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What is parallel connections of cells? Derive the equivalent equation of parallel connections of two cells. |
Answer» Solution :`rArr` If positive terminal of given CELLS are connected at one POINT and negative terminal of cell connected with second point then such connection is called parallel connection of cells.  `rArr` As shown in figure, two cells with emf `epsilon_(1) and epsilon_(2)` and respectively internal resistance `r_(1) and r_(2)` are connected in parallel between `B_(1) and B_(2)` . `rArr` Current in cell with emf `epsilon_(1) " is " I_(1) ` and with emf `epsilon_(2)" is " I_(2)` . ` rArr` Total current at `B_(1)` be, `I = I_(1) + I_(2) ""` .... (1) `rArr` Let potential at `B_(1) and B_(2)` be `V(B_(1)) and V(B_(2))`respectively. `rArr` Potential difference between two cell `V(B_(1)) - V(B_(2))` will be equal , `V(B_(1)) - V(B_(2)) - epsilon_(1) - I_(1) r_(1)` `V(B_(1)) - V(B_(2)) - epsilon_(2) - I_(2) r_(2)` `V(B_(1))- V(B_(2)) = V ` ` V = epsilon_(1) - I_(1) r_(1)` `I_(1) = (epsilon_(1)- V)/(r_(1)) ""` ....(2) SIMILARLY, `I_(2) = (epsilon_(2) - V)/(r_(2)) ""`....(3) `I = I_(1) + I_(2)` `= (epsilon_(1) - V) /(r_(1)) + (epsilon_(2) - V)/(r_(2))` `= (epsilon_(1))/(r_(1)) - (V)/(r_(1)) + (epsilon_(2))/(r_(2)) - (V)/(r_(2))` `therefore I = (epsilon_(1))/(r_(1)) + (epsilon_(2))/(r_(2)) - V((1)/(r_(1)) + (1)/(2)) ` `V ((r_(1) + r_(2))/(r_(1) r_(2))) = ((epsilon_(1) r_(2) + epsilon_(2) r_(1))/(r_(1) r_(2)) )- 1 ` `therefore V ((r_(1) + r_(2))/(r_(1) r_(2))) = ((epsilon_(1) r_(2) + epsilon_(2) r_(1))/(r_(1) r_(2)) )- 1 ` `therefore V = ((epsilon_(1) r_(2) + epsilon_(2) r_(1))/(r_(1) r_(2)) xx (r_(1) r_(2))/(r_(1) + r_(2)) ) - I ((r_(1) r_(2))/(r_(1) + r_(2)) )` `therefore V = (epsilon_(1) r_(2) + epsilon_(2) r_(1))/(r_(1) r_(2))- I ( (r_(1) r_(2))/(r_(1) + r_(2)) ) "" `....(4) Let cell with emf `E_(eq)` and internal resistance req is connected between `B_(1) and B_(2) `then, `V= epsilon_(eq) - Ir_(eq) "" ` ... (5) Comparing equation (4) and (5), `epsilon_(eq) = (epsilon_(1) r_(2) + epsilon_(2) r_(1))/(r_(1) r_(2)) "" ` ..... (6) `r_(eq) = (r_(1) r_(2))/(r_(1) + r_(2)) "" ` (7) By using equation (6) and (7), `(epsilon_(eq))/(r_(eq)) = (epsilon_(1))/(r_(1)) + (epsilon_(2))/(r_(2)) ` `(I)/(r_(eq)) = (1)/(r_(1)) + (1)/(r_(2))` here we have CONSIDERED that positive terminal of both cell are joined together. If negative terminal of second cell is connected to positive terminal of first cell then in equation `epsilon_(2) ` should be REPLACED by `- epsilon_(2)` . `rArr` If .n. cell are connected in parallel then, equivalent internal resistance, `(1)/(r_(eq)) = (1)/(r_(1)) + (1)/(r_(2)) + .... + (1)/(r_(n))` and `(epsilon_(eq))/(r_(eq) )= (epsilon_(1))/(r_(1)) + (epsilon_(2))/(r_(2)) + ..... + (epsilon_(n))/(r_(n))` `rArr`Special case : When two cell having equal emf and equal resistance are connected in parallel equivalent emf will be equal to individual cell (one cell). `epsilon_(eq)= (epsilon_(1) r_(1) + epsilon_(2) r_(2))/(r_(1) + r_(2))` Let `epsilon_(1)= epsilon_(2) = epsilon and r_(1) = r_(2) = r ` `thereforeepsilon_(eq) = (epsilon r + epsilon r)/(r+ r)` = `(2 epsilon r )/(2 r)` `therefore epsilon_(eq) = epsilon` |
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