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A 10 V battery with internal resistance1\Omega and a 15V battery with internal resistance 0.6\Omega are connected in parallel to a voltmeter

Answers (1)

As the two cells oppose each other hence, the effective emf in closed circuit is 15 - 10 = 5 V and net resistance is 1 + 0.6 = 1.6 Ω (because in the closed circuit the internal resistance of two cells are in series.

As the two cells oppose each other hence, the effective emf in closed circuit is 15 - 10 = 5 V 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,  \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 V 

V_{1}=E_{1}-I r_{1}=15-\frac{5}{1.6} \times 0.6=13.1 V

 

Posted by

Satyajeet Kumar

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