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Question

$\mathrm{S}_{2}$ is open and $\mathrm{S}_{1}$ is on position $-2$

$\mathrm{P}_{\mathrm{b}}$ is max

$\mathrm{P}_{\mathrm{c}}=\mathrm{P}_{\mathrm{

Vedclass · Accepted Answer

$a$ , $d$ , $c$ , $b$

Vedclass · Answer

$\mathrm{S}_{2}$ is open and $\mathrm{S}_{1}$ is on position $-2$ $\mathrm{P}_{\mathrm{b}}$ is max $\mathrm{P}_{\mathrm{c}}=\mathrm{P}_{\mathrm{d}}$ $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ $\mathrm{P}_{\mathrm{b}}>\mathrm{P}_{\mathrm{a}}>\mathrm{P}_{\mathrm{c}}=\mathrm{P}_{\mathrm{d}}$ $\mathrm{R}_{\mathrm{b}}>\mathrm{R}_{\mathrm{a}}>\mathrm{R}_{\mathrm{c}}=\mathrm{R}_{\mathrm{d}}$ $\mathrm{S}_{2}$ is closed and $\mathrm{S}_{1}$ is on position $-$ $1$ ${\rm{P}}_{\rm{b}}^\prime = \frac{{{\rm{V}}_0^2}}{{{{\rm{R}}_{\rm{b}}}}},{\rm{P}}_{\rm{c}}^\prime = \frac{{{\rm{V}}_0^2}}{{{{\rm{R}}_{\rm{c}}}}} \Rightarrow {\rm{P}}_{\rm{c}}^\prime > {\rm{P}}_{\rm{b}}^\prime $ $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$ $\mathrm{P}_{\mathrm{a}}^{\prime}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{a}}$ $\mathrm{P}_{d}^{\prime}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{d}};$ Since $\mathrm{R}_{\mathrm{a}}>\mathrm{R}_{\mathrm{d}}$ $\mathrm{P}_{\mathrm{a}}^{\prime}>\mathrm{P}_{\mathrm{d}}^{\prime}$ $\mathrm{P}_{\mathrm{c}}^{\prime}=\mathrm{I}_{2}^{2} \mathrm{R}_{\mathrm{c}}=\left[\mathrm{I}\left(\frac{\mathrm{R}_{\mathrm{b}}}{\mathrm{R}_{\mathrm{b}}+\mathrm{R}_{\mathrm{c}}}\right)\right]^{2} \mathrm{R}_{\mathrm{c}}$ $P_{c}^{\prime}=I^{2}\left[\left(\frac{R_{c} R_{b}}{R_{c}+R_{b}}\right)^{2} \frac{1}{R_{c}}\right]$ $\left(\frac{\mathrm{R}_{\mathrm{c}} \mathrm{R}_{\mathrm{b}}}{\mathrm{R}_{\mathrm{c}}+\mathrm{R}_{\mathrm{b}}}\right)^{2}\left(\frac{1}{\mathrm{R}_{\mathrm{c}}}\right)<\mathrm{R}_{\mathrm{c}}=\mathrm{R}_{\mathrm{d}}$ $\mathrm{P}_{\mathrm{c}}^{\prime}<\mathrm{P}_{\mathrm{d}}^{\prime}$ $\mathrm{P}_{\mathrm{a}}^{\prime}>\mathrm{P}_{\mathrm{d}}^{\prime}>\mathrm{P}_{\mathrm{c}}^{\prime}>\mathrm{P}_{\mathrm{b}}^{\prime}$

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