Saturday, August 6, 2011

compound ( series - parallel ) circuits

TTEC 4841 - Electrical and Electronics- Practical

5/8/2011 friday
start = 1.00pm finsh = 4.30pm



compound ( series - parallel ) circuits

To days task was how to  created a circuit that has a light bulb in series after two light bulbs in parallel. Then we measured the available voltage at different parts of the series and parallel the measured the volts for

Switch = 12.54v

before parallel light bulb 1 = 12.52v

before parallel light bulb 2 = 12.52v

after parallel light bulb 1 = 8.55v

after parallel light bulb 2 = 8.55v

before series light bulb = 8.55v

after series light bulb 0.00v

Then measured the voltage drop at different parts of the series and parallel to get the voltage drop which is

Across the parallel light bulb 1 = 3.94v

Across the parallel light bulb 2 = 3.94v

Across the series light bulb = 8.42v


Amps
After that we measured the current flowing throught the circuit which is

The switch = 0.60amps
parallel light bulb 1 = 0.17amps
parallel light bulb 2 = 0.43amps
series light bulb =  0.60amps

Watts
calculated watts used at each light bulb which is W=V*I
bulb 1 = 3.83v* 0.17amps = 0.65w
bulb 2 = 3.83v* 0.43amps =  1.64w
series bulb = 8.55v* 0.60amps = 5.13 w
the parallel has the same resistance and the same watts going in where the series has less watts and more resistance so Amps are different as amps depend on the resistance.Voltage is divided between the series bulb and the two parallel bulbs.

1 comment:

  1. good posting-- but what were your results i.e total resistance of the circuit-- how did you calculate it-- total amps and what happens in the compound circuit

    ReplyDelete