Author: Admin

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 .

Record your observation

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

You can also get Class XII Practicals on Biology, Physics, and Physical Education.

Aim : To determine resistance of a galvanometer by half deflection method and to find its figure of merit .

Apparatus : A Weston type of galvanometer , a voltmeter , a battery /battery eliminator , two 10,000 ohm and 200 ohm ) resistance boxes , two one way key , a rheostat , a screw gauge , a meter scale , an ammeter of given range , connecting wires and a piece of sand paper .

Theory

The resistance of the given galvanometer as found by half deflection method .

G= (RxS)/(R-S) (1)

Where R is the resistance connected in series with the galvanometer and S is the shunt resistance .

The figure of merit , K =E /[(R+G)Q] (2)

Where E is the end of the cell and Q is the deflection produced with resistance R.

The maximum current that can pass through the galvanometer , I_g=nk

Where n is the total number of dimension on the galvanometer scale on either side of zero .

Procedure

1. Resistance of galvanometer by half deflection method

• Make the connections accordingly as shown in circuit diagram .
• See that all plugs of the resistance boxes are tight
• Take out the high resistance (200 Ohm ) from the resistance box R and insert key K1 only .
• Adjust the value of R1 so that deflection is maximum , even in number and within the scale .
• Note the deflection . Let it be Q.
• Insert the key k2 also and without changing the value of R2 ,adjust the value of S , such that deflection in the galvanometer reduces to exactly half the value of resistance S.
• Note the value of resistance S .
• Repeat step 4 & 7 three times taking different R and adjusting .

2. Figure of merit

• Take one cell of the battery and find its emf by a voltmeter by connecting +ve of the voltmeter with +ve of cell and – ve of the voltmeter with -ve of cell . Let it be E.
• Take connection as in circuit diagram .
• Adjust the value of R to obtain a certain deflection Q when circuit is closed .
• Note the value of resistance R and deflection Q.
• Now change the value of R and note the galvanometer deflection again .
• Repeat the step with both cells of the battery with different voltages like 2,4,6 ,8 volts from battery eliminator.
• Find the figure of merit K using the formula

Observation table

Resistance (R)	Deflection (Q)	Shunt (S)	G =(RxS)/(R-S)
5000	10	5	5.005
10000	6	3	3
2000	20	10	10.05
1500	26	3	13.11
Average			G = 7.79 ohms

Resistance (R)	Deflection (Q)	K
5000	9	0.556
10000	5	0.60
2000	20	0.502
1500	25	0.524
Average		K = 0.5455 A/dn

Calculations

1. G = (5.005+3+10.05+13.11)/4 [E = 1.45V]
2. Figure of Merit : Current in the galvanometer per unit time
3. K = IQ = [(E)/(R+G)]x[1/Q] – SI unit = AD division = 10^-5 Al div

Result

Resistance of given galvanometer is 7.79 ohm .

Figure of merit of given galvanometer is 0.545 A/dn .

Precautions

All connections should be neat and tight .

The emf of cell / battery should be constant .

Source of error

The scene of the instrument may be loose .

The galvanometer division may not be of equal size.

Diagram

Resistance of Galvanometer

Figure of Merit

Viva question

What is the galvanometer ?

It is a device used for detecting feeble electric currents in circuit .

Define the figure of merit of a galvanometer ?

The quantity of current required to produce a deflection of one division in the galvanometer is called figure of merit of a galvanometer . It is represented by K . Its unit is ampere per division .

Why is this method called half deflection method ?

It is so because the deflection is made half by using a shunt resistance S.

Under what condition id the resistance of a galvanometer (E₁) equal to shunt resistance (S)?

E1=S , only when series resistance R is very high .

You can also get Class XII Practicals on Biology, Physics, and Physical Education.

Aim : To compare EMF of two given primary cells using potentiometer

Apparatus : potentiometer , a leclanche cell , a daniel cell , an ammeter , a voltmeter , a galvanometer , a battery , (battery eleminator), a rheostat , of low resistance , a resistance box , a one way key , a two way key , a jockey , a set square , connecting wire , a piece of sand paper .

