InterviewSolution
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InterviewSolution
| 1. |
Obtain an expression for the equivalent emf and internal resistance of two cells connected in parallel. |
Answer» Solution : LET `I_1,I_2` represent branch currents. Let V be the commonpotential ,so that , `I_1=(E_A-V)/r_1, I_2=(E_2-V)/r_2` hence, `I=(E_1-V)/r_1 + (E_2-V)/r_2` i.e., `I=(E_1/r_1 + E_2/r_2)-V (1/r_1+1/r_2)` Comparingthis with the terminalpotentialdifference, `V_E_(eq)-Ir_(eq)` i.e.,`V=E_(eq)-1/((1/r_(eq)))` For two CELLS in parallel COMBINATION , `(E_(eq))_p=(E_1/r_1 +E_2/r_2)/(1/r_1+1/r_2)=(E_1r_2 + E_2r_1)/(r_1+r_2)` `1/(r_(eq))_p =1/r_1+1/r_2`and main current=`i=E_(eq)/(R+r_(eq))` Note : (i) For .n. cells in parallel `V=[(sum_(r=1)^n E_i/r_i)/(sum_(i=1)^n 1/r_1)]-[I/(sum_(i=1)^n 1/r_1)]` , `(E_(eq))_"parallel" =[(sum_(r=1)^n E_i/r_i)/(sum_(i=1)^n 1/r_1)]-[I/(sum_(i=1)^n 1/r_1)]` and `(1/r_(eq))_(p)=sum_(i-1)^n (1/r_i)`. (ii) For .n. number of identicalcells E-emf of each CELL r-internalresistance of each cell. `(E_(eq))_p =(n(E/r))/(n(1/r))` i.e.,`(E_(eq))_p=E` and `(1/r_(eq))_p=n(1/r)` i.e. `(r_(eq))_p =r/n` Main current, `I=(nE)/(R+n/r)` and terminalpotential difference across cells V=IR. |
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