question_answer

Two batteries, one of emf 18 V and internal resistance \[2\,\,\Omega \] and the other of emf 12 V and internal resistance \[1\,\,\Omega ,\] are connected as shown. The voltmeter V will record a reading of:      [AIPMT (S) 2005]

A) 15 V                             

B)  30 V                

C) 14 V

D)                   18 V

Correct Answer: C

Solution :

[c] It is clear that the two cells oppose each other hence, the effective emf in closed circuit is \[182=6\text{ }V\] and net resistance is \[1+2=3\Omega \] (because in the closed circuit the inters. resistances of two cells are in series).

The current in circuit will be in direction of arrow shown in figure.

\[I=\frac{\text{effective}\,\text{emf}}{\text{total}\,\text{resistance}}=\frac{6}{3}=2\,A\]

The potential difference across V will be same as the terminal voltage of either cell.

Since, current is drawn from the cell of 18 volt, hence,

\[{{V}_{1}}={{E}_{1}}-i{{r}_{1}}\]

\[=18-(2\times 2)=18-4=14\,V\]

Similarly, current enters in the cell of 12 V, hence,

\[{{V}_{2}}={{E}_{2}}+i{{r}_{2}}\]

\[=12+2\times 1\]

\[=12+2=14\,V\]

Hence,  \[V=14\text{ }V\]