question_answer

A cell of emf 1.1 V and internal resistance \[0.5\,\Omega \] is connected to a wire of resistance \[0.5\,\Omega .\] Another cell of the same emf is connected in series but the current in the wire remains the same. Find the internal resistance of the second cell.

Answer:

When two cells are connected in series as shown in below figure, Then,    \[{{V}_{AB}}=({{E}_{1}}+{{E}_{2}})-i({{r}_{1}}+{{r}_{2}})\] or         \[{{E}_{eq}}={{E}_{1}}+{{E}_{2}}\] and       \[{{r}_{eq}}={{r}_{1}}+{{r}_{2}}\] Given, net emf of cell = 1.1 V Net resistance \[=r+R=0.5+0.5=1\,\Omega \] \[\therefore \] By Ohm's law, \[I=\frac{V}{R}=\frac{1.1}{1}=1.1\,A\] For second condition, Net emf = 1.1 + 1.1= 2.2 V Net resistance,    \[R=1+r\] where, r is internal resistance of given second cell. \[\therefore \] By Ohm's law,             \[I=\frac{2.2}{1+r}\Rightarrow 1.1=\frac{2.2}{1+r}\] or         \[1.1+1.1r=2.2\] or         \[1.1r=2.2-1.1=1.1\] or         \[r=1\,\Omega \]