He uses a resistance of \[5\Omega ,\]unknown resistance X and a battery of 5 V in secondary circuit. He touches the jockey\[{{\text{J}}_{\text{1}}}\]on potentiometer wire to get the point P, so that there is no deflection in \[{{G}_{1}}\] then he locates the point Q, so that \[{{G}_{2}}\] shows zero deflection. It is found that \[AP=\frac{AQ}{3}.\] Value of X is
A) \[5\,\Omega \]
B) \[15\,\Omega \]
C) \[10\,\Omega \]
D) This method won't works
Correct Answer: C
Solution :
Let, current I is flowing in secondary circuit (consisting of \[5\Omega \], X and 5V). Then, potential difference across \[5\Omega \] resistor is proportional to AP if at P, \[{{G}_{1}}\] shows zero deflection. ie, \[5I\infty AP\] Similarly, \[(5+X)I\propto AQ\] \[\Rightarrow \]\[\frac{5+X}{5}=\frac{AQ}{AP}=3\]\[\Rightarrow \]\[X=10\Omega \]
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