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Question-151832

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Question Number 151832 by DELETED last updated on 23/Aug/21
Answered by DELETED last updated on 23/Aug/21
57). R_s =2R+2R=4R (1/R_p )=(1/(2R))+(1/(4R))=(3/(4R)) R_p =((4R)/3) R_(total) =R+((4R)/3)=((7R)/3) I_(total) =(E/((7R)/3))=((3E)/(7R)) I_t =I_p =((3E)/(7R)) V_(paralel) =I_p ×R_p =((3E)/(7R))×((4R)/3)=((4E)/7) V_(2R) =((4E)/3) →I_(2R) =(((4E)/3)/(2R)) =((4E)/(6R))=((2E)/(3R))// o
$$\left.\mathrm{57}\right).\:\mathrm{R}_{\mathrm{s}} =\mathrm{2R}+\mathrm{2R}=\mathrm{4R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{R}_{\mathrm{p}} }=\frac{\mathrm{1}}{\mathrm{2R}}+\frac{\mathrm{1}}{\mathrm{4R}}=\frac{\mathrm{3}}{\mathrm{4R}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{R}_{\mathrm{p}} =\frac{\mathrm{4R}}{\mathrm{3}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{R}_{\mathrm{total}} =\mathrm{R}+\frac{\mathrm{4R}}{\mathrm{3}}=\frac{\mathrm{7R}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}_{\mathrm{total}} =\frac{\mathrm{E}}{\frac{\mathrm{7R}}{\mathrm{3}}}=\frac{\mathrm{3E}}{\mathrm{7R}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}_{\mathrm{t}} =\mathrm{I}_{\mathrm{p}} =\frac{\mathrm{3E}}{\mathrm{7R}} \\ $$$$\:\:\:\:\mathrm{V}_{\mathrm{paralel}} =\mathrm{I}_{\mathrm{p}} ×\mathrm{R}_{\mathrm{p}} =\frac{\mathrm{3E}}{\mathrm{7R}}×\frac{\mathrm{4R}}{\mathrm{3}}=\frac{\mathrm{4E}}{\mathrm{7}} \\ $$$$\:\:\:\:\:\mathrm{V}_{\mathrm{2R}} =\frac{\mathrm{4E}}{\mathrm{3}}\:\rightarrow\mathrm{I}_{\mathrm{2R}} =\frac{\frac{\mathrm{4E}}{\mathrm{3}}}{\mathrm{2R}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4E}}{\mathrm{6R}}=\frac{\mathrm{2E}}{\mathrm{3R}}// \\ $$$$ \\ $$$$\mathrm{o} \\ $$
Answered by DELETED last updated on 23/Aug/21
58). Q=m.c.ΔT=4000×4,2×22 =400×42×22 joule .....(1) W=i^2 .R.t=((V/R))^2 .R.t =((V^2 ×t)/R) =((220^2 ×t)/(55)) .....(2) (1)=(2) 400×42×22=((220×220×t)/(55)) t=((400×42×22×55)/(220×220)) = ((4×42×5×11)/(2×11)) =420 detik=((420)/(60))=7 menit
$$\left.\mathrm{58}\right). \\ $$$$\mathrm{Q}=\mathrm{m}.\mathrm{c}.\Delta\mathrm{T}=\mathrm{4000}×\mathrm{4},\mathrm{2}×\mathrm{22} \\ $$$$\:\:\:\:=\mathrm{400}×\mathrm{42}×\mathrm{22}\:\mathrm{joule}\:…..\left(\mathrm{1}\right) \\ $$$$\mathrm{W}=\mathrm{i}^{\mathrm{2}} .\mathrm{R}.\mathrm{t}=\left(\frac{\mathrm{V}}{\mathrm{R}}\right)^{\mathrm{2}} .\mathrm{R}.\mathrm{t} \\ $$$$\:\:\:\:\:=\frac{\mathrm{V}^{\mathrm{2}} ×\mathrm{t}}{\mathrm{R}}\:=\frac{\mathrm{220}^{\mathrm{2}} ×\mathrm{t}}{\mathrm{55}}\:…..\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)=\left(\mathrm{2}\right) \\ $$$$\mathrm{400}×\mathrm{42}×\mathrm{22}=\frac{\mathrm{220}×\mathrm{220}×\mathrm{t}}{\mathrm{55}} \\ $$$$\mathrm{t}=\frac{\mathrm{400}×\mathrm{42}×\mathrm{22}×\mathrm{55}}{\mathrm{220}×\mathrm{220}} \\ $$$$\:\:=\:\frac{\mathrm{4}×\mathrm{42}×\mathrm{5}×\mathrm{11}}{\mathrm{2}×\mathrm{11}} \\ $$$$\:\:=\mathrm{420}\:\mathrm{detik}=\frac{\mathrm{420}}{\mathrm{60}}=\mathrm{7}\:\mathrm{menit} \\ $$
Answered by DELETED last updated on 23/Aug/21
60). P=i^2 .R i^2 =(P/R) =((12×10^6 )/(40))=300.000 i=100(√(30)) A V=i.R=100(√(30)) ×40 =4000(√(3 )) volt
$$\left.\mathrm{60}\right).\:\mathrm{P}=\mathrm{i}^{\mathrm{2}} .\mathrm{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{i}^{\mathrm{2}} =\frac{\mathrm{P}}{\mathrm{R}}\:=\frac{\mathrm{12}×\mathrm{10}^{\mathrm{6}} }{\mathrm{40}}=\mathrm{300}.\mathrm{000} \\ $$$$\:\:\:\:\:\:\mathrm{i}=\mathrm{100}\sqrt{\mathrm{30}}\:\mathrm{A} \\ $$$$\:\:\:\:\:\:\mathrm{V}=\mathrm{i}.\mathrm{R}=\mathrm{100}\sqrt{\mathrm{30}}\:×\mathrm{40} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{4000}\sqrt{\mathrm{3}\:}\:\mathrm{volt} \\ $$

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