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  • In the given figure, there is a circuit of potentiometer of length . The resistance per unit length is <img alt="0.1 \Omega" src="/latex-image/?0.1%20

In the given figure, there is a circuit of potentiometer of length \mathrm{AB}=10 \mathrm{~m}. The resistance per unit length is 0.1 \Omega per \mathrm{cm}. Across \mathrm{AB}, a battery of emf E and internal resistance ' r ' is connected. The maximum value of emf measured by this potentiometer is :
Option: 1 5V
Option: 2 2.25V
Option: 3 6V
Option: 4 2.75V

Answers (1)

best_answer

 
primary\, circuit\, R_{AB}= \left ( 0\cdot 1 \right )\l _{AB}= 100\Omega
Current in the primary circuit is
I= \frac{6}{20+R_{AB}}= \frac{6}{20+\left ( 0\cdot 1 \right )\times 100}
I= \frac{1}{20}A
V_{AB}= I\left ( R_{AB} \right )= \frac{1}{20}\times \left ( 0\cdot 1 \right )\times 100
V_{AB}= 5V
The maximum value of emf measured by this potentiometer = V_{AB}= 5V
The current option is (1)

Posted by

vishal kumar

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