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(b) For no current flow between $C$ and $D$ ${\left( {\frac{Q}{t}} ight)_{AC}} = {\left( {\frac{Q}{t}} ight)_{CB}}$

==> $\frac{{{K_1}A({	he

Vedclass · Accepted Answer

$ {K_1}{K_4} = {K_2}{K_3} $

Vedclass · Answer

(b) For no current flow between $C$ and $D$ ${\left( {\frac{Q}{t}} ight)_{AC}} = {\left( {\frac{Q}{t}} ight)_{CB}}$

==> $\frac{{{K_1}A({	heta _A} - {	heta _C})}}{l} = \frac{{{K_2}A({	heta _C} - {	heta _B})}}{l}$

==> $\frac{{{	heta _A} - {	heta _C}}}{{{	heta _C} - {	heta _B}}} = \frac{{{K_2}}}{{{K_1}}}$...$(i) $

Also ${\left( {\frac{Q}{t}} ight)_{AD}} = {\left( {\frac{Q}{t}} ight)_{DB}}$

==>$\frac{{{K_3}A({	heta _A} - {	heta _D})}}{l} = \frac{{{K_4}A({	heta _D} - {	heta _B})}}{l}$

==>$\frac{{{	heta _A} - {	heta _D}}}{{{	heta _D} - {	heta _B}}} = \frac{{{K_4}}}{{{K_3}}}$ ..$.(ii)$ 

It is given that ${	heta _C} = {	heta _D},$ hence from equation $(i)$ and $(ii)$ we get $\frac{{{K_2}}}{{{K_1}}} = \frac{{{K_4}}}{{{K_3}}}$ ==> ${K_1}{K_4} = {K_2}{K_3}$

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