Cathy's Question: Motor on/off with LDR and Resistor?

AI Thread Summary
The discussion centers around a potential divider circuit involving a Light Dependent Resistor (LDR) and a fixed resistor, with the goal of determining if a motor activates under different light conditions. When the LDR is dark, the resistance values of 3000 Ω, 4000 Ω, and 5000 Ω are evaluated to see if they allow 4.0 V across the motor. In light conditions, similar evaluations are made for resistance values of 300 Ω, 400 Ω, and 500 Ω. Additionally, the impact of repositioning the motor across the LDR is considered, prompting a need for a specific resistance value for activation in light. The discussion emphasizes the importance of calculating the potential difference to ensure the motor receives adequate voltage.
CathyLou
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I've no idea how to start the following question so any help or guidance would be really appreciated.

A potential divider circuit is set up with a LDR and a fixed resistor, R. The LDR has a resistance of 200 Ω when it is light and 2500 Ω when it is dark. A motor is set so that it comes on if 4.0 V is applied across it. A 6.0 V battery provides the emf (assume it has no internal resistance).

When it is dark, does the motor come on if R is (a) 3000 Ω (b) 4000 Ω (c) 5000 Ω?

When it is light, does the motor come on if R is (a) 300 Ω (b) 400 Ω (c) 500 Ω?

If the motor were moved so that it is across the LDR instead, what would R have to be so that the motor came on in the light?


Thank you.

Cathy
 
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need to post at least some thought about this: maybe you don't have a diagram, and that is the issue: I read it as such. The diffference between
v1 and v3 is just the applied EMF of 6Volts.

V1-----R------v2------Rx------v3. where Rx=variable depending on whether dark or light. So for the potential difference between v1 and v2, to be >=4v,
what must R be. Does this help at all?
 
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