Finding the resistance of a given wire using meter bridge and hence determine resistivity of its material

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

metre bridge

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Internal Resistance Of a Given Primary Cell Using a Potentiometer

OBJECTIVE

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

APPARATUS

A potentiometer, a battery (or battery eliminator), two one-way keys, an id resistance, a galvanometer, a high resistance box, a fractional resistance box, an ammeter,  voltmeter, a cell (say Leclanche cell), a jockey, a set square, connecting wires and sandpaper.

CIRCUIT DIAGRAM

CIRCUIT DIAGRAM

THEORY

When Key K2 is open and K1 is closed,
Let null point be obtained at a distance l1 from A
Therefore E = Kl1 –> (1)
When Key K2 is closed and K1 is open,
Let null point be obtained at a distance l2 from A
Therefore V = Kl2 –> (2)
(1) divided by (2)
E/V = l1/l2
(S+r)/S  = l1/l2
r = S*(l1-l2)/l2

Where S : the shunt resistance in parallel with given cell.
l1 and l2 : Balancing lengths without and with shunt respectively.
r : internal resistance of the cell.

PROCEDURE

  1. Make the connections as shown in circuit diagram.
  2. Clean the ends of the connecting wires with sandpaper and make tight connections. Tighten the plugs of the resistance box.
  3. 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 won’t be obtained. (E’ > E)
  4. Take maximum current from the battery, making rheostat resistance small.
  5. Insert the plug in Key K1 and adjust the rheostat so that a null point is obtained or the fourth wire of potentiometer.
  6. Insert the 2000 Ohm plug in its position in resistance box and obtain a null point by slightly adjusting the jockey.
  7. Measure the balancing length L1.
  8. Take out the 2000 Ohm plug from the resistance box. Introduce the plugs in Key K1, as well as in Key K2. Take act a small resistance from the resistance box R connected in parallel with the cell. Slide the jockey along the potentiometer wire and obtain a null point. Insert the 2000 Ohm plug back in its position in R.B. and make a further adjustment for a sharp null point.
  9. Insert the 2000 Ohm plug back in its position in R.B. and make a further adjustment for a sharp null point.
  10. Measure the balancing length l2 from end P.
  11. Remove the plugs keys K1 and K2. Wait for some time and respect the activity for the same current.
  12. Record your observations.

OBSERVATIONS

  • Range of voltmeter
  •  Least count of voltmeter
  •  E.M.F. of battery (or battery eliminator)
  •  E.M.F. of cell

 

CALCULATIONS

 

RESULT

The internal resistance of the given cell is…………… ohm.

 

PRECAUTIONS

  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 a short time only while finding the null point.

 

 

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Determining resistance per cm of given wire by plotting a graph of potential difference versus current

AIM To determine the resistance per cm of a given wire by plotting a graph of potential difference versus current.

APPARATUS REQUIRED
Battery, key, rheostat, voltmeter, ammeter, resistance wire (unknown resistance), connecting wires, meter scale, sandpaper.

PRINCIPLE
This Experiment is based on OHM’S LAW
Ohm’s Law states that the electric current passing through a conductor is directly proportional to the potential difference applied across its ends.
Mathematically, V=IR
The resistance R of the wire depends on the material of the wire and its dimensions.

CIRCUIT DIAGRAM

internal resistance
Circuit Diagram – Determining Unknown Resistance

PROCEDURE

  1. Draw the circuit diagram as shown in figure above
  2. Arrange the apparatus as per the circuit diagram
  3. Clean the ends of the connecting wires with sandpaper and make them shiny.
  4. Make the connections as per circuit diagram. All connections must be neat and tight. Take care to connect the ammeter and voltmeter with their correct polarity. (+ve to +ve and -ve to -ve)
  5. Determine the zero error and least count of the ammeter and voltmeter and record them.
  6. Adjust the rheostat to pass a low current.
  7. Insert the key K and slide the rheostat contact to see whether the ammeter and voltmeter are showing deflections properly.
  8. Adjust the rheostat to get a small deflection in ammeter and voltmeter.
  9. Record the readings of the ammeter and voltmeter
  10. Take at least six sets of readings by adjusting the rheostat gradually
  11. Plot a graph with V along X axis and I along axis.
  12. The graph will be a straight line which verifies Ohm’s law
  13. Determine the slope of the V-I graph. The reciprocal of the slope gives the resistance of the wire.

OBSERVATIONS

  1. Range
    Range of the given Ammeter =  0-500m A
    Range of the given voltmeter =  0-5V
  2. Least Count
    Least Count of the given Ammeter = 10mA
    Least Count of the given voltmeter = 0.1V
  3. Zero Error
    Zero Error of the given Ammeter = 0A
    Zero Error of the given voltmeter = 0V
  4. Zero Correction
    Zero Correction of the given Ammeter = 0A
    Zero Correction of the given voltmeter = 0V
  5. Observation Table for Ammeter and Voltmeter Readings
S No Ammeter Observed (A) Ammeter Corrected (A) Voltmeter Observed (V) Voltmeter Corrected (V) Ratio (V/I) = R (ohm)
1 0.3 0.3 0.1 0.1 0.33
2 0.7 0.7 0.2 0.2 0.28
3 1.1 1.1 0.3 0.3 0.27
4 1.5 1.5 0.4 0.4 0.26
5 1.9 1.9 0.5 0.5 0.26

Graph

Voltage v/s Current Graph
Voltage v/s Current Graph

CALCULATIONS
Mean Value of V/I from observations, R = 0.28Ω
Length of resistance wire = 40.2cm
Value of slope of VI graph =0.27 Ω
Resistance per unit length = 0.675 Ωm-1

RESULT

  1. Ohm’s Law is verified as the I vs V graph is a straight line
  2. The resistance of the given wire = 0.28Ω
  3. The resistance per cm of given wire is 0.675 Ω m-1.

 PRECAUTIONS

  1. All the electrical connections must be neat and tight.
  2. Voltmeter and Ammeter must of proper range
  3. The key should be inserted only while taking readings.

VIVA VOCE

  • Give the mathematical form of Ohm’s law.
    V = RI,  is the mathematical representation of Ohm’s law.
  • State Ohm’s Law
    Ohm’s law states that the electric current ‘I’ flowing through a conductor is directly proportional to the potential difference V (voltage) across its ends (provided that the physical conditions, temperature, pressure, and dimensions of the conductor remain same).
  • What is the material chosen for rheostat wire and why?
    It is constantan alloy. It is chosen to make rheostat as its resistivity is very high and temperature coefficient of resistance is quite small.
  • What is the material used for making connection wires?
    Here, copper has been used.
  • What are the factors affecting resistance?
    The resistance depends upon length, the area of cross-section, nature of material and temperature of conductors.
  • What is electrical current? Define its S.I. unit?
    The flow of electric charge per unit time through a conductor is called the electric current. It’s S.I. unit is ampere (A). 1 Ampere is the amount of current flowing in a conductor which offers resistance 1 Ohm when 1 V potential difference is maintained across the conductor.
  • Define S.I. Unit of electric potential.
    S.I. unit of electric potential is Volt. 1 Volt is said to be the potential difference between two points if 1 joule of work is done in bringing 1 coulomb of charge from one point to the other.
  • Why is ammeter connected in series?
    To measure the current without any change in magnitude.
  • Why is voltmeter connected in parallel?
    A voltmeter is connected in parallel so that it can measure the potential drop without any change in its magnitude.

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