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Question-190216




Question Number 190216 by MWSuSon last updated on 29/Mar/23
Commented by MWSuSon last updated on 29/Mar/23
i really don't understand physics
Commented by JDamian last updated on 30/Mar/23
nor Google and Wikipedia
Commented by talminator2856792 last updated on 31/Mar/23
 dont you get given the formulas.
$$\:\mathrm{dont}\:\mathrm{you}\:\mathrm{get}\:\mathrm{given}\:\mathrm{the}\:\mathrm{formulas}. \\ $$
Answered by JDamian last updated on 29/Mar/23
D) 0.75 μF
$$\left.{D}\right)\:\mathrm{0}.\mathrm{75}\:\mu\mathrm{F} \\ $$
Commented by MWSuSon last updated on 29/Mar/23
detailed solution?
Answered by CElcedricjunior last updated on 30/Mar/23
(1/r_1 )=(1/3)+(1/6)+(1/2)=1  =>r_1 =1  (1/r_2 )=(2/4)=(1/2)=>r_2 =2 ■cedric  (1/R)=(1/(r_1 +r_2 ))+(2/2)=(1/3)+1=(4/3)  =>R=(3/4)=0.75𝛍F      ★junior
$$\frac{\mathrm{1}}{\boldsymbol{{r}}_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{1} \\ $$$$=>\boldsymbol{{r}}_{\mathrm{1}} =\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\boldsymbol{{r}}_{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}}=>\boldsymbol{{r}}_{\mathrm{2}} =\mathrm{2}\:\blacksquare{cedric} \\ $$$$\frac{\mathrm{1}}{\boldsymbol{{R}}}=\frac{\mathrm{1}}{\boldsymbol{{r}}_{\mathrm{1}} +\boldsymbol{{r}}_{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$=>\boldsymbol{{R}}=\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{0}.\mathrm{75}\boldsymbol{\mu{F}}\:\:\:\:\:\:\bigstar\boldsymbol{{junior}} \\ $$

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