A) \[R=\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{2}}-{{r}_{1}}}\]
B) \[R={{r}_{2}}\left( \frac{r{{+}_{1}}{{r}_{2}}}{{{r}_{2}}-{{r}_{1}}} \right)\]
C) \[R=\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{2}}+{{r}_{1}}}\]
D) \[R={{r}_{2}}-{{r}_{1}}\]
Correct Answer: D
Solution :
Let E be the emf of each source. When they are connected in series, then the current in the circuit is given by \[L=\frac{nh}{2\pi }=\frac{2h}{2\pi }=\frac{h}{\pi }\] \[\because \] So, potential drop across the cell of intermal resistance \[\lambda =\frac{h}{p}\] is \[\therefore \] Hence, \[\lambda \alpha \frac{1}{p}\] \[\Rightarrow \] So, \[\frac{\Delta p}{p}=-\frac{\Delta \lambda }{\lambda }\]You need to login to perform this action.
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