When you are asked measure the "Available Voltage/Supply Voltage" it means you take the common lead(black or "-") of the voltmeter attached to a good earth (ground or "-"). The positive lead (red or "+") of the voltmeter is attached to the point you want to measure. As shown in the circuit diagram below.
Measuring amps (current) flowing through the circuit:
To use a multimeter to measure the amperage flowing through a circuit you will need to move the positive lead (the red lead) into a different position on your meter. Take the positive out of the meter and refit in the DC amp position. There are some times 2 amp position 1 for milliamps and the other normally 10 or 20 amps
Now set your meter to the DC amp position and wire the meter in Series as shown in the diagram below. The (red) positive (+) lead goes to where the conventional current comung from and the (black) the negative (-) lead goes to the path to the negative also called earth or common.
Measure the Available Voltage in the circuit at these points:
Using a voltmeter measure the (supply) available voltage at he different points and list below:
The positive 12V supply (B+) 12.72V
Terminal before the switch 12.69V
Terminal after the switch 12.62V
Terminal before the light bulb 12.61V
Terminal after the light bulb 0V
The negative on the 12V supply (N-) 0V
Explain what happens to the voltage. If the voltage at the battery is 'available to do work', what happen to it as it progresses through different components of the circuit?
The voltage drop when the voltage progresses through different components of the circuit. The voltage drop was different, it's depend the components.
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Measure the Voltage Drop across the circuit parts:
Using a voltmeter measure the voltage drop across the different components of the circuit and list below:
From the B+ of the 12V supply to the input of the switch 0.03Vd
From the input of the switch to the output of the switch 0.03Vd
From the output of the switch to the input of the bulb 0.00Vd
From the input to the bulb to the output of the bulb 12.62Vd
From the output of the bulb to the N- of the 12V supply 0.00Vd
Which component tested has the largest voltage drop. Explain why the voltage drop is so large.
From the input to the bulb to the output of the bulb tested has the largest voltage drop, because it's got more consumer than another.
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Measure the amps (current) flowing through the circuit:
Remember the amp meter must be connected inseries. Record how many amps you find flowing through these points:
Wire before light bulb 0.34A
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Caculate the resistance of the light bulb:
You have measured the voltage and the amps. In the box below, calculate the resistance of the light bulb using Ohm Law.
Ohm Law：V=I*R;I=V/R;R=V/I
R=V/I I=0.34A V=12.70V
R=12.70/0.34=37.35Ω
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Calculate Watts used at the light bulb:
You have measure the voltage and the amps. In the box below, calculate the watts used at the light bulb using the Power Triangle.
Power Law: W=V*I;I=W/V;V=W/I
W=I*V I=0.34A V=12.62V
W=0.34*12.62=4.29w
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Create a circuit with a larger light bulb:
Using the battery or power supply, light bulb, switch, wires, and create a complete circuit for electricity to flow then turn the switch "on".
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Measure the Voltage Drop across the circuit parts:
Using a voltmeter measure the Voltage Drop across the different components of the circuit and list below:
Wire before the switch 0.1Vd
Switch 0.1Vd
Wire before light bulb 0.05Vd
Light bulb 12.38Vd
Wire after light bulb 0.05Vd
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Measure the amps (current) flowing through the circuit:
Measure the current flow through any part of the citcuit; remember the amp meter must be connected in series. Rcord your findings below:
Current flow through circuit 1.8amps
Make up a statement about what you can see happening to the amperage in this circuit compared with that with the small bulb?
The big bulb used more current than the small bulb.
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Calculate the resistance of the light bulb:
Calculate the resistance of the larger light bulb using Ohm Law.
R=V/I V=12.38V I=1.8A
R=12.38/1.8=6.87Ω
What does the resistance tell you about how current flow is affected through thecircuit when you compared with this result against the same calculation for the small bulb?
The resistance tell me the larger light bulb used less resistance and more current than the small bulb.
Calculate Watts used at the light bulb:
You have measured the voltage and the amps. In the box below, calculate the watts used at the light bulb using the Power Triangle.
W=I*V V=12.38V I =1.8A
W=12.38*1.8=22.28W
What does the watt tell you about the brightness of the light bulb compared with the result of the same calculation with the small bulb?
The big bulb was more brightness than the small bulb with the result of the same calculation with the small bulb, and the big used mor watt than the small one.
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Series Circuit:
Create a circuit with two light bulbs in series:
Use the same bulbs and make a circuit like before, but this time have the two light bulbs, one after another. There should be only one path for electricity to flow through, and it should flow first through one light bulb, then the other. Use the diagram below to help you design the circuit.
Measure Voltage Drop across the components:
Using a voltmeter measure the voltage drop across the different parts and list below:
Wire before the switch 0.01V
Switch 0.03V
Wire before light bulb 1 0.02V
Light bulb 1 6.13V
Wire between light bulb 1&2 0V
Light bulb 2 6.30V
Wire after light bulb 2 0V
Measure the voltage available at the battery 12.7V
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Explain what happens to the voltage in the different compoments in this circuit?
