I am reading here about pull-down resistors.
Question 1.
The circuit schematic of a light sensor using a voltage divider circuit is shown as:
(source: doctronics.co.uk)
LDR has a resistance of 0.5k\$ \Omega \$ in bright light, and \$ 200k\Omega\$ in the shade.
In the shade, \$V_{out}\$ will be: \$\dfrac{200}{210} \times9 \times \dfrac{k\Omega}{k\Omega}\times V =8.57V\$
In the bright light, \$V_{out}\$ will be: \$\dfrac{0.5}{10.5} \times 9 \times \dfrac{k\Omega}{k\Omega}\times V=0.43V\$
So by applying the voltage divider rule we find that the circuit gives a high output voltage in the shade and a low output in bright light. So if we use this circuit with a bulb connected at the output then in the night the bulb should glow.
There is a problem. The bulb which is to be connected has its own resistance which might be lower than \$200k\Omega\$. Let's say the bulb to be connected has resistance \$100 \Omega\$. Since the bulb is in parallel with the LDR, so in the night the equivalent resistance of this parallel combination is approximately \$100\Omega\$.
Applying the voltage divider rule we will find that the bulb will not glow in both, dark and bright light. So the circuit is impractical and is of no use.
My question is:
- Does the voltage divider circuit have no practical importance/use?
Question 2.
Here it is explained that we usually use a pull down resistor of very high resistance of nearly \$10k\Omega\$. Rather using a high resistance resister we can left open the terminals at which pull-down resister is connected. By doing so we will obtain \$\infty\$ resistance for pull-down resistor and whole of the \$V_{in}\$ will appear at \$V_{out}\$.
Why we use a resister of \$10k\Omega\$ not an open circuit for pulling whole of the \$V_{in}\$ at \$V_{out}\$.