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 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:
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Cells connected in parallel (same polarity) act as a single equivalent cell; the ideal voltmeter draws no current.
Updated On: Nov 11, 2026
- \(11.9\ \mathrm{V}\)
- \(13.1\ \mathrm{V}\)
- \(12.5\ \mathrm{V}\)
- \(24.5\ \mathrm{V}\)
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The Correct Option is B
Solution and Explanation
The two batteries are in parallel with the same polarity and the voltmeter draws no current, so a circulating current flows between the batteries.
Net emf driving the current = 15 - 10 = 5 V; total resistance = 1 + 0.6 = 1.6 \(\Omega\).
\(I=\frac{5}{1.6}=3.125\ \mathrm{A}\) (flowing out of the 15 V cell).
Voltmeter reading = terminal voltage of the 15 V cell = \(15-I(0.6)=15-1.875=13.125\ \mathrm{V}\approx 13.1\ \mathrm{V}\).
Answer: (2) 13.1 V
Net emf driving the current = 15 - 10 = 5 V; total resistance = 1 + 0.6 = 1.6 \(\Omega\).
\(I=\frac{5}{1.6}=3.125\ \mathrm{A}\) (flowing out of the 15 V cell).
Voltmeter reading = terminal voltage of the 15 V cell = \(15-I(0.6)=15-1.875=13.125\ \mathrm{V}\approx 13.1\ \mathrm{V}\).
Answer: (2) 13.1 V
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