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 ................ $V$
JEE MAIN 2015, Diffcult
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As the two cells oppose each other hence, the effective emf in closed circuit is $15-10=5 \,\mathrm{V}$ 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,
$I=\frac{\text { effective emf }}{\text { total resistance }}=\frac{5}{1.6} \,A$
The potential difference across voltmeter will be same as the terminal voltage of either cell.
Since the current is drawn from the cell of $15 \,\mathrm{V}$
$\therefore \quad V_{1}=E_{1}-I r_{1}$
$=15-\frac{5}{1.6} \times 0.6=13.1 \,\mathrm{V}$
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