Page 51 - CHAPTER-1 (Electricity)
P. 51

CHAPTER 1

            ELECTRICITY


           7.  Compare  the  power  used  in  the  2    resistor  in  each  of  the

                 following circuits :


                 (i) a 6 V battery in series combination with 1   and 2   resistors, and

                 (ii) a 8 V battery in parallel with 24   and 2   resistors.

                 Ans.     (i)   Since 6 V battery is in series with 1   and 2   resistors,

                 the current in the circuit,


                  I =    6V  =  6V  =2A
                     1Ω+2Ω     3Ω

                                                                         2
                                                             2
                   Power used in 2   resistor, P  = I R = (2A)  × 2   = 8W
                                                        1
                  (ii)   Since  8  V  battery  is  in  parallel  with  24    and  2    resistors,

                 potential drop across 2   resistor, V = 8V.


                  Power used in 4   resistor, P  =  V /R = 64/2 = 32
                                                               2
                                                        2
                  Clearly, P1/P2 = 8/32 = 1:4

           8.  Two lamps, one rated 220 W at 220 V, and the other 110 W at 220

                 V,  are  connected  in  parallel  combination  to  the  electric  mains

                 supply. Compute the current drawn from the line if the supply

                 voltage is 220 V?


                 Ans. Resistance of first lamp, r  = (220) /220 = 220  
                                                                      2
                                                           1
                 resistance of the second lamp, r  = (220) /110 = 440  
                                                                         2
                                                             2
                 Since  the  two  lamps  are  connected  in  parallel,  the  equivalent

                 resistance is given by


                      1
                  1
                          1
                   = + =       r +r 1
                               2
                 R p  r 1  r 2  r r
                                1 2
                 or     R  = 440*220/(440+220) = 440/3 = 146.67   
                          p
                 Current drawn from the line, i.e., I = V/ R  = 220/(440/3) = 3/2 = 1.5A
                                                                       p









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