A cell of emf ' E ' and internal resistance ...

CBSE Class 12 Physics 2015 Delhi Set 1 Paper

Question : 16 of 26

Marks: +1, -0

A cell of emf '

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

I

It is found that when

R=4\ \Omega

, the current is

1\ \mathrm{A}

and when

R

is increased to

9\ \Omega

, the current reduces to

0.5\ \mathrm{A}

. Find the values of the emf

E

and internal resistance

r

Solution:

Drawing of Plots of Part (i) & (ii)

Finding the values of emf and internal resistance

(i)

(ii)

(If the student just writes the relations

V=\varepsilon - I R

and

V=\frac{\varepsilon R}{R+r}

but does not draw the plots, award mark.)

I=\frac{E}{R+r}

I=\frac{E}{4+r}

\Rightarrow E=4+r

......(i)

\text{ Also, } 0.5=\frac{E}{9+r}

E=4.5+0.5r

......(ii)

From equation (i) & (ii)

4+r=4.5+0.5r

\therefore r=1\ \Omega

Using this value of

r

, we get

E=5\ \mathrm{V}

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