Two heaters A and B are in parallel across the supply voltage. Heater A produces 500 kJ in 20 minutes and B produces 1000 kJ in 10 minutes. The resistance of A is 100  Ω. If the same heaters are connected in series across the same voltage, then total heat produced in 5 minuteswill be

Two heaters A and B are in parallel across the supply voltage. Heater A produces 500 kJ in 20 minutes and B produces 1000 kJ in 10 minutes. The resistance of A is 100  Ω. If the same heaters are connected in series across the same voltage, then total heat produced in 5 minutes
will be

  1. A

    200 kJ

  2. B

    100 kJ

  3. C

    50 kJ

  4. D

    10 kJ

    Fill Out the Form for Expert Academic Guidance!l



    +91



    Live ClassesBooksTest SeriesSelf Learning



    Verify OTP Code (required)

    I agree to the terms and conditions and privacy policy.

    Solution:

    For heater A: 500×103=V2R120  ×60    …(i)

    Where  R1=100  Ω

    For heater B:

    1000×103=V2R210  ×60        ….(ii)

    From (i) and (ii), R2=25  Ω

    When heaters are connected in series: 

    Req=R1  +R2

    Heat produced:

    H=V2R1+R2  5  ×  60           …..(iii)

    From (i) and (iii), H = 100 kJ

    Chat on WhatsApp Call Infinity Learn

      Talk to our academic expert!



      +91


      Live ClassesBooksTest SeriesSelf Learning




      Verify OTP Code (required)

      I agree to the terms and conditions and privacy policy.