The figure shows three circuits I, II and III | vedclass

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Question

From the given circuit, net resistances ${{\rm{R}}_I} = 1\,\Omega ,{{\rm{R}}_{{\rm{II}}}} = 1/2\,\Omega ,{{\rm{R}}_{{\rm{III}}}} = 3/2\,\Omega $

It

Vedclass · Accepted Answer

$P_2 > P_1 > P_3$

Vedclass · Answer

From the given circuit, net resistances ${{\rm{R}}_I} = 1\,\Omega ,{{\rm{R}}_{{\rm{II}}}} = 1/2\,\Omega ,{{\rm{R}}_{{\rm{III}}}} = 3/2\,\Omega $ It is clear that $\mathrm{R}_{3}>\mathrm{R}_{1}>\mathrm{R}_{2}$ Hence, ${P_3} < {P_1} < {P_2}$ As Power $(P)$ $=\frac{\mathrm{V}^{2}}{\mathrm{R}} \Rightarrow \mathrm{P} \propto \frac{1}{\mathrm{R}}$

The figure shows three circuits $I, II$ and $III$ which are connected to a $3\,V$ battery. If the powers dissipated by the configurations $I, II$ and $III$ are $P_1 , P_2$ and $P_3$ respectively, then

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