A potentiometer arrangement is shown in figure. The driver cell has emf E and internal resistance r. The resistance of potentiometer wire A B is R . F is the cell of emf E / 3 and internal resistance r / 3. Balance point (J) can be obtained for all finite values of

Question

A potentiometer arrangement is shown in figure. The driver cell has emf E and internal resistance r. The resistance of potentiometer wire A B is R . F is the cell of emf E / 3 and internal resistance r / 3. Balance point (J) can be obtained for all finite values of (A) R >r / 2 (B) R< r / 2 (C) R >r / 3 (D) R< r / 3

Tardigrade Admin · Accepted Answer

Current in A B=I=(E/R+r) Potential difference across A B=I R=(E R/R+r) To obtain balanced point, (E R/R+r)>(E/3) or R >r / 2

Q. A potentiometer arrangement is shown in figure. The driver cell has emf $E$ and internal resistance $r$ . The resistance of potentiometer wire $A B$ is $R . F$ is the cell of emf $E /3$ and internal resistance $r /3$ . Balance point $(J)$ can be obtained for all finite values of

$R > r /2$

$R < r /2$

$R > r /3$

$R < r /3$

Current in $A B = I = \frac{E}{R + r}$
Potential difference across $A B = I R = \frac{ER}{R + r}$
To obtain balanced point, $\frac{ER}{R + r} > \frac{E}{3}$
or $R > r /2$

Q. A potentiometer arrangement is shown in figure. The driver cell has emf E and internal resistance r. The resistance of potentiometer wire AB is R.F is the cell of emf E/3 and internal resistance r/3. Balance point (J) can be obtained for all finite values of

R>r/2

R<r/2

R>r/3

R<r/3