Question

Describe the formula for the equivalent emf and internal resistance for the parallel combination of two cells with emf ε₁ and ​ε₂ and internal resistances r₁ and r₂ respectively. What is the corresponding formula for the series combination. Two cells of emf 1V, 2V and internal resistances 2 Ω and 1 Ω respectively are connected in (i) series (ii) parallel. What should be the external resistance in teh circuit so that the currnet through the resistance be the same in the two cases? In which case is more heat generated in the cells?

Answer 1

$$\begin{gathered} Equivalentemfofthecellwhenconnectedinseriesis \\ {E}_{equialent}=E_1+E_2 \\ {R}_{eqv}=r_1+r_2 \\ Equivalentemfofthecellwhenconnectedinparallelis \\ {E}_{equivalent}=\frac{E_1}{r_1}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}} \\ \frac{1}{R_{eqv}}=\frac{1}{r_1}+\frac{1}{r_2} \\ a)E_{equialent}=E_1+E_2=1+2=3V \\ {R}_{eqv}=r_1+r_2=1+2=3\Omega \\ b)E_{equivalent}=\frac{\frac{1}{2}+\frac{2}{1}}{\frac{1}{2}+\frac{1}{1}}=\frac{5}{3}V \\ \frac{1}{R_{eqv}}=\frac{1}{r_1}+\frac{1}{r_2}=\frac{1}{1}+\frac{1}{2}=\frac{3}{2}\Omega \\ {R}_{eqv}=\frac{2}{3}\Omega \\ Currentwhenaresistorisconnectedinseriescombinationofcellsis{i}_1=\frac{3}{3+R} \\ Currentwhenaresistorisconnectedinparallelcombinationofcellsis{i}_2=\frac{{\displaystyle \frac{5}{3}}}{R+{\displaystyle \frac{2}{3}}}=\frac{5}{3R+2} \\ Accordingtoquestion \\ \frac{3}{3+R}=\frac{5}{3R+2} \\ R=\frac{9}{4}\Omega \\ so{i}_1=\frac{3}{3+R}=\frac{3}{3+\frac{9}{4}}=\frac{12}{12+9}=\frac{12}{21}=\frac{4}{7}A \\ {i}_2=\frac{5}{3\times \frac{9}{4}+2}=\frac{20}{27+8}=\frac{20}{35}=\frac{4}{7}A \\ Heatproducedinseriescombinationofcellsis=(\frac{4}{7}{)}^2\times 3=\frac{48}{49}J \\ Heatproducedinparallelcombinationofcellsis=(\frac{4}{7}{)}^2\times \frac{2}{3}=\frac{32}{147}J \\ soheatproducedismoreinseriescombinationofcells. \\ Regards \end{gathered}$$Dear student