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 Ω

, the current is

1 {\A}

and when

R

is increased to

9 Ω

, the current reduces to

0.5 {\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}=ε-I R

and

V=\;{ε R}/{R+r}

but does not draw the plots, award mark.)

I\;=\;{E}/{R+r}

\;I=\;{E}/{4+r}

⇒ \;\; E\;=4+r

......(i)

\;\text " Also, "\; \;\; 0.5\;=\;{E}/{9+r}

E\;=4.5+0.5 r

......(ii)
From equation (i) & (ii)

4+r=4.5+0.5 r

∴ \;\; r=1 Ω

Using this value of

r

, we get

E=5 {\V}

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