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:

A)                            15 V     

B)                 30 V                      

C)                 14 V

D)                 18 V

Correct Answer: C

Solution :

                    It is clear that the two cells oppose each other hence, the effective emf in closed circuit is 18 ? 2 = 6 V and net resistance is 1 + 2 = 3\[\Omega \] (because in the closed circuit the inters. resistances of two ceils are in series).                 The current in circuit will be in direction of arrow shown in figure.                                 \[I=\frac{effective\,emf}{total\,resis\tan ce}=\frac{6}{3}=2\,A\]                 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 V