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Two-30-ohms-resistor-are-connected-in-parallel-what-should-be-the-resistance-to-be-connected-in-series-with-this-parallel-combination-such-that-the-power-in-each-30-ohms-is-1-4-th-of-total-power-




Question Number 14962 by tawa tawa last updated on 06/Jun/17
Two 30 ohms resistor are connected in parallel, what should be the resistance  to be connected in series with this parallel combination such that the power  in each 30 ohms is (1/4) th of total power.
$$\mathrm{Two}\:\mathrm{30}\:\mathrm{ohms}\:\mathrm{resistor}\:\mathrm{are}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{parallel},\:\mathrm{what}\:\mathrm{should}\:\mathrm{be}\:\mathrm{the}\:\mathrm{resistance} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{series}\:\mathrm{with}\:\mathrm{this}\:\mathrm{parallel}\:\mathrm{combination}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{power} \\ $$$$\mathrm{in}\:\mathrm{each}\:\mathrm{30}\:\mathrm{ohms}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{th}\:\mathrm{of}\:\mathrm{total}\:\mathrm{power}. \\ $$
Answered by ajfour last updated on 06/Jun/17
(P/4)=I^2 R  ,   P=2I^2 R+(2I)^2 x  ⇒  4I^2 R=2I^2 R+4I^2 x         x=((2I^2 R)/(4I^2 )) = (R/2) =((30Ω)/2) =15Ω .
$$\frac{{P}}{\mathrm{4}}={I}^{\mathrm{2}} {R}\:\:,\:\:\:{P}=\mathrm{2}{I}^{\mathrm{2}} {R}+\left(\mathrm{2}{I}\right)^{\mathrm{2}} {x} \\ $$$$\Rightarrow\:\:\mathrm{4}{I}^{\mathrm{2}} {R}=\mathrm{2}{I}^{\mathrm{2}} {R}+\mathrm{4}{I}^{\mathrm{2}} {x} \\ $$$$\:\:\:\:\:\:\:{x}=\frac{\mathrm{2}{I}^{\mathrm{2}} {R}}{\mathrm{4}{I}^{\mathrm{2}} }\:=\:\frac{{R}}{\mathrm{2}}\:=\frac{\mathrm{30}\Omega}{\mathrm{2}}\:=\mathrm{15}\Omega\:. \\ $$
Commented by tawa tawa last updated on 06/Jun/17
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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