The different components gets different voltage. The voltage drop from wire before the switch was 0.01V to switch was 0.03V, and a hurge voltage drop after bulb1 and bulb2, because they shared the same supply voltage.
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Measure amps in the circuit:
Use an amp meter to measure the amps at different parts of the circuit.
Record how many amps you find at these points:
Wire before the switch 0.22A
Wire before light bulb 1 0.22A
Wire between light bulb 1&2 0.22A
Wire after light bulb 2 0.22A
How is the amperage different from the amperage the individual circuit?
What is happening to the amps in this circuit if you compare it with the amps of both the individual circuit?What is going on and why?
In the individual circuit, it has one consumer and used supply voltage
In this circuit, the amps lesser than individual circuit, because in this circuit has two consumer to shared the same supply voltage.
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Calculate the total resistance of the circuit:
You have measured the voltage and the amps. In the box below, calculate the resistance of the circuit.(Using Ohm Law)
R=V/I V=12.7V;I=0.22A
R=12.7/0.22=57.73Ω
Calculate Watts used by each light bulb:
Since you have measured volts drops and amps, you can calculate watts. In the box below, calculate the watts used at each light bulb. (Using the Power Law)
W=I*V I=0.22A;V=6.13V
I=0.22A;V=6.3V
W=0.22*6.13=1.35W
W=0.22*6.3=1.39W
What do the watts caculated tell you about the brightness of the light bulbs in this series circuit and why has it happend?
In this series circuit, both of the bulbs used same wattage and the brightness is more darker than the small one,because both of the bulbs shared the same supply voltage.
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Create a circuit with three light bulbs in series:
Use the same bulbs and make a circuit like before, but add a third bulb, all one after another. There should be only one path for electricity to flow in, and it should flow first through one light bulb, then the next, then the last.
Measure Voltage Drop across the parts:
Using the voltmeter measure the voltage drop across the different components and list below:
Wire before the switch 0V
Switch 0.82V
Wire before light bulb 1 0V
Light bulb 1 3.79V
Wire between light bulb 1&2 0V
Light bulb 2 3.81V
Wire between light bulb 2&3 0V
Light bulb 3 4.34V
Wire after light bulb 3 0V
Measure the voltage available at the supply 12.71V
Explain what happens to the voltage:
What happened to the voltage drop caross the different components in this circuit when you compare it with the two bulb series circuit? Take special note of what happens to the voltage drop across the bulb, what has happened and why?
There are three main voltage drop in light bulb 1, bulb 2, bulb 3 at 3.79V, 3.81V and 4.31V respectively. The values of the voltage drop in the three bulbs are smaller than the two bulbs, because one bulb was added in three series circuit. In addition, the switch has a noticeable voltage drop which produces the resistance due to heating.
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Measure amps in the circuit:
Use an amp meter to measure the amps at different parts of the circuit. Record how many amps you find at these points: (remember amps have to be measured in series.)
Wire before the switch 0.18A
Wire before light bulb 1 0.18A
Wire between light bulb 1 &2 0.18A
Wire between light bulb 2 &3 0.18A
Wire after light bulb 3 0.18A
Explain what happens to the amperage?
What is happening to the amps in the circuit? How does this amp measurement differ from the two bulb series circuit? What is going on?
All the amps in this circuit are same. In this circuit use three bulbs and the amps lower than the two bulbs series circuit. The smallest bulb still going on but the brightness was more dark than two bulbs series circuit.
Calculate the total resistance of the circuit: You have measured the voltage and the amps. In the box below, calculate the resistance of the circuit. (Using Ohms Law)
R=V/I V=12.71V;I=0.18A
R=12.71/0.18=70.61Ω
Calculate Watts used at each light bulb:
In the box below, calculate the watts used at each of the light bulbs. (Using the Power Law)
W=V*I V1=3.79V;V2=3.81V;V3=4.34V;I=0.18A
W1=3.79*0.18=0.682W
W2=3.81*0.18=0.686W
W3=4.34*0.18=0.781W
What does this tell you about the brightness of the light bulbs compared with the two bulb series circuit?
In this series circuit, all of the bulbs use similar wattage, but it was lower than the two bulbs series circuit, and the brightness was more darker than the two bulbs series circuit, because all of the bulbs shared the same supply voltage likes the two bulbs series circuit.
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Use the "available voltage" method to measure voltage in the circuit:
Starting at the positive side of the battery, measure available voltage at the different parts of the circuit. Record the readings below:
Measure the voltage available at the:
Battery positive 12.71V
Input to the switch 12.71V
Output of the switch 11.80V (something wrong with switch)
Supply to light bulb 1 11.79V
Output of light bulb 1 7.94V
Input to light bulb 2 7.84V
Output of light bulb 2 4.12V
Input to light bulb 3 4.12V
Output of light 3 0V
At the negative of the supply 0V
Explain what the different reading are telling you when using Voltage Drop compared with Available Voltage:
The voltage drop show me how much voltage was used of each componet, for example: switch used 0.82V, light bulb 1 used 3.79V, light bulb 2 used 3.81V, light bulb 3 used 4.34V. The available voltage show me how much voltage can be used after each componer, for example: after battery positive 12.71V can be used, after Input to the switch 12.70V can be used, after output of the switch 11.80V can be used, etc.
