Sastry CBSE

To Determine The Internal Resistance of a Given Primary Cell Using Potentiometer

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

Apparatus
A potentiometer, a battery (or battery eliminator), two one-way keys, a rheostat of low resistance, a galvanometer, a high resistance box, a fractional resistance box, an ammeter, a voltmeter, a cell (say Leclanche cell), a jockey, a set square, connecting wires and a piece of sand paper.
Theory
The internal resistance of a cell is given by

where and l₂ are the balancing lengths without shunt and with shunt, respectively, and R is the shunt resistance in parallel with the given cell.

Circuit diagram

Procedure

1. Make the connections accordingly, as shown in circuit diagram.
2. Clean the ends of the connecting wires with sand paper and make tight connections according to the circuit diagram.
3. Tight the plugs of the resistance box.
4. Check the e.m.f. of the battery and cell and see that e.m.f. of the battery is more than that of the given cell, otherwise null or balance point will not be obtained (E’ > E).
5. Take maximum current from the battery, making rheostat resistance small.
6. To test the correctness of the connections. (Insert the plug in the key K₁ and note the ammeter reading. Take out 2000 Ω resistance plug from the resistance box. Place the jockey first at the end P of the wire and then at the end Q. If the galvanometer shows deflection in opposite directions in the two cases, the connections are correct).
7. Without inserting the plug in the key K₂ adjust the rheostat so that a null point is obtained on the fourth wire of potentiometer.
8. Insert the 2000 ohm plug back in its position in resistance box and by slightly adjusting the jockey near the previously obtained position of null point, obtain the null point position accurately, using a set square.
9. Measure the balancing length l₁ between this point and the end P of the wire.
10. Take out the 2000 ohms plug again from the resistance box R.B. Introduce the plugs in key K₁, as well as in key K₂. Take out a small resistance (1-5 Ω) from the resistance box R connected in parallel with the cell.
11. Slide the jockey along the potentiometer wire and obtain null point.
12. Insert 2000 ohms plug back in its position in R.B. and if necessary make further adjustment for sharp null point.
13. Measure the balancing length l₁ from end P.
14. Remove the plug keys at K₁ and K₂. Wait for sometime and for the same value of current (as shown by the ammeter), repeat the steps 7 to 13.
15. Repeat the observations for different values of R repeating each observation twice.
16. Record your observations as given below.

Observations

1. Range of voltmeter =………
   Least count of voltmeter = ………
   E.M.F. of battery (or battery eleminator) = ………
   E.M.F. of cell = ………
2.                                                                                           Table for Lengths
   

Calculations

1. For each set of observation find mean and l₂ and write in column 3c and 3f.
2. Calculate value of r for each set and write it in column 5.
3. Take mean of values of r recorded in column 5.

Result
The internal resistance of the given cell is………

Precautions
Same as in Experiment 4 and other precautions are as:

1. The e.m.f. of the battery should be greater than that of the cell.
2. For one set of observation the ammeter reading should remain constant.
3. Current should be passed for short time only, while finding the null point.
4. Rheostat should be adjusted so that initial null point lies on last wire of the potentiometer.
5. Cell should not be disturbed during experiment.
6. Jockey should not be rubbed against the potentiometer wire.

Viva Voce

Question. 1. What do you understand by the e.m.f. of a cell?
Answer. Electromotive force i.e., e.m.f. of a cell is the potential difference across the terminals of the cell when the cell is in an open circuit i.e., when no current is drawn from the cell.

Question. 2. What is a potentiometer?
Answer. It is an instrument used to measure potential difference or e.m.f. of a cell.

Question. 3. Why is it called a potentiometer? ,
Answer. Because it measures potential difference between any two points of electric circuits.

Question. 4. What is the principle of a potentiometer?
Answer. It works On the principle that for a constant current, fall of potential along a uniform wire is directly proportional to its length.

Question. 5. What is potential gradient?
Answer.It is the fall of potential per unit length of the potentiometer wire. K =V/l.

Question. 6. How does the potential gradient vary along the length of the wire from end P to end Q?
Answer. Potential gradient is same throughout if the wire has uniform cross-section and material density.

Question. 7. What kind of source of e.m.f. should be used as auxiliary battery?
Answer. The e.m.f. of the source must be steady. A freshly charged accumulator should be used for this purpose.

Question. 8. What should be the order of magnitude of the e.m.f. of the auxiliary battery?
Answer. The e.m.f. of the auxiliary battery should be slightly greater than the e.m.f. of the individual cells.
(With battery of lesser e.m.f., null point will not be obtained on the potentiometer wire).

