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A battery of emf 2.0 V and internal resi...

A battery of emf `2.0 V` and internal resistance `0.10 Omega` is being charged with a current of `5.0 A`. Find the potential difference between the terminals of the battery?

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Verified by Experts

The correct Answer is:
B
During charging,
`V=E+ir=2+(5)(0.1)=2.5V`
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DC PANDEY-CURRENT ELECTRICITY-Level 1 Subjective
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  3. A battery of emf 2.0 V and internal resistance 0.10 Omega is being cha...

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  8. Resistance R1and R2, each 60 Omega, are connected in series. The point...

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  9. A moving coil galvanometer of resistance 20 Omega gives a full scale d...

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  10. A cell of emf 3.4 V and internal resistance 3 Omega is connected to an...

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  11. (a) A voltmeter with resistance Rv is connected across the terminals o...

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  12. An ammeter with resistance RA is connected in series with a resistor R...

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  13. Each of three resistors in figure has a resistance of 2.4 Omega and ca...

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  14. A storage battery with emf 2.6 V loaded with external resistance produ...

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  15. In the circuit shown in figure E1=7V,E2=1 V,R1=2Omega, R2=2Omega and R...

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  16. In the circuit shown in figure find a) the rate of conversion of ...

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  17. Three resistors having resistances of 1.60 Omega, 2.40 Omega and 4.80 ...

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  18. The power of resistor is the maximum power the resistor can safely dis...

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  19. Find the equivalent resistance between points A and B in the following...

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  20. What will be the change in the resistance of a circuit between A and F...

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