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Two cells having an internal resistance of `0.2 Omega` and `0.4 Omega` are connected in parallel, the voltage across the battery is 1.5 V. If the emf of one cell is 1.2 V, then the emf of second cell isA. 2.1 VB. 2.7 VC. 3 VD. 4.2 V |
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Answer» Correct Answer - A `r_(1)=0.2 Omega, r_(2)=0.4 Omega` V=lR `R=(r_(1)r_(2))/(r_(1)+r_(2)), l=l_(1)+l_(2) implies V=(r_(1)r_(2))/(r_(1)+r_(2))(l_(1)+l_(2))` `l_(1)=E_(1)//r_(1) " and " l_(2)=E_(2)//r_(2)` `V=((E_(1))/(r_(1))+(E_(2))/(r_(2)))(r_(1)r_(2))/(r_(1)+r_(2))=((E_(2)r_(1)+E_(1)r_(2)))/(r_(1)r_(2))((r_(1)r_(2))/(r_(1)+r_(2)))` ` V=(E_(2)r_(1)+E_(1)r_(2))/(r_(1)+r_(2)) implies 1.5=(E_(2)xx0.21.2xx0.4)/(0.6)` `0.9=0.2 E_(2)+0.48 implies (0.42)/(0.2)=E_(2) implies E_(2)=2.1 V` |
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