If the power developed across 2 resistance is 8w the heat produced in 4 resistance in 5 minutes is

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The power developed in an unknown resistance R is same as the power developed in the resistance of 8 ohm when these are connected separately across a battery of internal resistance 4 ohm.The value of R is

Solution

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Text Solution

`2:4:3``1:4:27``4:8:27``8:4:27`

Answer : D

Solution : Current through `2Omega, I_(1)=(2I)/(3)` <br> Hence produced per second, `H_(1)=I_(2)^(2)xx=((2I)/(3))^(2)xx2=(8I^(2))/(9)` <br> Current through, `4Omega, I_(2)=(I)/(3)` <br> Hence produced per second `H_(2)=I_(2)^(2)xx4=((I)/(3))^(2)xx4=(4I^(2))/(9)` <br> Current through, `3Omega, I` <br> Heat produced, `H_(3)=I^(2)xx3=3I^(2)=(27I^(2))/(9)` <br> `therefore H_(1):H_(2):H_(3)=8:4:27`

Text Solution

Solution : The `6Omega and 3Omega` resistances are in parallel. So, their combined resistance is <br> `1/R=1/6+1/3=1/2` <br> or `R=2Omega` <br> The equivalent simple circuit can be drawn as shown. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DCP_V04_C23_S01_020_S01.png" width="80%"> <br> Current in the circuit `i=("netemf")/("total resistance")=20/(3+2+5)=2A` <br> `V=iR=(2)(2)=4V` <br> i.e. Potential difference across `6Omega and 3Omega` resistance are 4V. Now, <br> `H_(3Omega)` (which is connected in series) =`i^2Rt=(2)^2(3)(2)=24J` <br> `H_(6Omega)=V^2/Rt=(4)^(2)/(6) (2) = 16/3J` <br> `H_(3Omega)` (which is connected in parallel ) `=V^2/Rt=((4)^2(2))/3=32/3J` <br> and `H_(5Omega)=i^2Rt=(2)^2(5)(2)=40J`