
Two non-ideal batteries are connected in parallel. Consider the following statements:
(A) The equivalent emf is smaller than either of the two EMFs.
(B) The equivalent internal resistance is smaller than either of the two internal resistances.
A. both A and B are correct
B. A is correct but B is wrong
C. B is correct but A is wrong
D. both A and B are wrong
Answer
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Hint: The charges have the property through which there is movement from a direction of low voltage to high voltage. This also helps us to know when a short circuit arises and also when a very high amount of current flows. Hence we will be able to know whether the equivalent emf and the internal resistance are smaller.
Complete step by step solution:
When two non-ideal batteries are linked in parallel, their corresponding emf is as given below
\[{E_{eq}} = \dfrac{{\dfrac{{{E_1}}}{{{r_1}}} + \dfrac{{{E_2}}}{{{r_2}}}}}{{\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}}}}\]
$ \Rightarrow {E_{eq}} = \dfrac{{{E_1}{r_1} + {E_2}{r_2}}}{{{r_1} + {r_2}}}$
If emf of both the cells is equal, here we can take E
\[{E_{eq}} = \dfrac{{E{r_2} + E{r_1}}}{{{r_1} + {r_2}}}\]
Hence we can say that
\[{E_{eq}} = E\dfrac{{{r_2} + {r_1}}}{{{r_1} + {r_2}}} = E\]
so, equivalent emf is equivalent to emf of the single cell.
Likewise, when two non-ideal batteries are linked in parallel, their equivalent resistance is
\[{r_{eq}} = \dfrac{{{r_1}{r_2}}}{{{r_1} + {r_2}}}\]
Therefore, the equivalent internal resistance is smaller than both of the two internal resistances.
Hence, B, as well as A, is correct.
Note:
When two non-ideal batteries of unequal emf are connected, the current will start flowing between the two batteries from higher voltage battery into lower voltage battery. Once the voltage gets balanced, the current flow will stop. However, if there is a big voltage difference there can be severe damage, a large current may run through the battery possessing lower voltage. It may also overheat because of the resistance inside and may explode.
Complete step by step solution:
When two non-ideal batteries are linked in parallel, their corresponding emf is as given below
\[{E_{eq}} = \dfrac{{\dfrac{{{E_1}}}{{{r_1}}} + \dfrac{{{E_2}}}{{{r_2}}}}}{{\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}}}}\]
$ \Rightarrow {E_{eq}} = \dfrac{{{E_1}{r_1} + {E_2}{r_2}}}{{{r_1} + {r_2}}}$
If emf of both the cells is equal, here we can take E
\[{E_{eq}} = \dfrac{{E{r_2} + E{r_1}}}{{{r_1} + {r_2}}}\]
Hence we can say that
\[{E_{eq}} = E\dfrac{{{r_2} + {r_1}}}{{{r_1} + {r_2}}} = E\]
so, equivalent emf is equivalent to emf of the single cell.
Likewise, when two non-ideal batteries are linked in parallel, their equivalent resistance is
\[{r_{eq}} = \dfrac{{{r_1}{r_2}}}{{{r_1} + {r_2}}}\]
Therefore, the equivalent internal resistance is smaller than both of the two internal resistances.
Hence, B, as well as A, is correct.
Note:
When two non-ideal batteries of unequal emf are connected, the current will start flowing between the two batteries from higher voltage battery into lower voltage battery. Once the voltage gets balanced, the current flow will stop. However, if there is a big voltage difference there can be severe damage, a large current may run through the battery possessing lower voltage. It may also overheat because of the resistance inside and may explode.
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