A) \[\sqrt{{{r}_{1}}{{r}_{2}}}\]
B) \[{{r}_{1}}+{{r}_{2}}\]
C) \[{{r}_{1}}-{{r}_{2}}\]
D) \[\frac{{{r}_{1}}+{{r}_{2}}}{2}\]
Correct Answer: C
Solution :
There are two batteries with emf E each and the internal resistances \[{{r}_{1}}\]and \[{{r}_{2}}\]respectively. Hence we have \[I(R+{{r}_{1}}+{{r}_{2}})=2E,\] thus, \[I=\frac{2E}{2+{{r}_{1}}+{{r}_{2}}}\] Now the potential difference across the first cell would be equal to \[V=E-I{{r}_{1}}.\]From the question, V = 0, hence, \[E=I{{r}_{1}}=\frac{2E{{r}_{1}}}{R+{{r}_{1}}+{{r}_{2}}},\] thus, \[R+{{r}_{1}}+{{r}_{2}}=2{{r}_{1}},\] hence \[R={{r}_{1}}-{{r}_{2}}.\]You need to login to perform this action.
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