A potentiometer wire of 10m length and having a resistance of 1ohmm- 1 is connected to an accumulator of emf 2.2volt and a high resistance box. To obtain a potential gradient of 2.2 mili volt/ meter, the value of resistance used from the resistance box will be :

Question

A potentiometer wire of 10m length and having a resistance of 1ohmm- 1 is connected to an accumulator of emf 2.2volt and a high resistance box. To obtain a potential gradient of 2.2 mili volt/ meter, the value of resistance used from the resistance box will be : (A) 790ohm (B) 810ohm (C) 990ohm (D) 1000ohm

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-rde1iunwfv9jzh7p.jpg" /> RP m=10Ω Potential gradient =(VP M/L)=(I RP M/L)=(E RP M/(R + RP m) L)

Q. A potentiometer wire of $10 m$ length and having a resistance of $1 o hm m^{- 1}$ is connected to an accumulator of emf $2.2 v o lt$ and a high resistance box. To obtain a potential gradient of $2.2$ mili volt/ meter, the value of resistance used from the resistance box will be :

$790 o hm$

$810 o hm$

$990 o hm$

$1000 o hm$

$R_{P m} = 10Ω$
Potential gradient $= \frac{V _{PM}}{L} = \frac{I R _{PM}}{L} = \frac{E R _{PM}}{( R + R _{P m} ) L}$

Q. A potentiometer wire of 10m length and having a resistance of 1ohmm−1 is connected to an accumulator of emf 2.2volt and a high resistance box. To obtain a potential gradient of 2.2 mili volt/ meter, the value of resistance used from the resistance box will be :

790ohm

810ohm

990ohm

1000ohm