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Question Number 38939 by rahul 19 last updated on 01/Jul/18

Commented by rahul 19 last updated on 01/Jul/18

$$\mathrm{Pls}\mathrm{solve}\mathrm{this}\mathrm{problem}\mathrm{by}\mathrm{Nodal}\mathrm{Method}.$$$${\mathrm{Kirchhoff}}^{\prime }\mathrm{s}\mathrm{law}\mathrm{is}\mathrm{not}\mathrm{allowed}!$$

Commented by rahul 19 last updated on 01/Jul/18

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

Commented by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

Commented by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

$$plscheck...$$

Commented by rahul 19 last updated on 01/Jul/18

thank you sir ����

Commented by rahul 19 last updated on 01/Jul/18

pls check book sol. , is it correct ?

Commented by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

$$book{sol}^{n}iscorrect...ihavefoundmymistake$$$$andexpressed...$$

Commented by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

$$2i_1-5i_2-1=0$$$$2i_1-5\left(\frac{-1}{13}\right)-1=0Imademistakehereinthe$$$$previoussolution...plscheck$$$$2i_1=1-\frac{5}{13}=\frac{8}{13}so{i}_1=\frac{4}{13}$$$$pd=2(\frac{4}{13}-\frac{-1}{13})=\frac{10}{13}$$

Answered by ajfour last updated on 01/Jul/18

$$-2i_1+2-1-i_1=V_D-V_B=△V$$$$\Rightarrow i_1=\frac{1-△V}{3}$$$$2i_2=△V\Rightarrow i_2=\frac{△V}{2}$$$$-1-i_3-3i_3+3=△V$$$$\Rightarrow i_3=\frac{2-△V}{4}$$$$As{i}_2=i_1+i_3$$$$\frac{△V}{2}=\frac{1-△V}{3}+\frac{2-△V}{4}=0$$$$6△V=4-4△V+6-3△V$$$$\Rightarrow △\mathit{V}=\frac{10}{13}.$$

Commented by rahul 19 last updated on 01/Jul/18

Thank You sir ����

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