Darlington Pair with LDR

but changed the
10 k resistor - 1k resistor
10k pot - 1k pot

Darlington pair

 










470 instead of 560
68k instead of 100k

Heres what I made

LDRs and Potentiometers

Q1. How does the resistance on the LDR change with light illumination on the LDR?
Q2. Why do we need a protection resistor here?
Q3. Why do we always need a protection resistor for LEDs?
Q4. What are the three legs for on the 'Pot'?
Q5. What is the difference between a 10K pot and a 1K pot?
Q6. Why would a 10K pot be better for this LDR than a 1K pot?
Q7. What is a thermistor?
Q8. What is the resistance range of the LDR?

Resistor -
             Resistance 560 Ω
Potentiometer -
             Ressistance Code 120 Ω
             Ressistance - 1k

1) Normal light - 1.56 Ω
   light right on top - 0.30 Ω
    light very close - 0.56 Ω
    short distance - 0.63 Ω
    medium distance - 0.99 Ω 
    long distance -  1.44 Ω
    finger over the LDR - 3.25 Ω

2)  your using a protection resistor to protect the potentiometer . Because its small like you heat up quickly. and if it has a low ressistance and a high current so it'll blow.

3) to keep the current high so the resisstor doesnt blow. (to keep a high resiance so a low current)

4) there are 3 legs
- the top middle one is input
the bottom 2 (bottom left and bottom right) are both output (ground)
 
5) one has a bigger maximum resistance
 
6) to be able to have a bigger maximum resistance
 
7) A thermistor is a type of resistor whose resistance varies significantly with temperature, more so than in standard resistors.

8) the range is 20 - 20000k Ω

Questions from book

pg 54

9) 1.22 V
10) 3.050 V
11) 3.54 V
12) 22.5 V
13) 41.612 V
16) R1 - 180 Ω
      R2 - 180 Ω
17) R1 - 40 Ω
       R2 - 60 Ω
19) 24.05 Ω
20) 80 V

V=IR equation practice

Here's some practice

questions from text book

9) 1.22 V
10) 7.77
11) 3.54

Parallel

Here a photo of my parallel curcuit. 

Questions for today

1) 

R1 - 270 Ω

R2 - 470 Ω

R3 - 1k Ω

V0 - 6.05 V

V1 - 0.15 V 

V2 - 0.26 V

V3 - 5.63 V

To work out current...

V/R = I

6.05 / 1,740Ω = 0.00347701149...

To work out power...

VI = P

0.003... X 6.05 = 0.02103591954

2)

I1 - 

I2 - 0.6 mA

I3 - 

Some Equations

power = voltage*current

voltage = energy/charge

current = charge/time

Voltage experiment (Question 1)

1)

V0 = 5.59 

Predicted (V1) = 470/10470 X 5.59 = 0.2509

Predicted (V2) = 10000/10470 X 5.59 = 5.3390

V1 = 0.25 V

V2 = 5.34 V

R1 = 470 Ω

R2 = 10 k Ω

Voltage experiment

 Here is the experiment we are doing this lesson

 

Questions

1)60 Watts / 230 Volts = 0.26 Amps

2) Will finish next lesson (5 ish hours)

Power

Power = Voltage x Current

Here's a simple way of explaining

The Questions On Voltage

Here are the questions

Questions On Voltage

1) what are the voltage         (All to 3SF)
A)       
V1 - 0.82  9/11
V2 - 8.18  90/11

B)
V1 - 4.55  50/11
V2 - 5.45  60/11

C)
V1 - 2.22  20/9
V2 - 3.33  10/3
V3 - 4.44  4011