Question-219720

Question Number 219720 by Spillover last updated on 01/May/25

Answered by mr W last updated on 01/May/25

(R_2 /x)=((R_1 +R_2 )/a)=((R_2 +R_3 )/b)=k, say ⇒R_2 =kx ⇒R_1 +R_2 =ka ⇒R_1 =k(a−x) ⇒R_2 +R_3 =kb ⇒R_3 =k(b−x) ((R_2 −R_1 )/a)=((R_3 −R_2 )/b) ((kx−k(a−x))/a)=((k(b−x)−kx)/b) ((2x−a)/a)=((b−2x)/b) ((2x)/a)+((2x)/b)=2 ⇒(1/a)+(1/b)=(1/x) ✓

$$\frac{{R}_{\mathrm{2}} }{{x}}=\frac{{R}_{\mathrm{1}} +{R}_{\mathrm{2}} }{{a}}=\frac{{R}_{\mathrm{2}} +{R}_{\mathrm{3}} }{{b}}={k},\:{say} \\ $$$$\Rightarrow{R}_{\mathrm{2}} ={kx} \\ $$$$\Rightarrow{R}_{\mathrm{1}} +{R}_{\mathrm{2}} ={ka}\:\Rightarrow{R}_{\mathrm{1}} ={k}\left({a}−{x}\right) \\ $$$$\Rightarrow{R}_{\mathrm{2}} +{R}_{\mathrm{3}} ={kb}\:\Rightarrow{R}_{\mathrm{3}} ={k}\left({b}−{x}\right) \\ $$$$\frac{{R}_{\mathrm{2}} −{R}_{\mathrm{1}} }{{a}}=\frac{{R}_{\mathrm{3}} −{R}_{\mathrm{2}} }{{b}} \\ $$$$\frac{{kx}−{k}\left({a}−{x}\right)}{{a}}=\frac{{k}\left({b}−{x}\right)−{kx}}{{b}} \\ $$$$\frac{\mathrm{2}{x}−{a}}{{a}}=\frac{{b}−\mathrm{2}{x}}{{b}} \\ $$$$\frac{\mathrm{2}{x}}{{a}}+\frac{\mathrm{2}{x}}{{b}}=\mathrm{2} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{{x}}\:\checkmark \\ $$

Commented by Spillover last updated on 01/May/25