(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)
S safeer

(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}

Preparation Products

Knockout KCET 2021

An exhaustive E-learning program for the complete preparation of KCET exam..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout KCET JEE Main 2021

It is an exhaustive preparation module made exclusively for cracking JEE & KCET.

₹ 27999/- ₹ 16999/-
Buy Now
Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
Buy Now
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Buy Now
Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Exams
Articles
Questions