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Find the expession for the equivalent emf & internal resistanceof the series combination of cells. |
Answer» Solution :(i) Suppose n cells, each of emf `xi` volts and internal resistance r ohms are connected in series with an external resistance R.  (ii) The total emf of the battery `n xi` The total resistance in the CIRCUIT `NR+R` by Ohm's law, the currentin the circuit is `I=("total emf")/("total resistance")=(n xi)/(nr+R)` Case (a) if `r lt lt R`, then, `I=(n xi)/(R)~~nI_(1)` (iii) Where `I_(1)` is the current due to a single cell `(I_(1)=(xi)/(R))` Thus, if r is negligible when COMPAREDTO R the current supplied by the battery is n times that supplied by a single cell. Case (b) If ` r lt lt R, I=( n xi)/(nr)~~(xi)/(r)` (iv) It is the current due to a single cell. That is , current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells. (V) Thus series connection of cells is advantageous only when the effective internal resistanceof the cell is negaligibly small compared with R. |
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