Question: Q. 15. 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 $1 \mathrm{~A}$ 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 $r$.
A [Delhi I, II, III 2015]
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Solution:
Ans. (i)
(ii)
$1 / 2$
$$ \begin{aligned} & I=\frac{E}{R+r} \ & I=\frac{E}{4+r} \end{aligned} $$
$$ \Rightarrow \quad E=4+r $$
…(i) $1 / 2$
From equation (i) and (ii),
2 $4+r=4.5+0.5 r \quad 1 / 2$
Using this value of $r$, we get
$E=5 V$
[CBSE Marking Scheme 2015]
Detailed Answer :
From (1) & (2)
the pots are:
