A 10V battery with internal resistance 1Ω  and a 15V battery with internal resistance 0.6Ω are Connected in parallel to a voltmeter (see figure).The reading in the voltmeter will be close to in volts.

As the two cells oppose each other hence, the effective emf in closed circuit is 15-10=5V and net resistance is 1+0.6= 1.6Ω   (because in the closed circuit the internal resistance of two cells are in series)current  in the circuit=I= effective emf /  total resistance = 5/1.6 AThe potential difference across voltmeter will be same as the terminal voltage of either cellSince  the current I is drawn from the cell  of 15V ∴V1=E1−Ir1 =15−51.6×0.6=13.125V