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A parallel combination of two cells of emfs `epsilon_(1)` and `epsilon_(2)`, and internal resistance `r_(1)` and `r_(2)` is used to supply current to a load of resistance R. Write the expression for the current through the load in terms of `epsilon_(1), epsilon_(2),r_(1)` and `r_(2)`. |
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Answer» Equivalent resistance of two cells in parallel `r_(eq)=(r_(1) r_(2))/(r_(1) +r_(2))` Equivalent emf of two cells in parallel `r_(eq)=(epsilon_(1)r_(1) +epsilon_(2)r_(1))/(r_(1)+r_(2))` Current in circuit, `I=(epsilon_(eq))/(R+r_(eq))=((epsilon_(1)r_(2)+epsilon_(2)r_(1))/(r_(1)+r_(2)))/(R+(r_(1)r_(2))/((r_(1)+r_(2))))=(epsilon_(1)r_(2)+epsilon_(2)r_(1))/(R(r_(1)+r_(2))+r_(1)r_(2))` |
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