InterviewSolution
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InterviewSolution
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For what value of load resistance, the power transfer is maximum when two identical cells each of emf E and internal resistance r are connected (a) in series (b) in parallel. |
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Answer» (a) When cells are connecteed in series to the external load resistance R, then effective emf `= E + E = 2E`. Total resistance of the circuit `= R + r + r` `= R + 2r` Current in load, `I= (2E)/(R + 2r)` Power delivered to the external load is, `P = I^(2)R = ((2E)/(R + 2r))^(2) R` Power deliverd to the external load will be maximum if external resistance = internal resistance of cells, i.e., `R = 2r`. Then maximum power, `P_(max) = ((2E)/(R + 2r))^(2) xx 2r = E^(2)/(2r)` (b) When cells are connected in parallel to the external load resistance R, then effective emf = E. Total resistance of the circuit `R + (r xx r)/(r+r) = (R + r/2)` Current in load, `I = E/((R + r//2))` Power delivered to the external load, `P = I^(2)R = (E/(R + r//2))^(2)R` Power is maximum when `R = r//2`. Then `P_(max) = ((E)/(r//2 + r//2))^(2) r//2 = E^(2)/(2r)` Note that in both the cases, maximum power deliverd to the loas is the same. |
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