Question

A cell of E.M.F 'E' and internal resistance 'r' is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus

(i) R and

(ii) the current $\mathrm{i}$.

It is found that when R=4Ω, the current is 1A when R is increased to 9Ω, the current reduces to 0.5 A. Find the values of the E.M.F E and internal resistance r.

Answer 1

Step 1: Given

A cell of emf 'E' and internal resistance 'r' and a variable load resistor R.

when $R=4\Omega$ then current i is $1A$ and

when $R=9\Omega$ then current i is $0.5A$.

Step 2: Draw the graph

(i) Graph between the terminal voltage V and variable load resistor R-

(ii) Graph between the terminal voltage V and current i-

Step 3: find the emf E and internal resistance r

In condition 1, $R=4\Omega$ and $i=1A$

Terminal voltage, $V=E-ir$ and $V=iR$

so, $E-ir=iR$

put the value of R and i-

then, $E-r=4$ …(1)

In condition 2, $R=9\Omega$ and $i=0.5A$

and $E-ir=iR$

now, put the value of R and $\mathrm{i}$

then, $E-0.5r=4.5$ …(2)

Subtract equation 2 from equation 1-

$$\begin{gathered} E-r-E+0.5r=4-4.5 \\ -0.5r=-0.5 \\ r=1\Omega \end{gathered}$$

now put value of r in equation 1,

$$\begin{gathered} E-1=4 \\ E=5V \end{gathered}$$

Thus, the emf is 5V and internal resistance is $1\Omega$.

Hence, the values of the emf E and internal resistance r is $5\mathrm{V}$ and $1\Omega$ respectively.