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Two cells of same emf E but internal resistance r_(1) and r_(2) are connected in series to an external resistor R(figure). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero? |
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Answer» Solution :Applying OHM's law Effective resistance `=R+r_(1)+r_(2)` and effective emf of two CELLS `=E+E=2E`, so the electric current is given by `I=(E+E)/(E+r_(1)+r_(2))` The potential difference across the TERMINALS of the first cell and putting it equal to zero. `V_(1)=E-Ir_(1)=E-(2E)/(r_(1)+r_(2)+R)r_(1)=0` or `E=(2Er_(1))/(r_(1)+r_(2)+R)implies1=(2r_(1))/(r_(1)+r_(2)+R)` `r_(1)+r_(2)+R=2r_(1) impliesR=r_(1)-r_(2)` this is the required relation. |
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