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

Question

A 10 V battery with internal resistance 1 Ω and a 15 V battery with internal resistance 0.6 Ω are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to : (A) 12.5 V (B) 24.5 V (C) 13.1 V (D) 11.9 V

Tardigrade Admin · Accepted Answer

As the two cells oppose each other hence,
the effective emf in closed circuit is

15 - 10 = 5 V

and net resistance is

1 + 0.6 = 1.6 Ω

(because in the closed circuit the internal resistance of two cells are in series.
Current in the circuit,

\frac{effective emf}{total resistance} = \frac{5}{1.6} A

The potential difference across voltmeter will be same as
the terminal voltage of either cell. Since the current is
drawn from the cell of

15 V

∴ V_{1} = E_{1} - I r_{1}

= 15 - \frac{5}{1.6} \times 0.6 = 13.1 V

Q. 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 :

12.5 V

24.5 V

13.1 V

11.9 V

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