# (a) Derive a relation between the internal resistance, emf and terminal potential difference of a cell from which current I is drawn. Draw V vs I graph for a cell and explain its significance.(b) A voltmeter of resistance $998\; \Omega$  is connected across a cell of emf 2 V and internal resistance $2\; \Omega$. Find the potential difference across the voltmeter and also across the terminals of the cell. Estimate the percentage error in the reading of the voltmeter.

(a) Let us consider a battery of internal resistance r, battery emf E and external resistance R where (I) current is flowing.

From the diagram, we have

$E-IR-rI=0$        $(\because IR=V)$

So,

$E-V-Ir=0$

$E=V+Ir$

Hence, it is the relation between E,V and r.

• The Graph V versus I for a cell;-

It is used to find the emf and internal resistance of the cell.

(b) We have given,

$R=$ external resistance $=998\; \Omega$

$r=2\; \Omega$

$emf=2\; V$

We know that,

$E=V+Ir$

$V=E-Ir$            $(\because V=IR)$

$IR=E-Ir$

On putting the given values we have,

$998\times I=2-2\times I$

$998 I+2I=2$

$1000I=2$

$I=\frac{2}{1000}$

$I=0.002A$

We know that, $V=IR$

$\therefore V=0.002\times 998$

$V=1.996\; Volt$

Now., calculating the percentage error;

$\frac{0.004}{2}\times 100=\frac{0.400}{2}=20\; ^{o}/{o}$

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