Question. 9. Why do we use a rheostat in the battery circuit?
Answer. To vary the potential gradient.

Question. 10. What purpose is served by varying the potential gradient?
Answer. A lower potential gradient gives more length of wire upto null point. Accuracy becomes more.

Question. 11. On what factors does the potential gradient depend?
Answer. Potential gradient depends directly on the strength of the current and resistance per cm
of the wire. K = Ip/A.

Question. 12. What is the preferred material used for making potentiometer wires?
Answer. Manganin. It is characterised by a low temperature coefficient of resistance and a high resistivity.

Question. 13. Why do we want the material of the potentiometer wire to have a low temperature coefficient of resistance?
Answer. There is invariably some heating of the potentiometer wire when a current flows through it. A material with a low temperature coefficient ensures that its resistance does not change much because of this heating.

Question. 14. Why don’t we use a copper wire as a potentiometer wire?
Answer. Copper has a high temperature coefficient of resistance and low resistivity and hence a copper wire will have a low resistance. There would then be no appreciable potential drop across the ends of the potentiometer wire.

Question. 15. Which materials can be used for making potentiometer wire?
Answer. The alloys like manganin, constantan etc.

Question. 16. What do you mean with sensitivity of a potentiometer?
Answer. Sensitivity of a potentiometer is the smallest potential difference that it can measure.

Question. 17.Why is a ten-wire potentiometer more sensitive than a four-wire one?
Answer.The potential gradient, under same conditions, decreases with an increase in the length of the potentiometer wire. Hence, a 10-wire potentiometer (having a smaller potential gradient) is more sensitive than a 4-wire one.

Question. 18.How will you know that the apparatus can give a null point?
Answer.The jockey is put at the two ends of the potentiometer wire. The deflection in the galvanometer must be in opposite directions.

Question. 19.What will you conclude if the deflection of the galvanometer is in same direction at both the ends?
Answer.The reasons may be

1. the positive terminals of all the cells are not connected at one point.
2. the potential difference between the ends of the wire is less than the e.m.f. of the cell which is to be measured.
3. the e.m.f. of driving cell is less than the e.m.f. of each cells whose e.m.f. to be com¬pared or measured.

Question. 20. How are above situations corrected?
Answer.

1. Connections of positive terminals are checked.
2. Current in potentiometer wire is increased.
3. E > E₁ or E >E₂.

Question. 21. Under what conditions galvanometer will give no deflection when jockey is put on the wire?
Answer. The reason may be

1. the cell whose e.m.f. is being measured, is totally damaged to have infinite internal resistance.
2. connecting wire in the galvanometer circuit may be broken.

Question. 22. Under what conditions deflection in the galvanometer is shaky?
Answer. The reason may be

1. the e.m.f. of the battery or the cells may be fluctuating.
2. the circuit has a loose contact somewhere.

Question. 23. Why should we use a sensitive galvanometer?
Answer. A sensitive galvanometer will respond to even a small departure from the exact balance point and will hence enable us to locate the balance point with greater precision.

Question. 24. Why do we need a protective series resistance/shunt along with a sensitive galvanometer?
Answer. To prevent it from damage from the flow of excessive currents that may exist when the jockey is far from the balance point.

Question. 25. Does the use of a series protective resistance/shunt effect the location of the balance point?
Answer. No; however, it makes the galvanometer less sensitive. We therefore, remove it once we are near the balance point.

Question. 26. Why do we not want the balance point to be on the first wire, say?
Answer. The smaller is the balancing length, the greater is the relative uncertainty in its location.

Question. 27.What is the merit of a potentiometer over a voltmeter in measurement of e.m.f. of a cell?
Answer.E.M.F. measured by potentiometer is more accurate because the cell is in open circuit, giving no current.

Question. 28.How will you determine specific resistance of potentiometer wire material?
Answer.We measure V across a known length l of the wire. We measure diameter D of wire and

Question. 29.What do you mean by internal resistance of a cell?
Answer.It is the resistance offered by the electrolyte to the flow of ions to their respective electrodes.

Question. 30.Is there any change in the internal resistance of cell in open and closed circuit?
Answer.

