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(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.

Answers (1)

(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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Safeer PP

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