0.13 V 65.9

    3

    p 3 =Ω

    == R V

    I  

    Step  5:  The  result  is  reasonable  because  it  is  smaller  than  the  incoming  current,   I ,   and  both  methods  produce  the  same  answer.  

    21.2  ELECTROMOTIVE  FORCE:  TERMINAL  VOLTAGE  

    15.   Carbon-­‐zinc  dry  cells  (sometimes  referred  to  as  non-­‐alkaline  cells)  have  an  emf  of  1.54   V,  and  they  are  produced  as  single  cells  or  in  various  combinations  to  form  other   voltages.  (a)  How  many  1.54-­‐V  cells  are  needed  to  make  the  common  9-­‐V  battery   used  in  many  small  electronic  devices?  (b)  What  is  the  actual  emf  of  the   approximately  9-­‐V  battery?  (c)  Discuss  how  internal  resistance  in  the  series  connection   of  cells  will  affect  the  terminal  voltage  of  this  approximately  9-­‐V  battery.  

    Solution   (a)  To  determine  the  number  simply  divide  the  9-­‐V  by  the  emf  of  each  cell:  

    684.5V 54.1V 9 ⇒=÷  

    (b)  If  six  dry  cells  are  put  in  series  ,  the  actual  emf  is   V 24.96V 54.1 =×  

    (c)  Internal  resistance  will  decrease  the  terminal  voltage  because  there  will  be  voltage   drops  across  the  internal  resistance  that  will  not  be  useful  in  the  operation  of  the   9-­‐V  battery.  

    30.   Unreasonable  Results  (a)  What  is  the  internal  resistance  of  a  1.54-­‐V  dry  cell  that   supplies  1.00  W  of  power  to  a   Ω-0.15  bulb?  (b)  What  is  unreasonable  about  this   result?  (c)  Which  assumptions  are  unreasonable  or  inconsistent?  

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    Solution   (a)  Using  the  equation   A 0.258

    Ω 15.0 W1.00 have we,2 ====

    R PIRIP .  So  using  

    Ohm’s  Law  and   IrEV −=  we  have  

    Ω 9.04 Ω 15.0 A 0.258

    V 1.54 :or , −=−=−==−= R I ErIRIrEV  

    (b)  You  cannot  have  negative  resistance.  

    (c)  The  voltage  should  be  less  than  the  emf  of  the  battery;  otherwise  the  internal   resistance  comes  out  negative.  Therefore,  the  power  delivered  is  too  large  for  the   given  resistance,  leading  to  a  current  that  is  too  large.  

    21.3  KIRCHHOFF’S  RULES  

    31.   Apply  the  loop  rule  to  loop  abcdefgha  in  Figure  21.25.  

    Solution   Using  the  loop  rule  for  loop  abcdefgha  in  Figure  21.25  gives:    

    0 2233312132 =++−+− - Er I r I r I ERI  

    37.   Apply  the  loop  rule  to  loop  akledcba  in  Figure  21.52.  

    Solution   Using  the  loop  rule  to  loop  akledcba  in  Figure  21.52  gives:    

    0111115122222 R I - ErI R I R - Ir - I E =+++  

    21.4  DC  VOLTMETERS  AND  AMMETERS  

    44.   Find  the  resistance  that  must  be  placed  in  series  with  a   Ω-0.25  galvanometer  having   a   µA-0.50  sensitivity  (the  same  as  the  one  discussed  in  the  text)  to  allow  it  to  be  used   as  a  voltmeter  with  a  0.100-­‐V  full-­‐scale  reading.  

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    Solution   We  are  given   .A 0.50 and V, 200.0 , 0.25 µ==Ω= IVr  

    Since  the  resistors  are  in  series,  the  total  resistance  for  the  voltmeter  is  found  by   using   ...321s +++= RRRR .  So,  using  Ohm’s  law  we  can  find  the  resistance   R :  

    kΩ98.1 Ω1975 Ω0.25 A1000.5

    V100.0 that so 5tot r

    I VR,

    I VrRR - ==−×

    =−==+=  

    50.   Suppose  you  measure  the  terminal  voltage  of  a  1.585-­‐V  alkaline  cell  having  an   internal  resistance  of   Ω 100.0  by  placing  a   Ωk -00.1  voltmeter  across  its  terminals.   (Figure  21.54.)  (a)  What  current  flows?  (b)  Find  the  terminal  voltage.  (c)  To  see  how   close  the  measured  terminal  voltage  is  to  the  emf,  calculate  their  ratio.  

    Solution   (a)

        E r

    V

    I

     

    Going  counterclockwise  around  the  loop  using  the  loop  rule  gives:  

    ( ) A 10 58.1 A 10 5848.1 100.0 10 00.1 V 585.1

    or 0,

    3 3 3 ×=×=Ω+Ω×

    = +

    =

    =++−

    rR EI

    IR Ir E  

    (b)  The  terminal  voltage  is  given  by  the  equation   IrEV −= :  

    ( )( ) V 5848.1 100.0A 10 5848.1V 585.1 3 =Ω×−=−= IrEV  

    Note:  The  answer  is  reported  to  5  significant  figures  to  see  the  difference.  

    (c)  To  calculate  the  ratio,  divide  the  terminal  voltage  by  the  emf:  

    99990.0 V 585.1 V 5848.1 ==

    E V  

       

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    21.5  NULL  MEASUREMENTS    

    58.   Calculate  the   xemf of  a  dry  cell  for  which  a  potentiometer  is  balanced  when   Ω= 200.1xR ,  while  an  alkaline  standard  cell  with  an  emf  of  1.600  V  requires   Ω= 247.