Question Number 216016 by Tawa11 last updated on 25/Jan/25
Answered by som(math1967) last updated on 25/Jan/25
$${current}\:\mathrm{20}\Omega\:{when}\:{both}\:{switch} \\ $$$${oppen}=\frac{\mathrm{6}}{\mathrm{80}}=\frac{\mathrm{3}}{\mathrm{40}}{amp} \\ $$$${current}\:{flow}\:{cct}\:{when}\:{s}_{\mathrm{1}} \:{s}_{\mathrm{2}} \:{both} \\ $$$${close}\:\mathrm{6}\boldsymbol{\div}\left(\mathrm{50}+\frac{\mathrm{20}×{R}}{\mathrm{20}+{R}}\right) \\ $$$$\:=\mathrm{6}\boldsymbol{\div}\left(\frac{\mathrm{1000}+\mathrm{70}{R}}{\mathrm{20}+{R}}\right) \\ $$$$=\frac{\mathrm{6}\left(\mathrm{20}+{R}\right)}{\left(\mathrm{1000}+\mathrm{70}{R}\right)} \\ $$$${pd}\:\:{at}\:\mathrm{50}\Omega\:\frac{\mathrm{300}\left(\mathrm{20}+{R}\right)}{\mathrm{10}\left(\mathrm{100}+\mathrm{7}{R}\right)} \\ $$$$\:{pd}\:{at}\:\mathrm{20}\Omega\:\:\:\:\mathrm{6}−\frac{\mathrm{30}\left(\mathrm{20}+{R}\right)}{\mathrm{100}+\mathrm{7}{R}} \\ $$$$\:\therefore\:\mathrm{20}×\frac{\mathrm{3}}{\mathrm{40}}=\mathrm{6}−\frac{\mathrm{30}\left(\mathrm{20}+{R}\right)}{\mathrm{100}+\mathrm{7}{R}} \\ $$$$\:\Rightarrow\frac{\mathrm{30}\left(\mathrm{20}+{R}\right)}{\mathrm{100}+\mathrm{7}{R}}=\mathrm{6}−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{30}\left(\mathrm{20}+{R}\right)}{\mathrm{100}+\mathrm{7}{R}}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{400}+\mathrm{20}{R}=\mathrm{300}+\mathrm{21}{R} \\ $$$$\therefore{R}=\mathrm{100}\Omega \\ $$
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
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{help}. \\ $$$$\mathrm{Q216153} \\ $$