Laboromniavincit
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I have no idea how to do this please could someone helpName:  1DB5B5AC-FE45-4926-97BD-0F3278B4CCA6.png
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Athy_Obelia
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It wouldn't be very helpful if I straight out give you the answer, so I suggest you start off by looking at revision resources for that section.
One that's relatively short that I recommend is: https://www.physicsandmathstutor.com...%2520Notes.pdf
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Laboromniavincit
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Last edited by Laboromniavincit; 1 month ago
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Laboromniavincit
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How do I solve this?
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Callicious
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(Original post by Laboromniavincit)
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How do I solve this?
Make a new thread for each new question you have to ask (see posting guidelines.)

(Original post by Laboromniavincit)
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For future reference it's very good to correctly orientate your images that you upload- really helps people out. Also writing answers on a clean sheet helps keep things cleaner- there's not a lot of space on that one. Anyway, on we go...

For (d):
Putting aside the outside lamp circuit (you can't solve/deal with that- no information, so treat this as a series problem) ...

- You should know what an LDR is and how it responds to luminosity, specifically how with a reduction in luminosity, the resistance will change: this tells you if R increases/decreases. This should automatically tell you how V changes (see https://www.thestudentroom.co.uk/sho....php?t=7111648 and my post therein on how to use Kirchoff- the third equation will tell you how V depends on the "fractional contribution" of resistance of a series component to total resistance.)
- Once you find the way PD changes and R changes, you'll note that the total circuit resistance has changed. Has it gone up or down? Does that mean that the current (SERIES) has gone up or down?

For (e):

Again see https://www.thestudentroom.co.uk/sho....php?t=7111648 and if you understand Kirchoffs laws, this is just using the voltage/resistance ratio method I put in there. Pangol does a worked example on that post, too

For (f):

Literally V=IR with the total resistance of the circuit and the total PD of the circuit.


For the Google spiders:
A light dependent resistor (LDR) is used to turn on an outside lamp when it gets dark. Part of the circuit is shown in Figure 2. The light intensity decreases, what happens to the potential difference across the LDR and the current in the LDR?
Last edited by Callicious; 1 month ago
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Laboromniavincit
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#6
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(Original post by Callicious)
Make a new thread for each new question you have to ask (see posting guidelines.)


Name:  image (1).jpg
Views: 9
Size:  179.1 KB
For future reference it's very good to correctly orientate your images that you upload- really helps people out. Also writing answers on a clean sheet helps keep things cleaner- there's not a lot of space on that one. Anyway, on we go...

For (d):
Putting aside the outside lamp circuit (you can't solve/deal with that- no information, so treat this as a series problem) ...

- You should know what an LDR is and how it responds to luminosity, specifically how with a reduction in luminosity, the resistance will change: this tells you if R increases/decreases. This should automatically tell you how V changes (see https://www.thestudentroom.co.uk/sho....php?t=7111648 and my post therein on how to use Kirchoff- the third equation will tell you how V depends on the "fractional contribution" of resistance of a series component to total resistance.)
- Once you find the way PD changes and R changes, you'll note that the total circuit resistance has changed. Has it gone up or down? Does that mean that the current (SERIES) has gone up or down?

For (e):

Again see https://www.thestudentroom.co.uk/sho....php?t=7111648 and if you understand Kirchoffs laws, this is just using the voltage/resistance ratio method I put in there. Pangol does a worked example on that post, too

For (f):

Literally V=IR with the total resistance of the circuit and the total PD of the circuit.


For the Google spiders:
A light dependent resistor (LDR) is used to turn on an outside lamp when it gets dark. Part of the circuit is shown in Figure 2. The light intensity decreases, what happens to the potential difference across the LDR and the current in the LDR?
Omg you literally saved my life
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