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A cell supplies a current of 0.9 A through a 1 Omega resistor and a current of 0.3 A through a 2Omega resistor. Calculate the internal resistance of the cell. |
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Answer» Solution :Current from the CELL, `I_(1)=0.9A` Resistor, `R_(1)=2OMEGA` Current from the cell, `I_(2)=0.3A` Resistor, `R_(2)=7Omega` Internal resistance of the cell, `r=?` Current in the circuit, `I_(1)=(XI)/(r+R_(1))` `xi=I_(1)(r+R_(1)) ""...(1)` Current in the circuit, `I_(2)=(xi)/(r+R_(2))` `xi=I_(2)(r+R_(2))""...(2)` Equating equation (1) and (2), `I_(1)r+I_(1)R_(1)=I_(2)R_(2)+I_(2)r` `(I_(1)-I_(2))r=I_(2)R_(2)-I_(1)R_(1)` `r=(I_(2)R_(2)-I_(1)R_(1))/(I_(1)-I_(2))=((0.3xx7)-(0.9xx2))/(0.9-0.3)=(2.1-1.8)/(0.6)=(0.3)/(0.6)` `r=0.5Omega` |
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