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When a resistance of 12 ohm is connected across a cell, its terminal potential difference is balanced by 120 cm length of potentiometer wire. When the resistance of 18 ohm is connected across the same cell, the balancing length is 150 cm. Find the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell. |
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Answer» We have, Case I : `R=12Omegal_(2)=120cm` Case II : `R=18Omega,l_(2)=150cm` From first condition, Internal resistance (r) `=R((l_(1)-l_(2))/(l_(2)))` `r=12((l_(1)-l_(2))/(120))" "......(i)` From second condition, Internal resistance (r) `=R((l_(1)-l_(2))/(l_(2)))` `=18((l_(1)-150)/(150))" "......(ii)` From equations (i) and (ii) `12((l_(1)-120)/(120))=18((l_(1)-150)/(150))` `5(l_(1)-120)=6(l_(1)-150)` `5l_(1)-600=6l_(1)-900` `l_(1)=300cm.` Putting the value of `l_(2)` in equation (i) we, get `r=12((300-120)/(120))` `r=18Omega` Therefore, the balancing lenght is 300 cm and internal resistance is `18Omega`. |
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