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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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