Question

A resistance r is connected across a cell of emf e and internal resistance r. A potentiometer now measures the potential difference between the terminals of the cell as v. Write the expression for r in terms of e, v and r.

Answer 1

Answer:

e=I(R+r)  

V=IR

∴e/V=R+V/R

=1+r/R

=r/R

=(e/v−1)

∴r=R(e/V−1)

Explanation:

Answer 2

A potentiometer is three terminal resistor with sliding contact that forms an adjustable voltage divider. it act as rheostat or variable resistor.

Let internal resistance be r and R be resistance connected in series with potentiometer and voltage applied be v.

Let I be the current flowing through the main circuit in the given figure

Since R and r are in series

then I = $\frac{E}{R+r}$     ....(i)

$V_{AB}$ = (E + I.r) = I.R

Putting value of I from (i) in above equation we get

$V_{AB}$  =  $\frac{E}{R+r}$ . R

R + r = $\frac{E}{V_{AB} }$. R

r = $\frac{E.R}{V_{AB} }$ - R

Here $V_{AB}$ = V

r = $\frac{E-V}{V}$.R

r = ($\frac{E}{V}$-1).R