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36 cells each of internal resistance 0.5Omega and emf 1.5 V each are used to send current through an external circuit of 2Omega resistance. Find the best mode of grouping then for maximum current and the current through the external circuit. |
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Answer» Solution :LET, 36 cells are grouped in m ROWS, each row containing n cells . Then , MN = 36 Internal resistance in each row = nr , here, r = 0.5 `Omega` Internal resistance for m rows = `(nr)/(m)`. `therefore` Total resistance in the CIRCUIT = `(R + (nr)/(m))`, here, R = 2`Omega` The emf each row = nE = effective emf of m rows, here, E = 1.5 V `therefore` Current in external circuit, ` I = (nE)/(R + (nr)/(m)) = (mnE)/(mR + nr) = (mnE)/((sqrt(mR)- sqrt(nr))^(2) + 2 sqrt(mnRr))` AsmnRr = constant, I is maximum for `(sqrt(mR) - sqrt(nr))^(2) = `0 `therefore "" mR = nr or, (36)/(n) xx 2 = n xx 0.5` or, `"" n^(2) = (36 xx 2)/(0.5) = 144` Then, n = 12 and m = `(36)/(12) = 3 ` So, the best MODE of grouping is in 3 rows, each row having 12 cells. |
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