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Explain the effective internal resistance of cells connected in parallel combination. Compare the results to the external resistance. |
Answer» Solution :Cells in SERIES Several cells can be connected to form a battery. In series connection, the NEGATIVE terminal of one cell is connected to the positive terminal of the second cell, the negative terminal terminal of second cell is connected to the positive terminal of the third cell and so on. The free positive terminal of the first cell and the free negative terminal of the last cell become the terminals of the battery.  SUPPOSE n cells, each of emf `xi` volts and internal resistance r ohms are connected in series with an external resistance R. The total emf of the battery `=n xi` The total resistance in the circuit `=nr+R` By Ohm.s law, the current in the circuit is `I=("total emf")/("total resistance")=(n xi)/(n r +R)"" ...(1)` Case (a) If `r lt lt R`, then, `I=(n xi)/(R) nI_(1)=nI_(1)"" ...(2)` where, `I_(1)` is the current due to a single cell `(I_(1)=(xi)/(R))` Thus, if r is negligible when compared to R the current supplied by the battery is n TIMES that supplied by a single cell. Case (b) If `r gt gt R "" I=(n xi)/(nr)=(xi)/(R )"" ...(3)` 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. Thus series connection of cells is advantageous only when the effective internal resistance of the cells is negligibly small compared with R. |
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