Question. 31.On what factors does the internal resistance of a cell depend?
Answer.Internal resistance of a cell depends upon :

1. Distance between electrodes and is directly proportional to its
2. Facing surface area of the electrodes in electrolyte and is inversely proportional to it
3. Nature of electrolyte and is inversely proportional to its specific conductivity
4. Temperature increases, the internal resistance decreases and vice-versa.
5. Internal resistance increases with the use of cell.

Question. 32.Does the internal resistance depend on the current drawn from the cell?
Answer.Yes, the internal resistance usually increases as more current is drawn from the cell.

Question. 33.Can we find the internal resistance of an accumulator or secondary cell?
Answer.No. the internal resistance of an accumulator is so small (= 0.01 Q) that this method cannot be used.

Question. 34.Why a cell should not be disturbed during experiment?
Answer.Disturbing of the cell may change the factors (Q. 31 above) on which the internal resistance of the cell depends.

Question. 35.What other measurements can be made by a potentiometer?
Answer.A potentiometer can be used for measuring small thermo e.m.f. It can also be used for calibrating voltmeter and ammeter. It can be used to measure and control stress, temperature, radiation, pH, frequency etc.

Question. 36.Can you measure e.m.f. by a voltmeter?
Answer.No. The voltmeter measure the terminal potential difference of a cell because it draw some current
V = E -Ir, when  I≠0, then V < E.

Question. 37.Which voltmeters can be used to measure the e.m.f. of the cells?
Answer.Electric voltmeter. Vaccum tube volt meter (VTVM) afters nearly infinite resistance. So the current drawn is minimum, nearly zero. These two voltmeter are act as ideal voltmeter.

Question. 38. Is the terminal potential difference (V) and e.m.f. (E) of a cell different? Explain.
Answer.

Question. 39. Does the at position of balance point (null point) mean no current through the potentiometer?
Answer. No. the current always flow in potentiometer wire. These is no current in galvanometer f because there is no current drawn from the cell whose e.m.f. is to be measured or compared.

Question. 40. Does the potentiometer is used to determine the internal resistance of
(i) primary cell (ii) secondary cell?
Answer. The potentiometer is used to determine the internal resistance of primary cell only but not secondary cell because of very small resistance (0.02 Q).

Question. 41. What are the factors on which the e.m.f. of a cell depends?
Answer.

1. Nature of electrodes,
2. Nature of electrolyte,
3. concentration of electrolyte,
4. Temperature of electrolyte.

Question. 42. Why is a potentiometer preferred over a voltmeter for measuring the e.m.f. of cell?
Answer. A potentiometer draws no current from the cell whose e.m.f. is to be measured. On the other hand, the voltmeter always some current. Thus e.m.f. measured by voltmeter will be slightly less than the e.m.f. measured by potentiometer.
V = E -Ir

Question. 43. Why do we prefer a potentiometer with a longer bridge wire?
Answer. When the bridge wire is longer, the potential gradient is smaller. Smaller the potential gradient, more is the sensitivity of potentiometer wire.

Question. 44. What are the factors on which internal resistance of a cell depends.
Answer.

1. Nature of electrodes
2. Nature of electrolyte
3. Concentration of electrolyte
4. Temperature of electrolyte
5. Distance between the electrodes
6. The area of electrodes immered in electrolyte.

Question. 45. Can we consider the potentiometer as an ideal voltmeter?
Answer. Yes. At null point, the potentiometer does not draw any current. Hence it measure the emf. The potentiometer is equivalent to an ideal voltmeter.

Physics Lab ManualNCERT Solutions Class 12 Physics Sample Papers

To Compare The EMF of Two Given Primary Cells Using Potentiometer.

Aim
To compare the EMF of two given primary cells using potentiometer.

Apparatus
Potentiometer, a Leclanche cell, a Daniel cell, an ammeter, a voltmeter, a galvanometer, a battery (or battery eliminator), a rheostat of low resistance, a resistance box, a one way key, a two way key, a jockey, a set square, connecting wires and a piece of sand paper.

Theory

where, E₁ and E₂ are the e.m.f. of two given cells and l₁ and l₂  are the corresponding balancing lengths on potentiometer wire.

