- Introduction
- |
- ElectroMotive Force(emf)
- |
- Internal Resistance of Battery (or cell)
- |
- Electric Energy and Power
- |
- Kirchoff's Rules
- |
- The junction Rule (or point rule)
- |
- The Loop Rule (or Kirchoff's Voltage Law)
- |
- Grouping of the cell's
- |
- Meterbridge (slide wire bridge)
- |
- Potentiometer
- |
- Comparison of EMF's of two cells using potentiometer
- |
- Determination of internal resistance of the cell

- Consider the circuit arrangement of potentiometer given below used for comparison of emf's of two cells

- Positive terminals of two cells of emf's E
_{1}and E_{2}( whose emf are to be compared ) are connected to the terminals A and negative terminals are connected to jockey through a two way key K_{2}and a galvanometer

- Now first key K
_{1}is closed to establish a potential difference between the terminals A and B then by closing key Ksub>2 introduce cell of EMF E_{1}in the circuit and null point junction J_{1}is dtermined with the help of jockey.If the null point on wire is at length

l_{1}from A then

E_{1}=Kl_{1}

Where K -> Potential gradient along the length of wire - Similarly cell having emf E
_{2}is introduced in the circuit and again null point J_{2}is determined .If length of this null point from

A is l_{2}then

E_{2}=Kl_{2}

Therefore

E_{1}/E_{2}=l_{1}/l_{2}

This simple relation allows us to find the ratio of E_{1}/E_{2}

- if the EMF of one cell is known then the EMF of other cell can be known easily

- Potentiometer can also be used to determine the internal resistance of a cell

- For this a cell whose internal resistance is to be determined is connected to terminal A of the potentiometer across a resistance box through a key K
_{2}

- First close the key K
_{1}and obtain the null point .Let l_{1}be the length of this null point from terminal A then

E=Kl_{1}

- When key K
_{2}is closed ,the cell sends current through resistance Box (R).If E_{2}is the terminal

Potential difference and null point is obtained at length l_{2}(AJ_{2}) then

V=Kl_{2}

Thus

E/V=l_{1}/l_{2}

But E=I(R+ r) and V=IR

This gives

E/V=(r+R)/R

So (r+R)/R=l_{1}/l_{2}

giving

r=R(l_{1}/l_{2}-1)

- Using above equation we can find internal resistance of any given cell

Class 12 Maths Class 12 Physics

#### Recommended Physics books for class 12

- NCERT Solutions: Physics 12th
- NCERT Exemplar Problems: Solutions Physics Class 12
- Concepts of Physics - Vol. 2 HC Verma
- Dinesh New Millennium Physics Class -XII (Set of 2 Vols) (Free Complete Solutions to NCERT Textbook Problems & NCERT Exemplar Problems in Physics-XII)
- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- CBSE Chapterwise Questions-Solutions Physics, Class 12
- CBSE 15 Sample Question Paper: Physics for Class 12th

- ncert solutions for class 6 Science
- ncert solutions for class 6 Maths
- ncert solutions for class 7 Science
- ncert solutions for class 7 Maths
- ncert solutions for class 8 Science
- ncert solutions for class 8 Maths
- ncert solutions for class 9 Science
- ncert solutions for class 9 Maths
- ncert solutions for class 10 Science
- ncert solutions for class 10 Maths