# Determining the internal resistance of a given primary cell using cell using potentiometer

Aim : To determine the internal resistance of a given primary cell using cell using potentiometer .

Apparatus : a potentiometer , a battery , (or eliminator ) , two one way key , a rheostat of low resistance , a galvanometer , a high resistance box , a fractional resistance box , an ammeter , a voltmeter , a cell , a jockey , a set square , connecting wires , a piece of sand paper .

Theory

When Key K2 is open and K1 is closed ,

Let null point be obtained at a distance l1 from A

: E =Kl1 (1)

When key K2 is closed and K1 is open ,

Let null point be obtained at a distance l2 from A

V=Kl2 (2)

Where S is the shunt resistance in parallel with given cell .

l1 and l2 ; balancing length without & with shunt respectively .

r: internal resistance of the cell.

Procedure

Make the connection as shown in diagram .

Clean the ends of the connecting wires with sand paper and make tight connection , tighten the plug of the resistance box .

Check the emf of the battery and cell and see the emf of the battery is more than that of the given cell other wise null or balance point wont be obtained (E` >E).

Take maximum current from battery , making rheostat resistance small.

Insert the plug key k , and adjust the rheostat so that a null point is obtained on the fourth wire of the potentiometer .

Insert the 2000 ohm plug in its position in resistance box and obtain a null point by slightly adjusting the jockey .

Measure the balancing length l1.

Take out the 2000 ohm plug from the resistance box . introduce the plug in the key k1 as well in key k2 .Take out a small resistance from the resistance box R connected in parallel with cell.

Slide the jockey along the potentiometer wire and obtain a null point

Insert the 200 ohm plug back in its position in RB and make further adjustment for sharp null point .

Measure the balancing length l2 from end P.

Remove the plugs key k1 and K2 . wait for some time and repeat the activity for the same current .

Circuit Diagram

Observation

 Value of Shunt resistance (S in ohm) Balance Length l1 (K2 is open) without Shunt (cm) Balance length l2 with Stunt (K2 is closed) (cm) r = [(l1-l2)S]/l2 (in ohm) Mean ‘r’ 1.5 171.4 64 1.67 1.77 ohm 2 171.3 61.5 1.78 2.5 171.1 59.6 1.87

Calculation

Mean ‘r’ = (1.67+1.78+1.87)/3 = 1.77 ohm

Result

The internal resistance (R) of given cell is 1.77 OHM

Precaution

For one set of observation the ammeter reading should be constant .

Current should be passed for short time .

Jockey should be rubbed against potentiometer wire .

Sources of error

The emf of the battery is less then the cell (E`<E)

Cell is disturbed during the experiment .

Viva Questions

What do you mean by the internal resistance of cell ?

It is the resistance offered by electrodes

What are the factors on which internal resistance of a cell depends ?

1 Nature of electrodes

2 Nature of electrolyte

3 concentration of electrolyte

4 Temperature of electrolyte

5 Distance between electrodes

6 Area immersed

Does the internal resistance depends on current drawn from the cell ?

Yes the internal resistance usually increases as more current is drawn fom the cell

Can we find the internal resistance of a secondary cell ?

No the internal resistance of a secondary cell is so small (0.01 ohm ) that this method can’t be used .

What should a cell not be disturbed during the experiment ?

Disturbing the cell may change the factors on which internal resistance depends

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## 6 Replies to “Determining the internal resistance of a given primary cell using cell using potentiometer”

1. What is the Emf of the battery and the cell used to tabulate the above given readings?

2. Harsh says:

What is the least count of ammeter and voltmeter, and potential drop across the battery and the cell

3. Meshach says:

Please help am having pratical on..resistance of a pontientiometer wire, arrangement would be in series and paralle 1 to 7 resistor and draw table for that and graph
am to calculate standard error and proof , I+I=V^2/rR

4. Bhargav sompura says:

Least count is the minimum division of value that an instrument can measure .It is figured by counting no. of line divided by its value.

5. himanshu says:

Dude I think you forgot to multiply shunt resistance to calculate internal resistance of cell

6. Prakash Nath says:

Sir this experiment will be same . Determination of a internal resistance of a cell by potentiometer . This expt. I will write this above question ans will this above experiment i will confuse we can ans this and reply ……