Series circuit

 TTEC 4841 - Electrical and Electronics- Practical

Series Circuit

04/08/2011 thursday

start = 1.00pm finsh = 4.30pm

The task that had to do be done was to wire up a individual circuit on a circuit board. today i learend about  series circuit  how to create a circuit with two ligh bulbs.

 how to measure the voltage drop across the components using a multimeter.

wire before the switch = 0.01

switch = 0.01

wire before light bulb 1 = 0.02

light bulb 1 = 5.9 consummer

wire between light  bulb 1 & 2 = 0.00

light bulb 2 = 5.9 consummer

wire after light bulb 2 = 0.01

Then i measured the voltage at the battery which = to 12.00 v

Then i explained what happend to the voltage. well the voltage is getting used up as its going around the circuit.

Amps

Then we measured the amps in different parts of the circuit using a multimeter and the amps stays the same throghout the circuit.

wire before the switch = 0.49

wire before light bulb 1 = .049

wire between light bulb 1 & 2 = 0.49

 wire after light bulb 2. = 0.49

Resistance

Then we calculated the total resistance of the circuit which is R=V/I.
R=12v/0.49amps
R=24.48 ohms

Watts

 Then calculated the watts used by each light bulb and i used W=V*I.

bulb1= 5.9v * 0.49amps = 2.591 w

bulb 2= 5.95v*0.49amps = 2.93 w

so that tells us there is less amps going between the two light bulbs beacuse there is more resistance and less voltage going through the circuit.






we also created a circuit with three light bulbs in the series

I used a multimeter to measure the Voltage Drop across each part of the circuit with available voltage of 12 v. so bulb one has more resistance so the is not enough consumer going into the other two light bulbs to light them up. the first bulb always uses more voltage.
wire before the switch = 0.00v

switch 0.01v

wire before light bulb 1 = 0.02v

light bulb 1 = 1.27v

wire between light bulb 1 & 2 = 0.02v

light bulb 2 = 9.27v

wire between light bulb 2 & 3 = 0.00v

light bulb 3 = 1.31v

wire after light bulb 3 = 0.02v


Amps
then i measured the amps in the circuit by using an multimeter to measure the total amps.

wire before the switch = 0.64v

wire before light bulb 1 = 0.64v

wire between light bulb 1 & 2 = 0.64v

wire between light bulb 2 & 3 = 0.64v

wire after light bulb 3 = 0.64

Then i explained what happend to the amperage. well the amps stays the same through out the circuit. and the circuit is using less amps and more resistance.

Total Resistance
I calculated the total resistance which is  R=V/I
R=12v/0.64amps
R=18.75ohms

Watts

watts used at each light bulb which is W= V*I.

bulb 1 = 1.27v* 0.64amps = 0.812 w

bulb 2 = 9.27v* 0.64amps = 5.932 w

bulb 3 = 1.31v* 0.64amps = 0.838 w

so the circuit with three light bulbs its producing more watts then the two bulb circuit

the next method was to measure the voltage in the circuit to measure the available voltage.

Battery positive = 11.98v

input of the switch = 11.96v

out put of the switch = 11.95v

supply to the light bulb 1 = 11.94v

output of light bulb 1 = 1.96v

input  to light bulb 2 = 1.95v

output of light bulb 2 = 1.78v

input to light bulb 3 = 1.77v

output of light bulb 3 = 0.02

And at the negative of the suply. = 0.01

Different between voltage drop and availabe voltage?
voltage drop is how much voltage is been used up to do the work
available voltage is how much voltage available to do the work