Question

Two resistors R1 = 400Ω and R₂ = 20 n are connected in parallel to a battery. If heating the power developed in R₁ is 25 W. find the heating power developed in R₂

Answer 1

`Power across `R_1` = 25 W

`P = V^2/R`

`25 xx 20 = V^2`

`V^2 = 500`

`V = sqrt500 V`

Power across `R_2` `P = V^2/R` = `500/400 = 1.2 W`

`V = sqrt500` as potential in parallel remain constant.