Current Electricity Ques 58 — JEE PYQ Chapterwise

Answer:

Correct Answer: 58.(d)

Solution:

Formula:

Electrical Power:

58. $\quad P=\frac{V^{2}}{R}$ so, $R=\frac{V^{2}}{P}$

$ \therefore \quad R _1=\frac{V^{2}}{100} \quad \text { and } \quad R _2=R _3=\frac{V^{2}}{60} $

$ \text { Now, } \quad W _1=\frac{(250)^{2}}{\left(R _1+R _2\right)^{2}} \cdot R _1 $

$ \begin{gathered} W _2=\frac{(250)^{2}}{\left(R _1+R _2\right)} \cdot R _2 \quad \text { and } \quad W _3=\frac{(250)^{2}}{R _3} \\ W _1: W _2: W _3=15: 25: 64 \text { or } W _1<W _2<W _3 \end{gathered} $