Circuit diagram

Procedure

1. Arrange the apparatus as shown in circuit diagram figure.
2. Remove the insulation from the ends of the connecting copper wires with a sand paper.
3. Measure the e.m.f. (E) of the battery and the e.m.fs. (E₁ and E₂ ) of the cells. See that E > E₁  and also E > E₂ .
4. Connect the positive pole of the battery (a battery of constant e.m.f.) to the zero end (P) of the potentiometer and the negative pole through a one-way key, an ammeter and a low resistance rheostat to the other end (Q) of the potentiometer.
5. Connect the positive poles of the cells E₁ and E₂  to the terminal at the zero end (P) and the negative poles to the terminals a and b of the two way key.
6. Connect the common terminal c of the two-way key through a galvanometer (G) and a resistance box (R.B.) to the jockey J.
7. Take maximum current from the battery making rheostat resistance zero.
8. Insert the plug in the one-way key (K) in circuit and also in between the terminals a and c of the two-way key.
9. Take out a 2,000 ohms plug from the resistance box (R.B.).
10. Press the jockey at the zero end and note the direction of deflection in the galvanometer.
11. Press the jockey at the other end of the potentiometer wire. If the direction of deflection is opposite to that in the first case, the connections are correct. (If the deflection is in the same direction then either connections are wrong or e.m.f. of the auxiliary battery is less).
12.  Slide the jockey gently over the potentiometer wires till you obtain a point where galvanometer shows no deflection.
13. Put the 2000 ohms plug back in the resistance box and obtain the null point position accurately, using a set square.
14. Note the length l₁ of the wire for the cell E₁ Also note the current as indicated by the ammeter.
15. Disconnect the cell E₁  by removing the plug from gap ac of two-way key and connect the cell E₂  by inserting plug into gap be of two-way key.
16. Take out a 2000 ohms plug from resistance box R.B. and slide the jockey along potentiometer wire so as to obtain no deflection position.
17. Put the 2000 ohms plug back in the resistance box and obtain accurate position of null point for second cell E₂ .
18. Note the length l₂  of wire in this position for the cell E₂ . However, make sure that ammeter reading is same as in step 14.
19. Repeat the observations alternately for each cell again for the same value of current.
20. Increase the current by adjusting the rheostat and obtain at least three sets of observations in a similar way.
21. Record your observations as given below

Observations

Calculations

1. For each observation find mean l₁ and mean l₂  and record in column 3c and 4c.
2. Find E₁/E₂ for each set, by dividing mean l₁  (column 3c) by mean l₂  (column 4c).
3. Find mean E₁/E₂ .  

Result

Precautions

1. The connections should be neat, clean and tight.
2. The plugs should be introduced in the keys only when the observations are to be taken.
3. The positive poles of the battery E and cells E₁ and E₂  should, all be connected to the terminal at the zero of the wires.
4. The jockey key should not be rubbed along the wire. It should touch the wire gently.
5. The ammeter reading should remain constant for a particular set of observation. If necessary, adjust the rheostat for this purpose.
6. The e.m.f. of the battery should be greater than the e.m.f.’s of the either of the two
   cells.
7. Some high resistance plug should always be taken out from resistance box before the jockey is moved along the wire.

Sources of error

1. Same as in previous experiments.
2. The auxiliary battery may not be fully charged.
3. The potentiometer wire may not be of uniform cross-section and material density throughout its length.
4. End resistances may not be zero.

Physics Lab ManualNCERT Solutions Class 12 Physics Sample Papers

To verify the laws of combination (parallel) of resistances using a metre bridge.

Aim
To verify the laws of combination (parallel) of resistances using a metre bridge.

Apparatus
A metre bridge, a Leclanche cell (battery eliminator), a galvanometer, a resistance box, a jockey, two resistance wires or two resistance coils known resistances, a set square, sand paper and connecting wires.

Theory

where R is the resistance from the resistance box in the left gap and l is the length of the metre bridge wire from zero end up to balance point.

Circuit diagram

Procedure

1. Mark the two resistance coils as r₁ and r₂.
2. To find r₁ and r₂ proceed same way as in Experiment 1. (If r₁ and r₂ are not known.)
3. Connect the two coils r₁ and r₂ in parallel as shown in figure in the right gap of metre bridge and find the resistance of this combination. Take at least three sets of observations.
4. Record your observations.

Observations

Table for length(l) and Unknown resistance (x)

Calculations

Result

Within limits of experimental error, experimental and theoretical values of R_p are same. Hence, law of resistances in parallel is verified.

Precautions
Same as in Experiment 1.

Viva Voce

Electric current
Question.1.State Ohm’s Law.
Answer.Ohm’s law states that the electric current I flowing through a conductor is directly proportional to the potential difference (voltage) V across its ends (provided that the physical conditions—temperature, pressure and dimensions of the conductor remain same).

