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Question Number 33800 by tawa tawa last updated on 25/Apr/18

Answered by ajfour last updated on 25/Apr/18

$$I_3=I_{cell}=\frac{6V}{(\frac{5\times 4}{5+4}+3+0.5)}$$$$=\frac{6}{5.72}\approx 1.05A$$$$I_5=\frac{4}{9}\times 1.05=0.47A$$$$I_4=0.58A$$$$V_3=I_3\times 3=3.15V$$$$V_{4,5}=I_{cell}\times \frac{20}{9}=1.05\times \frac{20}{9}=2.33V.$$

Commented by tawa tawa last updated on 25/Apr/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by tawa tawa last updated on 25/Apr/18

$$\mathrm{please}\mathrm{sir}.\mathrm{why}\mathrm{is}\mathrm{it}\frac{4}{9}\times 1.05\mathrm{amd}\mathrm{please}\mathrm{how}\mathrm{is}{\mathrm{I}}_4=0.58.\mathrm{thanks}\mathrm{sir}.$$

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