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CURRENT ELECTRICITY


                                             2
                   Now using, Energy = W = I Rt                                      Electrical grounding
                   We get resistance as                                          The  Earth  is  a  fairly  good
                                                                                 electrical conductor. Hence, if
                                     2
                             3 = (0.5)  × R × 20
                                                                                 a charged object is connected
                             R = 3 ×1/20 × 1/0.25 = 3/5 = 0.6 Ω                  with  the  Earth  by  a  piece  of
                                                                                 metal, the charge is conducted
                   14.11  ELECTRIC POWER                                         away  from  the  object  to  the
                                                                                 Earth. This convenient method
                                                                                 of  removing  the  charge  from
                   The  amount  of  energy  supplied  by  current  in  unit  time  is   an  object  is  called  grounding
                   known as electric power.                                      the  object.  As  a  safety
                                                                                 measure,  the  metal  shells  of
                   Hence power P can be determined by the formula                electrical  appliances  are
                             Electric power P = electrical energy/time = W/t     grounded  through  special
                                                                                 wires that give electric charges
                   where W is the electrical energy given by                     in the shells paths to the Earth.
                                                   W = QV                        The round post in the familiar
                   Therefore, above equation becomes                             three-prong electric plug is the
                                                                                 ground connection.
                                       QV
                   Electric power   P =   = IV = I R
                                                 2
                                        t
                                                                                 Remembering power formula
                   When current I is passing through a resistor R, the electric
                                                                           2
                   power that generates heat in the resistance is given by I R.
                   The unit of electric power is watt which is equal to one joule
                                  -1
                   per second (1 Js ). It is represented by the symbol W. Electric
                   bulbs commonly used in houses consume 25 W, 40 W, 60 W,
                   75 W and 100 W of electric power.                                               I

                   Example 14.7: The resistance of an electric bulb is 500 Ω. Find   cover V to find V =   P
                                                                                                    I
                   the power consumed by the bulb when a potential difference
                   of 250 V is applied across its ends.


                   Solution: Given that,  R = 500 Ω, V = 250 V                         Do you know?
                                    Using the formula, I = V/R                   Although  the  light  intensity
                                                          I  = 250 V/ 500  = 0.5 AΩ  from a 60 W incandescent light
                                            2
                                                       2
                          and           Power  P = I R = (0.5 A) × 500 Ω = 125 W  bulb  appears  to  be  constant,
                                                                                 the  current  in  the  bulb
                   Kilowatt-Hour                                                 fluctuates  50  times  each
                                                                                 second  between  -0.71  A  and
                   Electric energy is commonly consumed in very large quantity    0.71 A. The light appears to be
                                                                                 s t e a d y   b e c a u s e   t h e
                   for  the  measurement  of  which  joule  is  a  very  small  unit.    fluctuations  are  too  rapid  for
                   Hence, a very large unit of electric energy is needed which is    our eyes to perceive.
                   called kilowatt-hour. It is defined as

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