Question.2.Give mathematical form of Ohm’s law.
Answer.Mathematical form of Ohm’s law is, V = RI.

Question.3.Define resistance.
Answer.The constant ratio of potential difference V across the ends of a conductor to the current I flowing through it, is called resistance of the conductor. It is represented by the symbol R.

Question.4.What are Ohmic and non-ohmic resistances?
Answer.Resistances which obey Ohm’s law, are called ohmic resistances e.g., metals like Cu, Al, Ag etc. at low temperature.
Resistances which do not obey Ohm’s law are called non-ohmic resistances e.g., diodes and transistors.

Question.5.Give common examples of non-ohmic resistances.
Answer.Vacuum tube diodes, semi-conductor diodes and transistors are non-ohmic resistances.

Question.6.What is effect of temperature on the resistance of a conductor?
Answer.The resistance of the conductors increases with increase in temperature.

Question.7.Name some substances whose resistance decreases with increase in temperature.
Answer.Resistance of semi-conductors (Si, Ge) decreases with increase in temperature.

Question.8.How do you conclude that the conductor used in experiment obeyed Ohm’s law?
Answer.It is done by two results :

1. The ratio of voltmeter reading (V) and the corresponding ammeter reading (J) comes to be constant.
2. A graph between V and I comes to be a straight line.

Question. 9. Why a large current is not allowed to be passed through the conductor during the experiment?
Answer. If a large current is passed (or even if a small current is passed unnecessarily for a long time), the conductor will become hot and its resistance will increase.- Then the graph will not remain a straight line.

Question.10. Why do we use thick connecting wires?
Answer. Thick connecting wires offer negligible resistance compared to given alloy wire whose resistance is to be determined.

Question.11. What is ohm? Define it.
Answer. Ohm is the S.I. unit of resistance. One ohm is the resistance offered by a conductor when one ampere current is flowing through it, when one volt P.D. is maintained across — its ends.

Question.12. What is a battery eliminator?
Answer. It is a rectifier. It converts high A.C. voltage (220 V) into low desired D.C. voltage such as 2 V, 4 V, 6 V, 8 V, 10 V, 12 V. It is a good substitute for a battery or a cell.

Question. 13. Why, unknown wire whose resistance is to be determined, is made of alloys such as manganin, Eureka?
Answer. Unknown resistance wire is made of alloy, but not of metals, because

1. The resistivity of alloys is greater than that of metals.
2. The temperature coefficient of resistance of alloys is negligible than that of metals.

Question.14. What is the effect of rheostat, range of voltmeter, ammeter on resistance of unknown wire?
Answer. No effect because resistance does not depend upon them.

Question.15. What material is chosen for rheostat wire and why?
Answer. It is constantan alloy. Because its resistivity is high and temperature coefficient of resistance is quite small.

Question.16. What is the material of the connecting wires used in the experiment?
Answer.Copper.

Question.17. Is there any advantage of battery eliminator over usual source of e.m.f.?
Answer. Main advantage of battery eliminator is that current can be drawn at desired voltages and it does not need any charging. It is easy to handle and maintain.

Question. 18. What are the factors affecting the resistance?
Answer. The resistance depends upon length, Area of cross-section, nature of material and temperature of the conductors.

Question.19. What is electric current? Define its S.I. unit.
Answer. The flow of electric charge per unit time through a conductor is called electric current. S.I. unit of current is ampere (A).
The one ampere is the amount of current flowing in a conductor which offers resistance 1 Ohm when one volts potential difference is maintained across the conductor.

Question.20. Define S.I. unit of electric potential.
Answer. Volt is the S.I. unit of electric potential. One volt is said to be the potential difference between two points if one Joule of work is done in bringing one coulomb of charge from one point to the other.

Question.21. Why is a large current not allowed to flow the circuit during the experiment.
Answer. If large current is passed or passed for a long time then wire become hot and its resistance increases. Therefore, the V-I graph will not be a straight line and Ohm’s law is not valid.

Question. 22. Can we take a metal wire in place of alloy wire whose resistance is to be measured?
Answer. No. Because large current will flow and battery will damage.

Question. 23. Why is the ammeter always connected in series to measure the current?
Answer.It is connected in series with circuit in order to measure the current without any change in magnitude.

Question. 24. Why is the voltmeter is connected in parallel?
Answer. So that it can measure the potential drop without any change in its magnitude.

Question.25. Why is the resistance wire on its self before it is wound on bobbin or reel?
Answer. To avoid induction effect.

