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Two batteries A and B whose emf is 2V are connected in series with external resistance `R=1Omega`. Internal resistance of battery A is `1.9Omega` and that of B is `0.9 Omega`. A. 2VB. 3.8VC. zeroD. None of these

Answer» Correct Answer - C
Total resistance `r = 1.9+0.9+1=3.8 Omega`
Total emf `E = 2+=4`
`therefore " " E = iR`
`" " i=(E )/(R )=(4)/(3.8)`
Potential difference across battery A
`V = E -ir`
`= 2-(4)/(3.8)xx1.9=0`


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