Theory

When we keep key (K₁) closed and (K₂) open , let the null point found be l_AJ₁

E₁=Kl_Aj₁ (1)

When we keep K₁ open and K₂ closed , let null point obtained by l_AJ₂ .

E₂=Kl_Aj₂ (2)

(1)/ (2)

E₁ / E₂ = Kl_AJ₁ / Kl_Aj₂

E₁/E₂ =l₁/l₂

Where E₁ and E₂ are the emf of two given cells .

Procedure

1. Arrange the apparatus as shown in the circuit diagram .
2. Connect the positive poles of the cells to the terminal and the negative poles to the terminal a and b of the two way key .
3. Insert the plug in the key K and also in between the terminals a and c of the two way key .
4. Slide the jockey gently over the potentiometer wires until you obtain a point of no deflection .
5. Note the length l₁ at the point.
6. Repeat this with E₂ by disconnecting E₁ and inserting plug into gap a and c of two way key .
7. Record l₂ at null point .
8. Repeat this different resistance .

Observation table

Balancing Lengths		E1 / E2 = lAJ1 / lAj2
L1 for cell E1 (cm)	L2 for cell E2 (cm)
327	376	0.86
323.5	371	0.87
321.5	369	0.87
312.5	352.5	0.88

Result

The ratio of emf E₁/E₂ ~0.87

Precaution

• • 1. The connections should be neat, clean and tight.
    2. The plugs should be introduced in the keys only when the observations taken.
    3. The positive poles of the battery E and cells E_x and E₂ should, all be connected to terminal at the zero of the wires.
    4. The jockey key should not be rubbed along the wire. It should touch the
    5. The e.m.f. of the battery should be greater than the e.m.f.’s of the either cells.
    6. Some high resistance plug should always be taken out from resistant the jockey is moved along the wire.

Sources of error

1. The auxiliary battery may not be fully charged.
2. The potentiometer wire may not be of uniform cross-section and material. throughout its length.
3. End resistances may not be zero.

Diagram

VIVA Question

What is emf of a cell?

Electromotive force of a cell is the potential difference across the terminals of the cells when it is in open circuit i.e. no current is drawn from the circuit .

What is potentiometer ?

An instrument to measure potentiometer difference or emf of a cell .

What is the principle of potentiometer ?

For a constant current , fall of potentiometer along a uniform wire is directly proportional to its length .

What is preferred for making potentiometer wire ?

Magnum , due to its low temperature coefficient of resistance and high resistivity .

What will you conclude if the deflection of the galvanometer is in the same direction ?

The positive terminals aren’t connected at one point

Potential difference between wires ends is less than emf of cells to be measured .

Emf of driving cells is less than that of cells to be measured .

You can also get Class XII Practicals on Biology, Physics, and Physical Education.

 

Aim : To verify the laws of combination (series) of resistance using a meter bridge (R=R₁ + R₂ )

Apparatus : a meter bridge , a leclanche cell (battery eliminator ) , a galvanometer , a resistance box , a jockey , two resistance wire or two resistance coils , set square , connecting wires .

Theory

1. The resistance (r) of a wire or coil is given by

Where R is the resistance from resistance box in the left gap and l is the length of the meter bridge wire from zero end upto balance point .

2. When two resistance r₁ & r₂ are connected in series , then their combined resistance

R₃=R₁+R₂

Procedure

1. Mark the two resistance coil as r₁ and r₂ .
2. To find the r₁ and r₂ proceed same way as in experiment 1.

(if r₁ and r₂ are not known )

1. Connect the two coils r₁ and r₂ in series as shown circuit diagram in the right gap of meter bridge and find the resistance of this combination. Take at least three sets of observation.
2. Record your observation.

Result

Within limits of experimental error , experimental and theoretical values of R  are same . Hence law of resistance in series is verified .

