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Explain 'Mixed Connection' of cells and derive an expression for its equivalent emf and current. |
Answer» Solution :n. cells of emf `epsilon_(1) , epsilon_(2) ....epsilon_(n) and r_(1) , r_(2) ....r_(n)`INTERNAL resistance are connected in a series. .m. rows of above series are joined in parallel toform a mixed connection as shown in figure. `rArr` Now total emf of each series is, `epsilon = sum_(i=l)^(n) epsilon_(i)` `rArr` If total (EQUIVALENT) internal resistance of mixed connection is r then `(1)/(r) = (1)/(r.) + (1)/(r.) + (1)/(r.) ` + ..... m times = `(m)/(r.) = (m)/(SUM r_(i)) ` `therefore r = (sum r_(i))/(m)` `rArr` All rows are parallel to each other, So total emf `epsilon = sum epsilon_(1)` `therefore` CurrentI =` (" Total emf" )/("Total resistance " )` `I = (sum epsilon_(i))/(R + (sum r_(i))/(m) ) ` where R is the series resistance given in the CIRCUIT. |
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