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  • A potentiometer wire of length 10 m and resistance  is connected in series with a 25 V battery and an ext

A potentiometer wire of length 10 m and resistance 20\Omega is connected in series with a 25 V battery and an external resistance 30\Omega. A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E in (volt) is \frac{x}{10}. The value of is______ 

Option: 1

2.5


Option: 2

--


Option: 3

--


Option: 4

--


Answers (1)

best_answer

P \rightarrow Balancing point (null poind)

\mathrm{\begin{aligned} & \mathrm{v}=25 \mathrm{v}\\ &R=30\Omega \\ &\mathrm{l_{AB}}=10m\\ &\mathrm{R_{AB}}=20\Omega \end{aligned}}     \begin{aligned} \phi &=\text { potential gradient } \\ \phi &=\frac{V_{A B}}{l_{A B}}=\frac{I R_{A B}}{l_{A B}} \end{aligned}

For p to be the null point

\mathrm{E=V_{A P}=I R_{A P}=\phi l_{A P}}

\mathrm{\begin{aligned} E &=\frac{I\left(R_{A B}\right)}{\left(l_{A B}\right)} \times\left(l_{A P}\right) \\ &=\left(\frac{V}{R+R_{A B}}\right) \times\left(\frac{R_{A B}}{l_{A B}}\right) \times l_{A P} \\ E &=2.5 V \end{aligned}}

the value of x is 2.5

Posted by

Rishabh

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