You measure the power delivered by a battery to be 1.15 W when it is connected in series with two equal resistors. How much power will the s

Question

You measure the power delivered by a battery to be 1.15 W when it is connected in series with two equal resistors. How much power will the same battery deliver if the resistors are now connected in parallel across it

in progress 0
Thiên Ân 3 years 2021-08-26T04:51:10+00:00 1 Answers 5 views 0

Answers ( )

  1. thienthi
    0
    2021-08-26T04:53:05+00:00
    Reply

    Answer:

    The power is  P_p = 4.6 \ W

    Explanation:

    From the question we are told that

       The power delivered is  P_{s} = 1.15 \ W

       Let it resistance be denoted as R

        The resistors are connected in series so the equivalent resistance is  

         R_{eqv} = R+ R = 2 R

    Considering when it is connected in series    

    Generally power is mathematically represented as

         P_s = V * I

    Here I is the current which is mathematically represented as

           I = \frac{V}{2R}

    The power becomes

         P_s = V * \frac{V}{2R}

         P_s = \frac{V^2}{2R}

    substituting value

        1.15 = \frac{V^2}{2R}

    Considering when resistance is connected in parallel

      The equivalent resistance becomes

        R_{eqv} = \frac{R}{2}

    So The current  becomes

           I = \frac{V}{\frac{R}{2} } = \frac{2V}{R}

    And the power becomes

         P_p = V * \frac{2V}{R} = \frac{2V^2}{R} = \frac{4 V^2}{2 R} = 4 * P_s

     substituting values

         P_p = 4 * 1.15

         P_p = 4.6 \ W

         

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )