Page 19 - 12-phy-13 Current Electricity
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13. Current Electricity                                                                              eLearn.Punjab



        The left side of this equation is the emf E of the cell which is equal to energy gained by unit charge
        as it passes through the cell from its negative to positive terminal. The right side of the equation

        gives an account of the utilization of this energy as the current passes the circuit.
        It states that, as a unit charge passes through the circuit, a part of this energy equal to Ir is dissipated

        into the cell and the rest of the energy is dissipated into the external resistance R. It is given by
        potential drop IR. Thus the emf gives the energy supplied to unit charge by the cell and the potential

        drop across the various elements account for the dissipation of this energy into other forms as the
        unit charge passes through these elements.

        The emf is the “cause” and potential difference is its “effect”. The emf is always present even when
        no current is drawn through the battery or the cell, but the potential difference across the conductor

        is zero when no current flows through it.



        Example  13.4:  The  potential  difference  between  the                                    Do You Know?
        terminals of a battery in open circuit is

        2.2 V. When it is connected across a resistance of 5.0
        W, the potential falls  to 1.8  V. Calculate  the current

        and the internal resistance of the battery.
        Solution:


               Given       E  =  2.2        = V,       5.0  W R  V  =,        1.8  V
               We are to calculate      I and  r.

               We have                 V  = IR
                                      V    1.8 V
               or                        I  =  =  = 0.36 A
                                      R    5.0  W
               Internal resistance     r can be calculated by using
                                                E  = V  + Ir

               or                       2.2 V=1.8 V+0.36 Ax r
               or                           r  = 1.11 VA −1  = 1.11 W


        Maximum Power Output





        In the circuit of Fig. 13.19, as the current I flows through

        the resistance R, the charges flow from a point of higher  A voltmeter connected across the terminals of
        potential to a point of lower potential and as such, they                  a cell measures (a) the emf of the cell on open

        loose  potential  energy.  If  V  is  the  potential  difference           circuit, (b) the terminal potential difference on
                                                                                   a closed circuit.
        across R, the loss of potential energy per second is VI.

        This loss of energy per second appears in other forms of energy and is known as power delivered
        to R by current I.
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