Experiment no.5 observation pictures

Picture no.1

100uF, 1k ohm resistor, 1v,

0.5seconds

Picture no.2

100uF, 0.1 ohm resistor, 1v,

0.05seconds

Picture no.3

100uF, 0.47k ohm resistor, 1v,

0.235seconds

Picture no.4

330 uF, 1k ohm resistor, 1v,

1.65seconds

NOTE:

The less resistance in the circuit the faster the current travels in time (ms) to complete the circuit,

Experiment no.5 ( capacitor )

Components:

1x resistor, 1x capacitor, 1x push button swicth (bridged wire)

Capacitor circuit:

Time taken to charge capacitor

Working:

no.1)
100Exp-6 x 1000 x 5
=0.5s x1000

= 500ms

no2)

100Exp-6 x 100 x 5

=0.05s x1000

= 50ms

no.3)

100Exp-6 x 470 x 5

=0.235 x1000

= 235ms

no.4)
330Exp-6 x 1000 x 5

= 1,65 x 1000

=1650ms

Experiment no.4 ( 5,1 zener diode )

Components: 1x resistor, 1x 5v1 400mW zener diode, 1x diode 1N4007

NOTE:

Readings are as such due to the zener diode being in reverse bias and the second diode in forward bias

Resulting in the 10 volt circuit to have a reading of 10.07 mA, and the 15volt circuit having 15.48 mA

Experiment no.3 ( 5.1 zener diode )

Components : 2x resistors,1x 5.1 400mW zener diode (Zd)


Vz = zener voltage
Vs = voltage supply

The value of Vz : 5volts
Vs now varys from 10 -15 v
What is Vz now: 5volts


NOTE:

Component voltage dos'nt change as this voltage amount is to operate the component, Zener diode has a reverse bias of 5.1 v and a forward bias 0.7v

Changing the supply voltage wont change a diodes voltage, only changing the diodes forward or reverse bias wil change its voltage drop.


Zener diodes are good for circuits containing different voltage components.


Reversing the polarity of the zener diode:
0.846 was measured in this reversed polarity giving a forward bias, which should read between 0.6-0.7, the reading 0.8 would mean there was a leak in the diode exceding its forward bias voltage.

experiment no.2 ( diodes )

Diodes are symbolised by a white ring on the diode, closet to the cathode end indicating the negative end of the diode.

When the cathode end of the diode is connected to the earth of the power supply the diode is in forward bias
Reverse bias is when the diode is conected back to front

calculate the value of current flowing through the diode:

calculated : i=v/r = 5.02-0.7/1000 = 0.00432 = 4.32 ma

measured : 4.35 ma

To find the amount of current flowing through the diode, i

used the formula i = v/r , dividing the voltage by the resistance giving me the current .

calculate the voltage drop across the diode :

calculated : v= i x r =

measured : 0.65 v

maxium value of current flow through the diode:

Ta=1.0@75degreesceluis

maxium value of supply voltage (Vs) at which the diode is safe to run:

v=ixr= 1x1000 = 1000@75degreesceluis

Diode is replaced by LED :

calculated: 5.02 -1.8/1000 = 3.22/1000 = 0.00322 ma

measured: 3.3 ma

NOTE: LED circuit uses more volts but less amperage than the diode.

Experiment no.1 (resistors)

NOTE:

In a series circuit resistance is added together to giving a prpportional decrease in the flow of current ,in addtion a lower resistance reading.

resistance in a parallel circuit
-

Working:

Resistor 1 : 98900 ohms

Resistor 2 : 389 ohms

Both in series :
calculated : 99289 ohms

measured : 99200 ohms

Both in parallel :

calculated : 387.48 ohms

measured : 388 ohms