Precaution

1. The key should be inserted only while taking observation .
2. Connections should be neat and tight .

Sources of error

1. The key in resistance box may be loose .
2. The galvanometer may not show opposite deflection .

 

Circuit diagram

Observation table

Resistance in R.B. (ohm)	Balance l (cm)	(100-l) (cm)	[(100-l)*R]/l	Mean r1 (cm)
0.5	24	76	1.583	1.616
1	38	62	1.631
2	55	45	1.636

Resistance in R.B. (ohm)	Balance l (cm)	(100-l) (cm)	[(100-l)*R]/l	Mean r2 (cm)
0.5	33	67	1.015	1.015
1	50	50	1
2	66	34	1.030

Resistance in R.B. (ohm)	Balance l (cm)	(100-l) (cm)	[(100-l)*R]/l	Mean rs (cm)
0.5	16	84	2.625	2.615
1	28	72	2.571
2	43	57	2.651

Calculation

R₁ + R₂ = 1.616+1.015 = 2.631 ohm
R_s = 2.615 ohm

Therefore, experimental error = [(2.615-2.631)/2.615]*100 = 0.6%

Viva question

How does resistance change in series combination ?

Ans. Resistance increase in series combination

Why does resistance increase in series combination ?

Ans. Effective length of resistor increases . As R*l , resistance increases .

What is resistance ?

Ans. it is the opposition offered by the material of two wire to flow of electric current .

What is the cause of resistance ?

Ans. it is the opposition offered by the material of wire to flow of electric current .

What is the effect of temperature In the resistance of conductor ?

Resistance of all conductor increases with increase in temperature .

You can also get Class XII Practicals on Biology, Physics, and Physical Education.

Aim: To verify laws of combination (parallel) of resistance using meter bridge .

Apparatus: a meter bridge , a leclanche cell , a galvanometer , a jockey , a jockey , a resistance box , two resistance wire , connecting wires , set square .

Theory

The resistance (r) of a resistance wire or coil is given by r=(100-l)/L *R where R is the resistance from resistance box in the left gap and L is the length of meter bridge wire from O end upto balance point.

When r₁ and r₂ are connected in parallel , then their combined resistance

Rp = r₁r₂ /r₁+r₂

Procedure

1. Mark the two resistance wires as r₁ and r₂
2. Connect the two coils r₁ and r₂ in parallel as shown in fig. in the right gap of meter bridge and find the resistance of this combination . Take at least three set of observation
3. Record your observation .

Result

Within limit of experimental error , experimental and theoretical values of R_p are same .

Hence law of resistance in parallel are verified .

Precaution

The key should be inserted only while taking observation .

Connection should be neat and tight .

Source of error

The galvanometer may not show opposite deflection

The key in the resistance box may be close .

OBSERVATION TABLE

Value of r₁

Resistance of R.B. (ohm)	Balance L (cm)	(100-l) (cm)	[(100-l)*R]/L	Mean r1
0.5	59	41	0.347	0.6393
1	37	63	0.587
2	33	67	0.985

Value of r₂

Resistance of R.B. (ohm)	Balance L (cm)	(100-l) (cm)	[(100-l)*R]/L	Mean r2
0.5	83.2	16.8	0.1009	1.729
1	80.8	19.2	4.208
2	69.4	30.6	0.881

Value of r_p

Resistance of R.B. (ohm)	Balance L (cm)	(100-l) (cm)	[(100-l)*R]/L	Mean rp
0.1	27	73	0.27	0.36
0.2	32	68	0.42
0.5	56	44	0.39

Calculations

Rp = (R₁R₂)/R₁+R₂ = (0.693×1.729)/(0.693+1.729) = 0.4667 ohm

Experimental Error = [(0.36-0.4667)/(0.36)]*100 = 29.6%

 

Viva question

1. How does resistance change in parallel combination ?

Resistance decrease in parallel combination

2. explain the decrease of resistance in parallel combination ?

Effective area of cross section increases as r x 1/A , resistance decreases .

3. Why should jockey not be pressed too hard on wire while sliding ?

Sliding the jockey with a hard press will scratch the wire and make its thickness non-uniform .Resistance per unit length of wire will be inconsistent as resistance depends on area of cross section .

4. What is the resistance of an open key ? Explain.

Open key has infinite resistance as it makes current zero (R=V/I)

5. Define unit of resistance .

SI unit of resistance is OHM or volt per ampere . 1 Ohm is the resistance of a conduction carrying 1 ampere current when potential difference maintained across its end is 1 volt .