Resistance

Question.1. Define resistance.
Answer.It is the opposition offered by the material of wire to the flow of electric current, R = V/1

Question.2.How can it measured?
Answer.It can be measured by (i) Ohm’s law (ii) Metre bridge (iii) Multimeter.

Question.3.Define the unit of resistance.
Answer.The S.I. unit of resistance is Ohm (Ω) or volt per ampere. One ohm is the resistance of a conductor carrying current one ampere when unit p.d. is maintained across its ends.

Question.4.What is the cause of resistance?
Answer. Collisions of drifting electrons with the atoms.

Question.5.What is the conductance?
Answer.The reciprocal of resistance is called conductance. It is denoted by G.
G = 1/R S.I. of G is mho, or Siemen.

Question.6.How does resistance depend upon the length of a conductor?
Answer.The resistance is directly proportional to the length of a conductor (provided its area of cross-section remains constant) i.e., R ∝l. It means if length of conductor increases, then resistance increases and vice-versa.

Question.7.How does resistance depend upon the area of cross-section of a conductor?
Answer.The resistance is inversely proportional to the area of cross section of a conductor
(provided its length remains constant) i.e., R ∝=1/A. It means, if area of cross section increases then resistance decreases and vice-versa.

Question.8.What is the resistance of an open key? Explain it.
Answer.

Question.9.What is the length of resistance wire used between the gap of resistance box marked INFINITE? Explain it.
Answer.Infinite marked plug has no wire.
Explanation. We have R = Pl/A We have made R infinite by making A = 0 (rather than by making l infinite).

Question.10.What is the conductance?
Answer.

Question. 11. Define resistivity or specific resistance of the material of conductor
Answer.

Hence, resistivity is defined as the resistance of a conductor of unit length and unit cross-sectional area. The unit of resistivity is ohm-metre (Ω-m).

Question.12.Define electrical conductivity
Answer.It is reciprocal of resistivity. It is represented by the symbol σ =1/P. The unit of electrical conductivity is Siemen per metre (S m^-1).

Question.13.What is the order of magnitude of resistivity of conductors?
Answer. Resistivity of the conductors is of the order of 10^-8 Ω-m.

Question.14.What is effect of temperature on the resistance of a conductor?
Answer.Resistance of all conductors increases with increase in temperature of the conductor.

Question.15.Define temperature coefficient of resistance.
Answer.

Hence, temperature coefficient is defined as the increase in resistance of a conductor of unit resistance due to unit increase in temperature.

Question.16.Give unit of temperature coefficient of resistance.
Answer.The unit of temperature coefficient is per °C (or K) (°C^-1 or K^-1).

Question. 17. How does resistance change in series combination?
Answer. Resistance increases in series combination.

Question. 18. Explain increase of resistance in series combination.
Answer.In series combination, the effective length of resistor increases. As R∝ l, resistance increases in series combination.

Question.19. How does resistance change in parallel combination?
Answer. Resistance decreases in parallel combination.

Question. 20. Explain decrease of resistance in parallel combination.
Answer. In parallel combination, the effective area of cross-section increases. As R∝ 1/A resistance decreases in parallel combination.

Question. 21. What is Wheatstone bridge?
Answer. It is the arrangement of four resistance in quadrilateral form to determine one unknown resistance in term of other three resistances.

Question. 22. What is a metre bridge?
Answer. It is the practical form of Wheatstone bridge to determine the unknown resistance and resistivity of a given alloy wire.

Question. 23. Why is constantan used in the bridge wire?
Answer.

1. The resistivity (49 x 10^-8 Ω-m) of the constantan is high.
2. The temperature coefficient of resistance (a) is very small (0.40 x 10^-4 ) (°C^-1 ).

Question.24. How are the coils wound in a post office box or resistance box?
Answer. The resistance coil is doubly wound to avoid electromagnetic induction.

Wheatstone’s bridge

Question. 25. When is a Wheatstone’s bridge said to be balanced?
Answer. A Wheatstone’s bridge is said to be balanced, when no current flows through the galva¬nometer and it gives zero deflection.

Question. 26. What is the condition for a Wheatstone’s bridge to become balanced?
Answer.

Question. 27. Will the interchange of positions of cell and galvanometer effect the balance condition?
Answer. No. The condition of balanced Wheatstone bridge remains satisfied.

Question.28. When is a Wheatstone’s bridge most sensitive?
Answer. The bridge is most sensitive when all the four resistances P, Q, R and S are of same order of magnitude.

