Resistance of a Wire
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Determining Voltage, Resistance and Current in a Parallel, Series and Series-Parallel Circuit.
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Introduction
Practical 4
Resistance and Ohm’s Law
Aim
To determine the voltage, resistance and current in a parallel, series and series-parallel circuit.
Theory
In everyday life, electricity is being used for the purpose of things like powering electronics. Electricity itself is a flow of electric charge (Millikan 2000). The movement of the electrically charge particles such as electron are best defined as the electric current. For the current to flow, it requires a pathway, in which a circuit is required as it provides that particular pathway (Millikan 2000).
There are different types of electric circuit as they can all be connected in many different ways. The two most common and simplest one are series and parallel (Young 2011) circuit. In a series circuit the components connected to that circuit are connected along a single pathway, sharing voltage. However for parallel circuit, the components connected individually receives the exact same voltage.
In the experiment, the components used to conduct the electric currents are light bulb, where the source of the voltage is received from AA batteries. In order to determine the current of the light bulb, Ohm’s Law must be
Middle
- Measure the voltage and resistance of each bulb.
- Measure the current in the circuit
Results
Series Resistance
Measurement Using Multimeter
Bulb 1 Resistance = 3.90 Ω Bulb 2 Resistance = 3.80 Ω ∴Total Resistance = 3.90 + 3.80 = 7.70 Ω | Power Supply Voltage = 6.12V Voltage of Bulb 1 = 3.00V Voltage of Bulb 2 = 2.95V ∴Total Voltage of the Bulbs = 3.00 + 2.95 = 5.95V |
Calculating the Current of Bulb 1 and bulb 2
Using Ohm’s Law, given V = I x R,
I = 
Bulb 1 ∴ I = = 0.769amps Measured current in the series circuit Current in Circuit = 0.159amps Parallel Resistance Measurement Using Multimeter | Bulb 2 ∴ I = = 0.776amps |
Bulb 1 Resistance = 2.70 Ω Bulb 2 Resistance = 2.60 Ω ∴Total Resistance = 2.70 + 2.60 = 5.30Ω | Power Supply Voltage = 5.73V Voltage of Bulb 1 = 5.62V Voltage of Bulb 2 = 5.59V |
Equivalent Resistance of Parallel Circuit using two Bulbs:
Req = 
Req =
Req = 1.32 Ω
Series –Parallel Resistance
Measurement Using Multimeter
Bulb 1 Voltage = 2.80V Bulb 2 Voltage = 2.83V ∴Total Voltage of Bulb 1 & 2 = 2.80 + 2.83 = 5.63V | Power Supply Voltage = 5.78V Bulb 3 Voltage = 5.16V |
Bulb 1 Resistance = 3.90 Ω Bulb 2 Resistance = 3.80 Ω Bulb 3 Resistance = 2.65 Ω |
Conclusion
This practical had many errors involved such as the fluctuation of the multimeter. A possible improvement to the practical is to reduce the fluctuation on the reading of the multimeter. Since the charge of the batteries drops over time, this causes the multimeter to fluctuate. It is more preferably to use a battery back such as a transformer, as this would lessens the chance of charges dropping, hence less fluctuation of the multimeter reading.
Conclusion
From following the method, it was found that when a series-circuit is used the bulbs share the voltage from the power supply, however when a parallel circuit was used, the bulb receives nearly full voltage from the power supply. It was also observed that the bulb shines brighter in the parallel circuits, which also gives indication that more voltage is flowing to the bulb. Nevertheless there were errors involved during the practical, which caused some calculated, and measurement values to have a less level of accuracy. An improvement was suggested which would increase the possibility of observing a more accurate data.
Reference
Young, HD 2011, College Physics, 9th edn, Addison-Wesley, Boston, US
Millikan, RA 2000, Elements of Electricity, American Technical Publisher, Orland Park, US
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