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In the secondary circuit of a potentiometer , a cell of internal resistance 1.5 Omega gives a balancing length of 52 cm. To get a balancing length of 40 cm , how much resistance is to be connected across the cell ? |
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Answer» Solution :`R= 1.5 Omegal_1= 52 CM , l_2 = 40 cm` `r/R = (l_1 - l_2)/(l_2)` Resistance `R=(r.l_2)/((l_1 - l_2)) =(1.5xx40)/((52-40))` `=(1.5xx40)/(12)= 5 Omega` |
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