Question.29. What are applied forms of a Wheatstone’s bridge?
Answer. The applied forms of a Wheatstone’s bridge are :

1.  Metre Bridge or Slide Wire Bridge.
2. Post Office Box.

Metere bridge 

Question. 30. Why is a metre bridge so called?
Answer. Since the bridge uses one metre long wire, it is called a metre bridge.

Question. 31. Why is a metre bridge also called a slide wire bridge?
Answer. Since a jockey is slided over the wire (during the experiment), it is also called a slide- wire bridge.

Question. 32. Why the jockey should not be pressed too hard on the wire when sliding over it?
Answer. Sliding the jockey with a hard press, will scratch the wire and make its thickness non¬uniform. Then the resistance per unit length of the wire will not remain constant because resistance depend upon area of cross-section.

Question. 33. What is null point?
Answer. It is a point on the ware, keeping jockey at which, the galvanometer gives zero deflec¬tion.

Question. 34. Why is it advised to keep null point between 45 cm and 55 cm?
Answer.It is done to minimise the effect of neglecting of end resistances in calculation and Wheatstone bridge is most sensitive when all four arms have same order of resistances.

Question.35. What are end resistances?
Answer. The resistances of thick copper strips which keep the two ends of the wire pressed, are called end resistances.

Question.36. What is an ideal value of null point and why?
Answer. Null point at 50 cm is an ideal null point. It makes, P/Q = 1. This ratio is not affected by neglecting end resistance of equal values at the two ends.

Question.37. How can a null point be obtained near 50 cm?
Answer. It can be done by keeping value of R very near the value of X.

Question.38. Why copper strips, used to pressed the ends of wire, are thick?
Answer. Thick Cu strips have negligible resistance over the resistance of alloy metre bridge wire and minimise effect of end resistances.

Question. 39. Why should the bridge wire have uniform thickness and material density throughout?
Answer. Because only then, the resistance per unit length (σ) will be same throughout. Then P = σ l and Q =σ (100 –l) will be correct.

Question. 40. Why the bridge method for resistance measurement is better than Ohm’s Law?
Answer. It is so because the bridge method is a null method (at null point, there is no current flowing in galvanometer) and more sensitive.

Question. 41. Under what conditions, the metre bridge is most sensitive (and hence result most accurate)?
Answer. The bridge is most sensitive when all the four resistances are of equal value. It brings null point automatically at 50 cm.

Question. 42. Why the metre bridge is suitable for measuring moderate resistances?
Answer. Because, Wheatstone bridge is suitable for moderate values of resistances. Therefore, meter bridge is more sensitive for moderate values.

Question. 43. When the sensitivity of the bridge becomes less?
Answer. Bridge has poor sensitivity when resistances P, Q, R and S (or X) are of different order.

Question. 44. Why should current be passed for a short time?
Answer. Continuous current will cause unnecessary heating and affecting values of resistances used.

Question.45. Why is Wheatstone bridge (or metre bridge) method considered unsuitable for the measurement of very low resistance?
Answer. For measuring low resistance, all resistances and resistance of galvanometer should be f low. The end resistance and connecting wires become comparable to the resistance being
measured and introduce error in the result.

Question. 46. Why is Wheatstone bridge (or metre bridge) method considered unsuitable for
the measurement of very high resistance?
Answer. The resistance forming the bridge should be high and the current in the galvanometer
reduces and it become insensitive.

Question. 47. What are the advantages of a Wheatstone bridge method of measuring
resistance over other methods?
Answer.

1. It is a null method, hence the result does not get affected from extra resistances,
2. It is easier to detect a small change in deflection than to read a deflection directly.

Question. 48. What do you mean by sensitiveness of a Wheatstone bridge?
Answer. A Wheatstone bridge is said to be sensitive if it produces more deflection in the galva- nometer for a small change of resistance in resistance arm.

Physics Lab ManualNCERT Solutions Class 12 Physics Sample Papers

To Verify the Laws of Combination (Series) of Resistances Using a Metre Bridge.

Aim
To verify the laws of combination (series) of resistances using a metre bridge.

Apparatus
A metre bridge, a Leclanche cell (battery eliminator), a galvanometer, a resistance box,
a jockey, two resistance wires or two resistance coils known resistances, a set square, sand paper and connecting wires.

