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Figure shows a potentiometer circuit for comprasion of two resistances. The balance point with a standard resistor `R = 10.0 Omega` is found to be 58.3 cm, while that with the unknown resistance X is 68.5 Cm. determine the value of X. What might you do if you failed to find a balance point with the given cell of emf `epsilon` ? |
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Answer» Here `L_(1) = 58.3 cm, L_(2) = 68.5 cm, R = 10 Omega, 1 X = ?` Let bet the current in the potentiometer wire and `epsilon_(1)` and `epsilon_(2)` be the potential drops across R and X respectively when connected in circuit by closing respective key. Then `(epsilon_(2))/(epsilon_(1)) = (1X)/(1R) = (x)/(R )` or `X = (epsilon_(2))/(epsilon_(1)) R` But `(epsilon_(2))/(epsilon_(1)) = (L_(2))/(L_(1))` From (i) `X = (L_(2))/(L_(1)) R = (68.3)/(58.3) xx 10.0 = 11.75 Omega` If there is no balance point with given cell of emf it means potential drop across R or X greater than the potential drop across the potentiometer wire. AB in order to obtain the balance point, the potential drops across R and X are to reduced which is possilbe by reducing the current in R and X for that either suitable resistance should be put in series with R and X or cell of smaller emf E should be used. Another possilbe way is to increase the potential drop across the potentiometer wire by increasing the voltage of dirver cell. |
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