Solutions A,B,and C of the same strong electr | vedclass

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(A) Let the volumes be $v_{1}, v_{2}$,and $v_{3}$.The relationship for the equivalent conductance of a mixture is given by $\frac{v_{1}}{R_{1}} + \fra

Vedclass · Accepted Answer

$67.3$

Vedclass · Answer

(A) Let the volumes be $v_{1}, v_{2}$,and $v_{3}$.The relationship for the equivalent conductance of a mixture is given by $\frac{v_{1}}{R_{1}} + \frac{v_{2}}{R_{2}} + \frac{v_{3}}{R_{3}} = \frac{v_{1} + v_{2} + v_{3}}{R_{eq}}$.Given $R_{1} = 50 \ \Omega, R_{2} = 100 \ \Omega, R_{3} = 150 \ \Omega$.The volumes are in reciprocal proportion to their resistances: $v_{1} = \frac{1}{50}, v_{2} = \frac{1}{100}, v_{3} = \frac{1}{150}$.Substituting these values into the equation:$\frac{(1/50)}{50} + \frac{(1/100)}{100} + \frac{(1/150)}{150} = \frac{(1/50) + (1/100) + (1/150)}{R_{eq}}$.$\frac{1}{2500} + \frac{1}{10000} + \frac{1}{22500} = \frac{(6+3+2)/300}{R_{eq}}$.$\frac{36 + 9 + 4}{90000} = \frac{11/300}{R_{eq}}$.$\frac{49}{90000} = \frac{11}{300 	imes R_{eq}}$.$R_{eq} = \frac{11 	imes 90000}{300 	imes 49} = \frac{3300}{49} \approx 67.3 \ \Omega$.

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