A $$10V$$ battery with internal resistance $$1$$Ω and a $$15V$$ battery with internal resistance $$0.6$$Ω are Connected in parallel to a voltmeter (see$$ $$figure).The reading in the voltmeter will be close to in volts.

Question

A $$10V$$ battery with internal resistance $$1$$Ω  and a $$15V$$ battery with internal resistance $$0.6$$Ω are Connected in parallel to a voltmeter (see$$ $$figure).The reading in the voltmeter will be close to in volts.

Infinity Learn · Accepted Answer

As the two cells oppose each other hence, the effective emf in closed circuit is 15-10=5V and net resistance is 1+0.6= $1.6 Ω$ (because in the closed circuit the internal resistance of two cells are in series)
current in the circuit=I= effective emf / total resistance = 5/1.6 A
The potential difference across voltmeter will be same as the terminal voltage of either cell
Since the current I is drawn from the cell of 15V
$\begin{array}{l} ∴ V_{1} = E_{1} - I r_{\begin{array}{l} 1 \end{array}} \\ = 15 - \frac{5}{1.6} \times 0.6 = 13.125 V \end{array}$

Sources of emf and Kirchhoff's laws

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A 10V battery with internal resistance $1 Ω$ and a 15V battery with internal resistance 0.6 $Ω$ are Connected in parallel to a voltmeter (see figure).The reading in the voltmeter will be close to in volts.

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Sources of emf and Kirchhoff's laws

Remember concepts with our Masterclasses.

A 10V battery with internal resistance 1Ω and a 15V battery with internal resistance 0.6Ω are Connected in parallel to a voltmeter (see figure).The reading in the voltmeter will be close to in volts.

Ready to Test Your Skills?

free study material

A 10V battery with internal resistance $1 Ω$ and a 15V battery with internal resistance 0.6 $Ω$ are Connected in parallel to a voltmeter (see figure).The reading in the voltmeter will be close to in volts.