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Question Number 216016 by Tawa11 last updated on 25/Jan/25

Answered by som(math1967) last updated on 25/Jan/25

$$current20\mathrm{\Omega }whenbothswitch$$$$oppen=\frac{6}{80}=\frac{3}{40}amp$$$$currentflowcctwhen{s}_1{s}_{2}both$$$$close6\mathbf{÷}(50+\frac{20\times R}{20+R})$$$$=6\mathbf{÷}\left(\frac{1000+70R}{20+R}\right)$$$$=\frac{6(20+R)}{(1000+70R)}$$$$pdat50\mathrm{\Omega }\frac{300(20+R)}{10(100+7R)}$$$$pdat20\mathrm{\Omega }6-\frac{30(20+R)}{100+7R}$$$$\therefore 20\times \frac{3}{40}=6-\frac{30(20+R)}{100+7R}$$$$\Rightarrow \frac{30(20+R)}{100+7R}=6-\frac{3}{2}$$$$\Rightarrow \frac{30(20+R)}{100+7R}=\frac{9}{2}$$$$\Rightarrow 400+20R=300+21R$$$$\therefore R=100\mathrm{\Omega }$$

Commented by Tawa11 last updated on 25/Jan/25

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$$$\mathrm{I}\mathrm{really}\mathrm{appreciate}.$$

Commented by Tawa11 last updated on 28/Jan/25

$$\mathrm{Sir},\mathrm{please}\mathrm{help}.$$$$\mathrm{Q}216153$$

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