Lab Report: Resistors In Parallel
In this lab , the theoretical relationship about resistors in parallel has been discussed
. Students can copy paste this report in their own school/college lab report or they can tweak it a little bit so as to make sure that they are not penalized for plagiarism. Data analysis has been intentionally left for the reader to fill. But an example has been given o give them an idea.
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Title: Parallel Circuit
Introduction: The purpose of this experiment is to prove the theoretical relationship about resistors in Parallel viz 1/R=1/R1+1/R2.
This result is valid for resistors which are ohmic in nature. In normal circumstances all the resistors are Ohmic in nature provided their temperature does not increase too much from normal values.
Equipments used:
1. Two digital multi-meters, DMM
2. Two 100 ohms resistors
3. Basic electricity laboratory
4. D cell battery
5. Patch cords with banana plugs
6. Decade box
Theory:
When two resistors are connected in Parallel, then the volatge in the two resistors is the same because the same battery is used to supply the potential difference to both.
Also using Ohms law, we know that, V=IR,
So voltage drop across first resistor is V=I1*R1 So I1=V/R1
And voltage drop across resistor 2 is V=I2*R2 So I2=V/R2
Since the sum of these two voltages must be equal to the total battery voltage, namely V, so
I= I1+I2
V/R=V/R1+V/R2
So 1/R=1/R1+1/R2
So we see that if two resistors are connected in Parallel,
Then the Reciprocal of Net resistance is obtained by simply adding the reciprocal of resistances of the two resistors.
Data:
First of all the individual resistors were measured by DMM and noted down.
R1=20.4 Ohms
R2=42.8 ohms
Next the voltage of the D cell was measured using the DMM set into V mode.
V=2.9 Volts
Then the two resistors were put in Parallel and the current I in the circuit was measured. For measuring current, the DMM was set to the current measuring mode.
I (P1)= 220 mA
Then the two resistors were replaced by a single resistor of value 1/R=1/R1+1/R2.
1/R=1/20.4+1/42.8 So R=13.82 Ohms
The current was measured and noted down.
I (S2)=222 mA
Calculations:
Assuming a one percent error in the value of resistances measured, and also in the value of current and voltages measured, we can write the data as below.
R1 ohmsR2 ohmsV voltsI (P1) milli ampsI (P2) milli amps
20.4 +- 0.242.8 +- 0.42.9 +- 0.03220222
1/R=1/R1+1/R2
So R=13.82 +- 0.6
Conclusion:
As expected the current was found to be approximately the same in both the cases. This shows that two resistors of resistance R1 and R2 in parallel can be replaced by a Net resistance of 1/R=1/R1+1/R2. This proves our theory about resistors in parallel.
Some of the sources of error are
1.Precision of the instrument or the least count error
2.Change in resistance due to heating
To get a better data, we can do the experiment at a good speed so that it wont get too much time for the equipments to get heated.
by: myphysicsbuddy
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