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

Q: 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 ................ $V$

c 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 $15 \,\mathrm{V}$ $\therefore \quad V_{1}=E_{1}-I r_{1}$ $=15-\frac{5}{1.6} \times 0.6=13.1 \,\mathrm{V}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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 ................ $V$

A$12.5$
B$24.5$
C$13.1$
D$11.9$

JEE MAIN 2015, Diffcult

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

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