Theory

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

Circuit diagram

Procedure
1. Mark the two resistance coils as r₁ and r₂.
2. To find r₁ and r₂ proceed same way as in Experiment 1. (If r₁ and r₂ are not known.)
3. Connect the two coils r₁ and r₂ in series as shown in figure in the right gap of metre bridge and find the resistance of this combination. Take at least three sets of observations.
4. Record your observations as follows.

Observations
Table for length (1) and unknown resistance (X)

Calculations
1. Calculation for r₁ only, r₂ r₂only, r₁ and r₂ in series.
Same as in Experiment 1.
2. Calculation for verification of laws Experimental value of R_s = ……
Theoretical value of R_s = r₁ + r₂ = ……
Difference (if any) = ……

Result
Within limits of experimental error, experimental and theoretical values of Rs are same. Hence, law of resistances in series is verified.

Precautions

1. The connections should be neat, clean and tight.
2. Thick copper wires should be used for the connections after removing the insulations near their ends by rubbing with sand paper.
3. Voltmeter and ammeter should be of proper range.
4. A low resistance rheostat should be used.
5. The key should be inserted only while taking observations to avoid heating of resistance (otherwise its resistance will increase).

Physics Lab ManualNCERT Solutions Class 12 Physics Sample Papers

To Find Resistance of a Given Wire Using Metre Bridge And Hence Determine The Resistivity (Specific Resistance) of its Material.

Physics Lab ManualNCERT Solutions Class 12 Physics Sample Papers

Aim
To find resistance of a given wire using metre bridge and hence determine the resistivity (specific resistance) of its material.

Apparatus
A metre bridge (slide wire bridge), a Leclanche cell (Battery eliminator), a galvanometer, a resistance box, a jockey, a one way key, a resistance wire, a screw gauge, a metre scale, a set square, connecting wires and a piece of sand paper.

Theory
(i) The unknown resistance X is given by

where, R is known resistance placed in the left gap and unknown resistance X in the right gap of metre bridge. I cm is the length of metre bridge wire from zero end upto balance point.
(ii) Specific resistance (p) of the material of the given wire is given by

where, L is the length and D is the diameter of the given wire.

Circuit diagram

Procedure
For Resistance

1. Arrange the apparatus as shown in arrangement diagram.
2. Connect the resistance wire whose resistance is to be determined in the right gap between C and B. Take care that no part of the wire forms a loop:
3. Connect resistance box of low range in the left hand gap between A and B.
4.  Make all the other connections as shown in the circuit diagram.
5. Take out some resistance (say 2 ohm) from the resistance box, plug the key K.
6. Touch the jockey gently first at left end and then at right end of the bridge wire.
7. Note the deflections in the galvanometer. If the galvanometer shows deflections in opposite directions, the connections are correct. If the deflection is one side only, then there is some fault in the circuit. Check or take help of your teacher and rectify the fault.
8. Move (slide) the jockey gently along the wire from left to right till galvanometer gives zero deflection. The point where the jockey is touching the wire is null point D.
9. Choose an appropriate value of 12 from the resistance box such that there is no deflection in the galvanometer when the jockey is nearly in the middle of the wire (i.e.,between 45 cm to 55 cm).
10. Note position of point D (with the help of a set square) to know length AD = l.
11. Take at least four sets of observations in the same way by changing the value of 12 in steps.
12. Record your observations.
    
    For Specific Resistance
13. Cut the resistance wire at the points where it leaves the terminals, stretch it and find its length by using a metre scale.
14. Measure the diameter of the wire at least at four places, in two mutually perpendicular directions at each place with the help of screw gauge.
15. Record your observations as given in tables.

Observations
1.Length of given wire L = cm.
1. Length of given wire L = cm.
2. Table for unknown resistance (X)

3. Least count of the screw gauge

4. Table for diameter (D) of the wire

Calculations

Result

1. The value of unknown resistance X =………
2. The specific resistance of the material of the given wire =………
3. Percentage error =……….

Precautions

1. The connections should be neat, clean and tight.
2. All the plugs in the resistance box should be tight.
3. Move the jockey gently over the bridge wire and do not rub it.
4. The plug in key K should be inserted only when the observations are to be taken.
5. Null point should be brought between 45 cm and 55 cm.
6. Set square should be used to note null point to avoid error of parallax.
7. At one place, diameter of wire should be measured in two mutually perpendicular directions.
8. The wire should not make a loop.

Sources of error

1. The instrument screws may be loose.
2. The plugs may not be clean.
3. The wire may not have uniform thickness.
4. The screw gauge may have faults like back lash error and wrong pitch.