question_answer

Two batteries, one of emf 18V 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 - 12 = 6 V and net resistance is \[1+2=3\,\Omega \](because in the closed circuit the internal resistances of two cells are in series). The current in circuit will be in direction of arrow shown in the figure. \[I=\frac{\text{effective}\,\text{emf}}{\text{total}\,\text{resistance}}=\frac{6}{3}=2A\] The potential difference across V will be the same as the terminal voltage of either cell. Since, current is drawn from the cell of 18 volt, hence, \[{{V}_{1}}={{E}_{1}}-i{{r}_{1}}=18-2(2\times 2)=18-4=14\,V\] Similarly, current enters in the cell of 12 V, hence, \[{{V}_{2}}={{E}_{2}}+i{{r}_{2}}=12+12\times 1=12+2=14\,V\] \[V=14\,V\]