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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-




Question Number 14963 by tawa tawa last updated on 06/Jun/17
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
$$\mathrm{A}\:\mathrm{resistor}\:\mathrm{R}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{in}\:\mathrm{series}\:\mathrm{with}\:\mathrm{a}\:\mathrm{parallel}\:\mathrm{combination}\:\mathrm{of}\:\mathrm{two}\:\mathrm{resistors} \\ $$$$\mathrm{of}\:\mathrm{24}\:\mathrm{and}\:\mathrm{8}\:\mathrm{ohms}\:.\:\mathrm{The}\:\mathrm{total}\:\mathrm{power}\:\mathrm{disipated}\:\mathrm{in}\:\mathrm{the}\:\mathrm{circuit}\:\mathrm{is}\:\mathrm{64}\:\mathrm{watt}\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{applied}\:\mathrm{voltage}\:\mathrm{is}\:\mathrm{24}\:\mathrm{volt}.\mathrm{Find}\:\mathrm{R} \\ $$
Answered by ajfour last updated on 06/Jun/17
 P=(V^2 /R_(eq) ) , R_(eq) =R+((24×8)/(32))Ω =R+6Ω ;  ⇒ 64W=(((24V)^2 )/((R+6Ω)))  or,      R+6Ω=((24×24)/(64)) Ω               R = 9Ω−6Ω =3Ω  .
$$\:{P}=\frac{{V}^{\mathrm{2}} }{{R}_{{eq}} }\:,\:{R}_{{eq}} ={R}+\frac{\mathrm{24}×\mathrm{8}}{\mathrm{32}}\Omega\:={R}+\mathrm{6}\Omega\:; \\ $$$$\Rightarrow\:\mathrm{64}{W}=\frac{\left(\mathrm{24}{V}\right)^{\mathrm{2}} }{\left({R}+\mathrm{6}\Omega\right)} \\ $$$${or},\:\:\:\:\:\:{R}+\mathrm{6}\Omega=\frac{\mathrm{24}×\mathrm{24}}{\mathrm{64}}\:\Omega \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{R}\:=\:\mathrm{9}\Omega−\mathrm{6}\Omega\:=\mathrm{3}\Omega\:\:. \\ $$
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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