Twelve cells,each having emf E volts,are conn | vedclass

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(D) Let the number of wrongly connected cells in the $12$-cell battery be $m$. Each wrongly connected cell cancels the emf of one correctly connected

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Let the number of wrongly connected cells in the $12$-cell battery be $m$. Each wrongly connected cell cancels the emf of one correctly connected cell. Thus,the effective emf of the $12$-cell battery is $E_{12} = (12 - m)E - mE = (12 - 2m)E$. When the $12$-cell battery and the $2$-cell battery (emf $2E$) aid each other,the total emf is $E_{total} = (12 - 2m)E + 2E = (14 - 2m)E$. The current is $i_1 = \frac{(14 - 2m)E}{R} = 3 	ext{ A}$ ...$(i)$ When they oppose each other,the total emf is $E_{total} = (12 - 2m)E - 2E = (10 - 2m)E$. The current is $i_2 = \frac{(10 - 2m)E}{R} = 2 	ext{ A}$ ...(ii) Dividing equation $(i)$ by (ii): $\frac{3}{2} = \frac{14 - 2m}{10 - 2m}$ $3(10 - 2m) = 2(14 - 2m)$ $30 - 6m = 28 - 4m$ $2 = 2m$ $m = 1$. Therefore,the number of wrongly connected cells is $1$.

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