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InterviewSolution
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Figure shows a potentiometer circuit for comparison of tworesistances. The balance point with a standard resistor R = 10.0 Ωis found to be 58.3 cm, while that with the unknown resistance X is68.5 cm. Determine the value of X. What might you do if you failedto find a balance point with the given cell of emf ε ? |
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Answer» Resistance of the standard resistor, `R = 10.0 Omega` Balance point for this resistance, `l_(1)=58.3` cm Current in the potentiometer wire `= l` Hence, potential drio across `R, E_(1)=iR` Resistance of the unknown resistor = X Balance point for this resistor, `I_(2)= 68.5` cm Hence, potential drop across `X, E_(2)= iX` The relation connecting emf and balance point is, `(E_(1))/(E_(2))=(l_(1))/(l_(2))` `(iR)/(iX) = (l_(1))/(l_(2))` `(iR)/(iX)=(l_(1))/(l_(2))` `X = (l_(1))/(l_(2))xxR` `= (68.5)/(58.3) xx 10 = 11.789 Omega` Therefore, the value of the unknown resistance, X, is `11.75 Omega` If we fall to find a balance point with the given cell emf `epsilon`, then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained. |
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