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Standard Grade Physics. Unit 4 Electronics. Exercise. Label the following signals as analogue or digital. ( a ) ( b ) ( c ). analogue. analogue. digital. analogue. digital. analogue.
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Standard Grade Physics Unit 4 Electronics
Exercise Label the following signals as analogue or digital. (a) (b) (c) analogue analogue digital
analogue digital analogue (d) (e) (f)
Label following devices as analogue or digital. (a) (b) analogue digital (c) (d) digital analogue
sound digital sound analogue kinetic(rotation) analogue light digital (analogue with variable R) light digital light digital kinetic digital kinetic (in straight line) digital
The solenoid.Set up the circuit as shown below: Solenoid Unit 5 V Touch lead here “flying” lead 0 V
Negative sign on its side. LED only allows current to flow and light up when “negative connected to negative”.
on off on For a current to flow there has to be a difference in voltage. on off
10 VS– VLED 0∙015 12 – 2 = 10 V 667 Ω
VR = I R VS– VR VR = 20 × 10-3 × 140 5 – 2∙8 = 2∙2 V VR = 2∙8 V
0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1
T.U.R.D. temperature up resistance down L.U.R.D. light up resistance down
sound to electrical light to electrical heat to electrical
Capacitor Investigation V Potential divider 5 V resistor component investigations 0 V capacitor switch With power off, discharge capacitor using switch so that V = 0 V. When power is switched on, start timer.
Conclusions: The time to charge a capacitor depends on the values of the capacitance and the series resistor. If the value of R and/or C are increased, the time taken to reach the final voltage also increases. Comparison with waterCopy diagrams from P.T.A. page 107
Discharging a capacitor The quickest and easiest way to discharge a capacitor is to place a wire across both ends of it. 5V 0V charged discharged
Set up the apparatus as shown below: Potential divider 5 V V1 resistor R1 V2 0 V resistor R2 switch Complete the table by measuring the voltage V1 and V2 for each pair of resistors.
V/R = 12/36 (leave as fraction to avoid rounding off) 12/36 x 24 = 8 V 12/36 x 12 = 4 V 8 244 = 12 = 2 (check!)
Step 1. RT = 1k + 5k = 6 kΩ Step 2. I = V/R = 4∙5/6000 Step 3. V = I R V = 4∙5/6000x 5000 V = 3∙75 V
Voltage across R= 6 – 2 = 4 V V1 R1V2= R2 4 R2 = 4 2 x R = 4 x 4 R = 16/2 = 8 Ω
Adjust the knob on the potentiometer until Vbe = required value. Now measure the corresponding Vout. 5 V Potential divider Transistor Vout 0 V 4∙7 kpotentiometer Vbe
Now draw a graph of your results. Vout (V) ON OFF Vbe (V) 0∙7 V
A current cannot flow through the collector unless a current flows through the base. For a current to flow through the base, Vbe≥ 0∙7 V.
RT = 1800 + 200 = 2 kΩ I = V/R = 5/2000 V1 = Vbe = I R Vbe = 5/2000x 200= 0∙5 V Off 5 – 0∙5 = 4∙5 V If temperature increases, resistance of thermistor decreases V across thermistor decreases and V across 200Ω increases V2 decreases and V1increases (V2 + V1 = 5 V)
0 V 5 V 2∙5 V
Remember:When the voltage divides, the resistor with the biggest value will take the biggest share of the voltage. high high low low low high
A temperature sensor.Set up the circuit as shown below: Potential divider Transistor 5 V thermistor 0 V 4∙7 k pot. Adjust potentiometer until LED is just off. Now warm thermistor by rubbing with your finger.
decreases V across thermistor decreases V across variable R increases Vbe increases Vbe≥ 0∙7 V, transistorswitches ON LED is ON reducing
high high low low low high dark
A light sensor.Set up the circuit as shown below: Potential divider Transistor 5 V 4∙7 k pot. 0 V LDR Adjust potentiometer until LED is just off. Now cover the LDR with your finger.
increases voltage across LDR increases Vbe increases Vbe≥ 0∙7 V transistor switches on LED is on swap the positions of the LDR and variable resistor.
The moisture unit.Set up the circuit as shown below: 200 k Ωsetting Rain sensing unit Ω Observe what happens to the reading on the ohmmeter when water is added to the moisture unit.