1) Aim
2) Apparatus
3) Diagram
4) Formula with explanation/law
5) Theory
6) Observations
7) Observation Table
8) Calculation
9) Result/Conclusion
10) Precautions
11) Questions
12) Graph
आता आपण Experiment चे Solutions पाहण्यास सुरुवात करुया :-
Class 12 Physics Practical No 8 : Law of Resistance using meter bridge Solutions Maharashtra Board
[Page No 66]
Aim :
To verify the laws of combination (series and parallel) of resistance using a metre bridge.
Apparatus :
Meter bridge, two resistance, galvanometer, resistance box, rheostat, cell, jockey, plug key, connecting wires.
Circuit Diagram :
Formula :
1) X = R*(Lx/Lr) where, Lx and Lr are balancing lengths corresponding to X and R.
2) For series combination, Xs = X1 + X2
3) For parallel combination, Xp = X1 * X2 / X1 + X2
[Page No 67]
Observations :
Resistance (X1) = 10 Ω
Resistance (X2) = 20 Ω
1) For series combination :
Resistance Position |
Obs. No. |
Resistance from Resistance Box (Ω) R |
Lx(cm) | Lr(cm) | Xs = R(Ls/Lr)Ω |
---|---|---|---|---|---|
Series in left gap |
1 | 10 | 75 | 25 | 30 |
2 | 20 | 60 | 40 | 30 | |
3 | 30 | 50 | 50 | 30 | |
Series in right gap |
1 | 10 | 75 | 25 | 30 |
2 | 20 | 60 | 40 | 30 | |
3 | 30 | 50 | 50 | 30 |
Mean Xs = 30 Ω
2) For parallel combination :
Resistance Position |
Obs. No. |
Resistance from Resistance Box (Ω) R |
Lx(cm) | Lr(cm) | Xs = R(Ls/Lr)Ω |
---|---|---|---|---|---|
Parallel in left gap |
1 | 10 | 20 | 80 | 2.5 |
2 | 20 | 25 | 75 | 6.665 | |
3 | 30 | 30 | 70 | 12.85 | |
Parallel in right gap |
1 | 10 | 20 | 80 | 2.5 |
2 | 20 | 25 | 75 | 6.665 | |
3 | 30 | 30 | 70 | 12.85 |
Mean Xs = 2.338 Ω
3) For series combination :
Left gap | Right gap | |||||
---|---|---|---|---|---|---|
Obs. No | 1 | 2 | 3 | 1 | 2 | 3 |
Log R = a | 1.0000 | 1.3010 | 1.4771 | 1.0000 | 1.3010 | 1.4771 |
Log lx = b | 1.8751 | 1.7782 | 1.6990 | 1.8751 | 1.7782 | 1.6990 |
c = a + b | 2.8751 | 3.0732 | 3.1761 | 2.8751 | 3.0792 | 3.1761 |
Log Lr = d | 1.3980 | 1.6021 | 1.6990 | 1.3980 | 1.6021 | 1.6990 |
c - d = e | 1.4771 | 1.4771 | 1.4771 | 1.4771 | 1.4771 | 1.4771 |
Antilog e = Xs | 30.00 | 30.00 | 30.00 | 30.00 | 30.00 | 30.00 |
4) For parallel combination :
Left gap | Right gap | |||||
---|---|---|---|---|---|---|
Obs. No | 1 | 2 | 3 | 1 | 2 | 3 |
Log R = a | 1.0000 | 1.3010 | 1.4771 | 1.0000 | 1.3010 | 1.4771 |
Log Lx = b | 2.3010 | 1.3979 | 1.4771 | 2.3010 | 1.3979 | 1.4771 |
c = a + b | 2.3010 | 2.6989 | 2.9542 | 2.3010 | 2.6989 | 2.9542 |
Log Lr = d | 1.3031 | 1.8751 | 1.8451 | 1.3031 | 1.8751 | 1.8451 |
c - d = e | 0.3979 | 0.8751 | 1.1091 | 0.3979 | 0.8751 | 1.1091 |
Antilog e = Xs | 2.500 | 6.665 | 12.85 | 2.500 | 6.665 | 12.85 |
[Page No 68]
Calculations :
1) Xs = R (Ls/Lr)
Mean Xs = 30 Ω
(experiment value)
2) Xp = R(Ls/Lr)
Mean Xp = 2.338 Ω
(experiment value)
Result :
1) In series
a) Experimental value of Xs = 30 Ω
b) Theoretical value of Xs = X1 + X2 = 30 Ω
2) In Parallel
a) Experimental value of Xp = 2.338 Ω
b) Theoretical value of Xp = X1 * X2/X1 + X2 = 6.667 Ω
Hence law's of series and parallel is verified.
Precautions :
i) All the connections and plugs in the resistance box should be tight.
ii) Null point is obtained in the middle (1/3)rd portion of the meter bridge wire.
iii) Do not slide the jockey on the wire.
iv) Remove the plug key after completion of experiment.
Questions :
1) What is a balance point in meter bridge?
Ans : In a meter bridge, the balance point is the position on the wire where there is no potential difference (zero current) between the two ends, indicating that the resistance of the unknown resistor is equal to the resistance of the known resistor. It allows us to find the unknown resistance using the principle of Wheatstone bridge.
2) Resistance of a wire is x Ω, The wire is stretched to triple it's length, then the resistance becomes?
Ans : When the wire is stretched to triple its length, its resistance becomes 9 times (3^2) the original resistance (x Ω) since resistance is directly proportional to the length of the wire for a given material and cross-sectional area.
अशा प्रकारे मित्रांनो आपण इयत्ता बारावी Physics Practical Number 8 - Law of Resistance using meter bridge याचे Solutions पहिले.
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