You can also get Class XII Practicals on Biology, Physics, and Physical Education.

Aim – To find the resistance of a given wire using meter bridge and hence determine resistivity of its material .

Apparatus

A meter bridge (slide wire bridge ) , a lecranche cell, a galvanometer , a resistor , a jockey , a one way key , a resistance wire , a screw gauge , a meter scale , a set of square , connecting wire , a piece of sand paper .

Theory :

The unknown resistance ‘X” is given by

1. Where ‘R” is the known resistance placed in the left gap & unknown resistance ‘X” is the right gap of meter bridge .’L’ is length of meter bridge wire from zero end upto balance .
2. Resistivity of the material of the given wire is given by

Where ‘L’ is the length & D is the diameter of the given wire .

Procedure

1. Arrange the apparatus as shown in the arrangement diagram .
2. Connect the resistance wire whose resistance is to be determine in the right gap b/w C& B .Take care that no point /part of the wire forms a loop.
3. Connect resistance box of low range in the left gap b/w A & B .
4. Make all other connection as shown in the circuit diagram .
5. Take out some resistance from the resistance box , ping the key ‘K’
6. Touch the jockey gently first at length end & then right end of the bridge wire .
7. Note the deflection in the galvanometer. If the galvanometer shows deflection in the galvanometer reading in opposite direction the correction are correct. If the deflection is on one side only then there is fault in the circuit . Check & rectify the fault .
8. Move the jockey gently along the wire from left to right till gives zero deflection . The point where the jockey is touching the wire is null point ‘D’.
9. Choose an appropriate value of ‘R’ from the box such that there id no defection in the galvanometer when the jockey is nearly in the middle of the wire .
10. Note position of point ‘D’ to known the length AD=l
11. Take atleast 4 sets of observation in the same way by changing the value of R in the steps .
12. Record your observation .
13. For specific resistance
    1. Cut the resistance wire at the point where it leaves the terminal , started it & find its length by using a meter scale .
    2. Measure the diameter of the wire at least at 4 places in two mutually perpendicular direction at each place with the of screw gauge .
    3. Record your observation as given in the table .

Observation

1. Length of a given wire L = 66 cm =0.66 m
2. Table for unknown resistance (X)

Resistance from box, R (Ohm)	Length AB =l (cm)	Length BC = (100-l) (cm)	Unkown Resistance X = [R(100-l)]/L (ohm)
0.5	58.3	41.7	0.35
0.7	60.7	39.3	0.45
1	61.9	38.1	0.61
1.5	61.1	38.9	0.95
			Mean = 0.59

3. Least count of screw gauge

Pitch of screw gauge =0.01

Total no of division on the circular scale =

LC of screw gauge = Pitch /No of the circular scale

Zero error (e) =(0)

Zero connection =(e)=0

Radius of the resistance wire

Main Scale Reading (mm)	Circular Scale Reading	Total Reading (diameter) (mm)	Mean D (mm)	Mean radius (D/2) (mm)
0	43	0.43	0.42	0.21
0	41	0.41

Result

1. The value of the unknown resistance X =0.5 ohm
2. The specific resistance of material of wire = 0.104 x10^-3 ohm m
3. Percentage error

Precaution

The connection should be neat, clean & tight.

Source of error

Plug may not be clean

The wire may not be of uniform thickness.

Viva questions

1. Why is the meter bridge so called?

Since the bridge uses one meter long wire, it is called a meter bridge .

1. What is a null point ?

It is a point on the wire , keeping jockey at which galvanometer gives 0 deflection .

1. Why is the bridge method better that ohms law for measurement ?

It is so because the bridge method is a null method (at null point , no current is flowing in galvanometer ) and more sensitive .

1. Why copper strips used to press ends of wire are thick ?

Thick Cu strips have negligible resistance over the resistance of alloy meter bridge wire and minimize affect of end resistance .

Circuit diagram

You can also get Class XII Practicals on Biology, Physics, and Physical Education.