A-resistor-R-is-connected-in-series-with-a-parallel-combination-of-two-resistors-of-24-and-8-ohms-The-total-power-disipated-in-the-circuit-is-64-watt-when-the-applied-voltage-is-24-volt-Find-R-

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Question Number 14963 by tawa tawa last updated on 06/Jun/17

$$\mathrm{A}\mathrm{resistor}\mathrm{R}\mathrm{is}\mathrm{connected}\mathrm{in}\mathrm{series}\mathrm{with}\mathrm{a}\mathrm{parallel}\mathrm{combination}\mathrm{of}\mathrm{two}\mathrm{resistors}$$$$\mathrm{of}24\mathrm{and}8\mathrm{ohms}.\mathrm{The}\mathrm{total}\mathrm{power}\mathrm{disipated}\mathrm{in}\mathrm{the}\mathrm{circuit}\mathrm{is}64\mathrm{watt}\mathrm{when}\mathrm{the}$$$$\mathrm{applied}\mathrm{voltage}\mathrm{is}24\mathrm{volt}.\mathrm{Find}\mathrm{R}$$

Answered by ajfour last updated on 06/Jun/17

$$P=\frac{V^2}{R_{eq}},R_{eq}=R+\frac{24\times 8}{32}\mathrm{\Omega }=R+6\mathrm{\Omega };$$$$\Rightarrow 64W=\frac{{\left(24V\right)}^2}{(R+6\mathrm{\Omega })}$$$$or,R+6\mathrm{\Omega }=\frac{24\times 24}{64}\mathrm{\Omega }$$$$R=9\mathrm{\Omega }-6\mathrm{\Omega }=3\mathrm{\Omega }.$$

Commented by tawa tawa last updated on 06/Jun/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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