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Parallel Circuits:
Measure the available voltage at each light bulb in the parallel circuit and record below:
Light bulb 1 12.64V
Light bulb 2 12.64V
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Measure voltage drop across each light bulb in the parallel circuit and record below:
Light bulb 1 12.57Vd
Light bulb 2 12.63Vd
Describe what happens to the volts in the parallel circuit (What rule of electricity applies here and how is it different from a series circuit.)
The available voltage at each light bulb were same. The voltage drop across each light bulb were similar and bigger than series circuit. In this circuit each consumer therefore has own input and output and used same supply voltage, but in series circuit each consumer shared the same supply voltage.
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Measure the current flow （amps) through the parallel circuit with the two light bulbs using an amp meter and record:
Current flow through light bulb 1 circuit 0.34 Amps
Current flow through light bulb 2 circuit 0.33 Amps
Total current flow through both circuit 0.67 Amps
Describe what happens to the amps in the parallel circuit. (What rule of electricity applies here and how is it different from a series circuit.)
The current flow through light bulb 1 and 2 at 0.34A & 0.33A each and the current was divided by each bulbs in this parallel circuit, but in the series circuit, the amps were all same values any steps.
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Calculate the total resistance of each bulb in the circuit:
R=V/I V1=12.57V;V2=12.63V;I1=0.34A;I2=0.33A
R1=12.57/0.34=36.97Ω
R2=12.63/0.33=38.27Ω
Calculate the total resistance of the circuit using the following formula:
Show the full working in the box below 1/Rt=1/R1+1/R2
1/Rt=1/R1+1/R2 R1=36.97Ω；R2=38.27Ω
1/Rt=1/36.97+1/38.27
Rt=18.8Ω
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Calculate total watts used in the parallel circuit while both lights are on:
W=V*I V=12.70V;I=0.67A
W=12.7*0.67=8.51W
Calculate watts for each indibidual bulb
W=V*I V1=12.57V;V2=12.63V;I1=0.34A;I2=0.33A
W1=12.57*0.34=4.27W
W2=12.63*0.33=4.17W
How does these watts compare with watts used when the bulbs were in series? Why has this happened?
In the parallel circuit, these watts higher than the watts used when the bulbs were in series because in the series circuit the voltage drop value were lower than parallel circuit value and the current value in the series circuit also lower than parallel circuit's value.
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Create a parallel circuit that now has 3 lights bulbs:
Now measure the current (amps) flow through each of the 3 lights bulbs circuit, using an amp meter and record:
Current flow through light bulb 1 circuit 0.34Amps
Current flow through light bulb 2 circuit 0.34Amps
Current flow through light bulb 3 circuir 0.32Amps
Total current flow through all the circuit 1 Amps
What happened to the total amps flowing through the circuit when the third light bulb was added? (Did the current flow through each of the other light bulbs change? If so why?
The current flow didn't through each of the othe light bulbs change, when the third light bulb was added, because the third light bulb used same supply voltage, it doesn't need to share the supply voltage from other light bulbs.
What rule or concept does this show about amps in a parallel circuit?
In a parallel circuit each consumer has its own input and output so when we added more consumer, the amp for each consumer never be changed.Current draw for this circuit is high!
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Measure the available volts in the circuit now that there are 3 light bulbs in the circuit and record:
Light bulb 1 12.63V
Light bulb 2 12.63V
Light bulb 3 12.63V
Measure the voltage drop across each of the lights now that there are three light bulbs in the circuit
Light bulb 1 12.61Vd
Light bulb 2 12.61Vd
Light bulb 3 12.60Vd
What hanppened to the available voltage in the circuit when the third light bulb was added? Why did this happen?
The available voltage in the circuit didn't has any change when the third bulb was added, because the third light bulb has its own input and output and without share the supply voltage.
What happened to the voltage drops in the circuit when the third light bulb was added? Why did this happen?
The voltage drops in the circuit was similar when the third light bulb was added. In parallel circuit each consumer has its own input and output, depending on the number of electrical consumer.
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Calculate the resistance of each bulb:
R=V/I V1=12.61V;V2=12.61V;V3=12.60V;I1=0.34A;I2=0.34A;I3=0.32A
R1=12.61/0.34=37.09Ω
R2=12.61/0.34=37.09Ω
R3=12.60/0.32=39.38Ω
Calculate the total resistance of the circuit using the following formula:
1/Rt=1/R1+1/R2+1/R3 R1=37.09Ω;R2=37.09Ω;R3=39.38Ω
1/Rt=1/37.09+1/37.09+1/39.38
Rt=12.61Ω
Calculate total watts used in the parallel circuit with three lights working or "on"
W=V*I v=12.68V;I=1A
W=12.68*1=12.68W
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