InterviewSolution
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InterviewSolution
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Two batteries of emf `epsi_(1) and epsi_(2)(epsi_(2) gt epsi_(1))` and internal resistance `r_(1) and r_(2)` respectively are connected in parallel as shown in figure.A. The equivalent emf `epsi_(eq)` of the two cells is between `epsi_(1) and epsi_(2)`, i.e., `epsi_(1) lt epsi_(eq) lt epsi_(2)`B. the equivalent emf `epsi_(eq)` is smaller than `epsi_(1)`C. The `epsi_(eq)` is given by `epsi_(eq)=epsi_(1)+epsi_(2)` always.D. `epsi_(eq)` is independent of internal resistance `r_(1) and r_(2)`. |
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Answer» Correct Answer - A The equivalent emf of this combination is given by `epsi_(eq)=(epsi_(2)r_(1)+epsi_(1)r_(2))/(r_(1)+r_(2))` This suggest that the equivalent emf `epsi_(eq)` of the two cells is given by `epsi_(1) lt epsi_(eq) lt epsi_(2)`. |
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