Two batteries, one of emf 18V and internal resistance `2Omega` and the other of emf 12 and internal resistance `1Omega` , are connected as shown. The voltmeter V will record a reading of

Options

A. 14 V
B. 15 V
C. 18 V
D. 30 V

Answer

A

Explanation

It is clear that the two cells oppose each other hence th effective emf in closed circuit is 18-12=6V and net resistance is `1+2=3omega`(because in the closed circuit the internal resistance of two cells are in series)
The current in circuit will be in direction of arrow shown in figure
`I=(effective emf)/(total resistacne)=6/3=2A`
The potential difference across V will be same as the terminal volage of either cell

since current is drawn from the cell of 18V hence
`V_(1)=E_(1)-Ir_(1)=18-(2xx2)=18-4=14V`