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A potentiometer wiere has a length of 4 and resistance of 20 Omega. It is connected in series with resistance of 2980 Omegaand a cell of emf 4 V. Calcaulte the potential along the wire. |
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Answer» SOLUTION :Length of the potential WIRE, `l=4m` Resistance of the wire, `r=20Omega` Resistance connected series with potentiometer wire, `R=2980Omega` Emf of the cell, `xi=4V` Effective resistance, `R_(s)=r+R=20+2980=3000Omega` Current FLOWING through the wire, `I=(xi)/(R_(s))=(4)/(3000)` `I=1.33xx10^(-3)A` Potential drop across the wire, `V=Ir` `=1.33xx10^(-3)xx20` `V=26.6xx10^(-3)V` Potential gradient `=("Potential drop across the wire")/("length of the wire")=(V)/(l)=(26.6xx10^(-3))/(4)` `=6.65xx10^(-3)` Potential gradient `=0.66xx10^(-2)Vm^(-1)` |
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