# PhysicsWatch

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#1
I am not sure how to do this question.
Last edited by average_human; 2 weeks ago
0
2 weeks ago
#2
Part A:
Is a series circuit so voltage will be shared between components according to there resistance. Ratio of 5:1 therefore the LDR will take 5/6 of the total voltage = 5V
0
#3
(Original post by seabasssss)
Part A:
Is a series circuit so voltage will be shared between components according to there resistance. Ratio of 5:1 therefore the LDR will take 5/6 of the total voltage = 5V
Thank you! Do you know how to work out parts b and c
0
2 weeks ago
#4
You got b correct
0
2 weeks ago
#5
For part C, i think the answer would be 0 because with no resistance the LDR isn't taking any share of the voltage so it is effectively not there.
Hope this helps but i'm not completely certain!
0
2 weeks ago
#6
(Original post by seabasssss)
For part C, i think the answer would be 0 because with no resistance the LDR isn't taking any share of the voltage so it is effectively not there.
Hope this helps but i'm not completely certain!
0
2 weeks ago
#7
LDR goes from 500ohms when dark to 0 when bright. 100 ohm one doesn’t change. Series so RT= sum of resistances. V=IR so when dark I = V/(100+500) and when bright I=V/100 I think fixed resistor was 100ohms if not replace its resistance into equation. As light intensity increases current increases and resistance decreases. Need anymore help, just ask
0
2 weeks ago
#8
Also V across each component = IR where R is resistance of that component so when bright LDR takes no voltage as R=0ish
0
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