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Question-40918




Question Number 40918 by Tawa1 last updated on 29/Jul/18
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Jul/18
i am using another mdthod to find current through R_L   let i_1  current confined infirst mesh  i_2   in second mesh  i_3  in third mesh  l  10−6i_1 +6i_2 −3i_1 =0  9i_1 −6i_2 =10....eqn 1  9−4i_2 +4i_3 −6i_2 +6i_1 −10=0  6i_1 −10i_2 +4i_3 =1...eqn2  4−3i_3 −4i_3 +4i_2 −9=0  4i_2 −7i_3 =5...eqn 3  i_3 =((4i_2 −5)/7)  putting this value in eqn 2  6i_1 −10i_2 +4(((4i_2 −5)/7))=1  42i_1 −70i_2 +16i_2 −20=7  42i_1 −54i_2 =27  14i_1 −18i_2 =9...eqn4  9i_1 −6i_2 =10...eqn1  ×3  14i_1 −18i_2 =9  27i_1 −18i_2 =30  substruct...  −13i_1 =−21  i_1 =((21)/(13))A  is the required current passing thrlugh  R_L ....pls check
$${i}\:{am}\:{using}\:{another}\:{mdthod}\:{to}\:{find}\:{current}\:{through}\:{R}_{{L}} \\ $$$${let}\:{i}_{\mathrm{1}} \:{current}\:{confined}\:{infirst}\:{mesh} \\ $$$${i}_{\mathrm{2}} \:\:{in}\:{second}\:{mesh} \\ $$$${i}_{\mathrm{3}} \:{in}\:{third}\:{mesh} \\ $$$${l} \\ $$$$\mathrm{10}−\mathrm{6}{i}_{\mathrm{1}} +\mathrm{6}{i}_{\mathrm{2}} −\mathrm{3}{i}_{\mathrm{1}} =\mathrm{0} \\ $$$$\mathrm{9}{i}_{\mathrm{1}} −\mathrm{6}{i}_{\mathrm{2}} =\mathrm{10}….{eqn}\:\mathrm{1} \\ $$$$\mathrm{9}−\mathrm{4}{i}_{\mathrm{2}} +\mathrm{4}{i}_{\mathrm{3}} −\mathrm{6}{i}_{\mathrm{2}} +\mathrm{6}{i}_{\mathrm{1}} −\mathrm{10}=\mathrm{0} \\ $$$$\mathrm{6}{i}_{\mathrm{1}} −\mathrm{10}{i}_{\mathrm{2}} +\mathrm{4}{i}_{\mathrm{3}} =\mathrm{1}…{eqn}\mathrm{2} \\ $$$$\mathrm{4}−\mathrm{3}{i}_{\mathrm{3}} −\mathrm{4}{i}_{\mathrm{3}} +\mathrm{4}{i}_{\mathrm{2}} −\mathrm{9}=\mathrm{0} \\ $$$$\mathrm{4}{i}_{\mathrm{2}} −\mathrm{7}{i}_{\mathrm{3}} =\mathrm{5}…{eqn}\:\mathrm{3} \\ $$$${i}_{\mathrm{3}} =\frac{\mathrm{4}{i}_{\mathrm{2}} −\mathrm{5}}{\mathrm{7}}\:\:{putting}\:{this}\:{value}\:{in}\:{eqn}\:\mathrm{2} \\ $$$$\mathrm{6}{i}_{\mathrm{1}} −\mathrm{10}{i}_{\mathrm{2}} +\mathrm{4}\left(\frac{\mathrm{4}{i}_{\mathrm{2}} −\mathrm{5}}{\mathrm{7}}\right)=\mathrm{1} \\ $$$$\mathrm{42}{i}_{\mathrm{1}} −\mathrm{70}{i}_{\mathrm{2}} +\mathrm{16}{i}_{\mathrm{2}} −\mathrm{20}=\mathrm{7} \\ $$$$\mathrm{42}{i}_{\mathrm{1}} −\mathrm{54}{i}_{\mathrm{2}} =\mathrm{27} \\ $$$$\mathrm{14}{i}_{\mathrm{1}} −\mathrm{18}{i}_{\mathrm{2}} =\mathrm{9}…{eqn}\mathrm{4} \\ $$$$\mathrm{9}{i}_{\mathrm{1}} −\mathrm{6}{i}_{\mathrm{2}} =\mathrm{10}…{eqn}\mathrm{1}\:\:×\mathrm{3} \\ $$$$\mathrm{14}{i}_{\mathrm{1}} −\mathrm{18}{i}_{\mathrm{2}} =\mathrm{9} \\ $$$$\mathrm{27}{i}_{\mathrm{1}} −\mathrm{18}{i}_{\mathrm{2}} =\mathrm{30} \\ $$$${substruct}… \\ $$$$−\mathrm{13}{i}_{\mathrm{1}} =−\mathrm{21} \\ $$$${i}_{\mathrm{1}} =\frac{\mathrm{21}}{\mathrm{13}}{A}\:\:{is}\:{the}\:{required}\:{current}\:{passing}\:{thrlugh} \\ $$$${R}_{{L}} ….{pls}\:{check} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 30/Jul/18
God bless you sir. thanks.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{thanks}.\: \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 30/Jul/18
thank  you God shower its blessing to all...
$${thank}\:\:{you}\:{God}\:{shower}\:{its}\:{blessing}\:{to}\:{all}… \\ $$

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