determine the resistivity of 2/3 wire by plotting graph for potential difference vs current . full experiment
Answers
"GIVEN":
To determine the resistivity of 2/3 wire by plotting graph for potential difference vs current
Apparatus Required:
- A wire of unknown resistance
- battery
- voltmeter
- milliammeter
- rheostat
- plug key
- connecting wires
- piece of sandpaper
Theory:
- According to ohm's law, the electric current flowing through a conductor is directly proportional to the potential difference across its end.
- The physical state(pressure, temperature ,and dimensions) of the conductor remains unchanged.
- If I is the current flowing through the conductor and V is the potential difference across its end, then.
- I∝V
- V∝I
₍or₎ - ohms law
- V=RI
- V/I = constant(R)
- R∝L/A⇒ R=ρ L/A
- ρ= RA/L
- When R is the constant of electrical resistance of the conductor. Resistance R depends of the dimensions and material of the conductor.
- The resistance of a material and its length and area of the cross-section is given by the formula
- R=ρ I/A
- Where ρ is the specific resistance or resistivity .
Procedure:
- Take the sand paper clean the ends of connecting wire.
- Connect the resistance , rheostat , battery, voltmeter , and ammeter.
- The voltmeter and milliammeter coincide with the zero mark on measuring scale in pointer . If not, adjust the pointer to zero mark base using a screwdriver.
- Note the range and the least count of the voltmeter and milliammeter. Insert the key(k) slide the rheostat end the current flow is minimum.
- Note the voltmeter and milliammeter readings. Remove the key k and wire cool. Insert the key and increase the voltage by moving rheostat . Note the milliammeter and voltmeter reading.
Observation:
- Range of Voltmeter=0-2v
- Range of Ammeter=0-2A
- Least count of Voltmeter=0.05V
- Least count of Ammeter=0.05A
- Length of each wire=100 cm
- Zero error=Nill
OBSERVATION TABLE:
Wire 1:
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s.no Ammeter reading (I)A Voltmeter reading (v) v R=V/I ohm
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1 0.1 0.25 2.5
2 0.2 0.50 2.5
3 0.3 1.00 2.5
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Mean=2.5ohm
Wire 2:
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s.no Ammeter reading(I)A Voltmeter reading(v) v R=V/I ohm
-------------------------------------------------------------------------------------------------------------
1 0.05 0.25 5
2 0.10 0.50 5
3 0.15 0.75 5
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Mean=5 ohm
Observation table for diameter by using screw guage :
- Pitch=1mm
- No .of divisions on circular scale=100
- Least count =1/100 =0.01mm=0,001 cm
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s.no Main scale reading No .of circular n*L.C Diameter
(N mm) divisions coincide(n) N + n(L.C)
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Wire 1 0 50 50*0.001 0+0.05
cm D1=0.05cm
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Wire 2 0 44 44*0,001 0+0.044
cm D2=0.044cm
--------------------------------------------------------------------------------------------------------------
Calculation:
- From graph .resistance
wire 1=2.5 ohm
wire 2=5 ohm
Length of each wire=100cm
- Diameter D1=0.05 cm, D2=0.044cm
- Resistivity:- ρ1=R₁πD₁²/4L
= 2.5*3.14*0.05*0.05/4*100
= 4.9*10⁻⁵ohm cm
- Resistivity :-ρ2=R₂πD₂²/4L
=5*3.14*0.044*0.044/4*100
= 7.5*10⁻⁵ohm cm
Answer:
uppar hai
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