CBSE Class 12 Physics 2015 Delhi Set 1 Paper

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Question : 16
Total: 26
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=4Ω , the current is 1A and when R is increased to 9Ω , the current reduces to 0.5A . 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−IR and V=‌
εR
R+r
but does not draw the plots, award mark.)
I‌=‌
E
R+r

‌I=‌
E
4+r

⇒‌‌E‌=4+r ......(i)
‌ Also, ‌‌‌0.5‌=‌
E
9+r

E‌=4.5+0.5r ......(ii)
From equation (i) & (ii)
4+r=4.5+0.5r
∴‌‌r=1Ω
Using this value of r, we get
E=5V
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