JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2015
- Hard
- MCQ
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 ................ \(V\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(13.1\)
Step-by-step Solution
Detailed explanation
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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