Question

A $10\, V$ battery with internal resistance $1\, \Omega$ and a $15\, V$ battery with internal resistance $0.6\, \Omega$ 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

Answer 1

Answer: (C)

13.1 V

Answer 2

As the two cells oppose each other hence,
the effective emf in closed circuit is $15-10=5 V$
and net resistance is $1+ 0.6=1.6\, \Omega$
(because in the closed circuit the internal resistance of two cells are in series.
Current in the circuit, $\frac{\text { effective emf }}{\text { total resistance }}=\frac{5}{1.6} A$
The potential difference across voltmeter will be same as
the terminal voltage of either cell. Since the current is
drawn from the cell of $15 \,V$
$\therefore V_{1}=E_{1}-I r_{1}$
$=15-\frac{5}{1.6} \times 0.6=13.1 \,V$