Question

Three batteries of emf \[1V\] and internal resistance \[1\Omega \] each are connected as shown. Effective emf of combination between the points \[PQ\] is
  
A. zero
B. \[1V\]
C. \[2V\]
D. \[\left( {2/3} \right)V\]

Answer 1

Hint: Before we proceed into the problem, it is important to know the definitions of cells each of emf.
The total emf will be the same as that of individual emf. Therefore we can say that the equivalent emf in a parallel circuit is equal to the individual emf of the battery. Cells are said to be connected in series when they are joined end to end so that the same quantity of electricity must flow through each cell.



Formula used:
Effective emf of combination between the points \[PQ\] is:
\[I = \frac{E}{r}\]

Equivalent emf is given by
\[{E_{net}} = {E_1}{r_2}-{\rm{ }}{E_2}{r_1}/{\rm{ }}{r_1} + {r_2}\]



Complete answer:
Since the internal resistances are connected in parallel combination with each other, the equivalent internal resistance of the combination is given by
\[\;{E_{net}} = {\rm{ }}2 - 2{\rm{ }}/{\rm{ }}2 + 1{\rm{ }} = {\rm{ }}0\]
\[{E_{eq}} = \;2 \times 1{\rm{ }}/{\rm{ }}2 + 1\]
\[ = {\rm{ }}2/3\]
We have,
\[{E_{eq}}/{\varepsilon _{eq}}\; = {\rm{ }}{E_1}/{\varepsilon _1} = {\rm{ }}{E_2}/{\rm{ }}{\varepsilon _2}\]
\[{\varepsilon _{eq}}/{\rm{ }}\left( {2/3} \right){\rm{ }} = {\rm{ }}2/2{\rm{ }}-{\rm{ }}1/1\]
\[{E_{eq}} = {\rm{ }}0\]volt.
Hence, the correct option is A

Note:If one of the cells in a series grouping of identical cells is connected incorrectly, the effect of two cells will be negated. If there are " \[x\] " incorrectly connected cells in a group of " \[n\] '' identical cells, each of which has an internal resistance " \[r\] " and an emf " \[E\]," the equivalent EMF is given by: EMF (equivalent) \[ = \left( {n - 2x} \right)\]Internal resistance equals external resistance, which equals nr. Total resistance equals \[nr\] plus \[R\], and total current equals \[nr + r \times \left( {n - 2x} \right){\rm{ }}E/nr + R\].