InterviewSolution
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InterviewSolution
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(a) Two cells of emfe and € have their internal resistances epsi_(1) and epsi_(2) respectively. Deduce an expression for the equivalent emf and internal resistance of their parallel combination when connected across an external resistance R. Assume that the two cells are supporting each other. (b) In case the two cells are identical, each of emf e= 5V and internal resistance r= 2Omega, calculate the voltage across the external resistance R= 10Omega. |
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Answer» Solution :(a). Consider a parallel combination of two cells. Let `epsi_(1) and epsi_(2)` be the emfs of two cells and `r_(1) and r_(2)` their respective INTERNAL RESISTANCES. `I_1 and I_2` are the currents leaving the positive ELECTRODES of two cells and at Junction `B_1`, these currents are added up so that net CURRENT `l=l_(1)+l_(2)` ..........(i) Let `V(B_1) and V(B_2)` be the potentials at `B_1 and B_2,` respectively, then considering first cell, we have `V=V(B_(1))-V(B_(2))=epsi_(1)-l_(1) r_(1) rArr l_(1)=(epsi_(1)-V)/(r_(1))` ............(ii)  and considering second cell, we have `V=V(B_(1))-V(B_(2))=epsi_(2)-l_(1) r_(2) rArr l_(2)=(epsi_(2)-V)/r_(2) ..........(iii)` `l=l_(1)+l_(2) =(epsi_(1)-V)/(r_(1))+(epsi_(2)-V)/(r_(2)) =(epsi_1/r_1+epsi_2/r_2)-V(1/r_(1)+1/r_(2))` `rArr V=[(epsi_1/r_1+epsi_2/r_2) -I]//(1/r_(1)+1/r_(2))=((epsi_(1)r_(2)+epsi_(2)r_(1))/(r_(1)+r_(2))-I (r_(1)r_(2))/(r_(1)+r_(2))` .......(iv) If instead of two cells we join a single cell having equivalent emf `e_(eq)` and equivalent internal resistance `e_(eq)` then we have `V=epsi_(eq)-I r_(eq)` .........(v) Thus, comparing (iv) and (v), we get `e_(eq)=(epsi_(1)r_(2)+epsi_(2)r_(1))/(r_(1)+r_(2)) and r_(eq)=(r_(1)r_(2))/(r_(1)+r_(2))` Alternately, we can write `1/r_(eq)=1/r_(1)+1/r_(2) and (epsi_(eq))/(r_(eq))=epsi_1/r_1+epsi_(2)/r_(2)` (b) As per question emf of cells `epsi_(1)=epsi_(2)=5V`, internal resistance of cells `r_(1)=r_(2)=2 Omega` and external resistance R=10 V. Effective emf of cell combination `epsi=(epsi_(1)r_(2)+epsi_(2)r_(1))/(r_(1)+r_(2))=(5 xx 2 +5 xx 2)/(2+2)=5V` and effective internal resistance `r=(r_(1)r_(2))/(r_(1)+r_(2))=(2 xx 2)/(2+2) =1 Omega` Current in the circuit `I=epsi/(R+r)=(5)/(10+1)=5/(11) A` Voltage across external resistance, `V=IR=(5)/(11) xx 10=4.55V` |
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