Electricity question help

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maxcovel39
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#1
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#1
Could someone help me calculate and explain how the voltage across each lamp is calculated??? I'm so confuseddd. i attached a link below.

https://imgur.com/a/8slXq2y
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Arkcano
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#2
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Okay so since the lamps are in parallel their resistance is calculated as 1/R1 + 1/R2
then you take the reciprocal of that value. eg you got the answer 1/12 + 1/12 = 1/6 you flip it and you get 6. So 6 ohms is the resistance. The fixed resistor is in series with the lamps so you need to realize that voltage will be split in a series circuit. the ratio is 12 ohms (resistor) : 6 ohms ( both lamps in parallel)

so both lamps will get 6/(6+12) * 15 volts = 5 volts
therefore each lamp gets 5/2 = 2.5 volts

Please let me know if this is correct
Last edited by Arkcano; 4 weeks ago
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JinTech
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#3
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(Original post by Arkcano)
Okay so since the lamps are in parallel their resistance is calculated as 1/R1 + 1/R2
then you take the reciprocal of that value. eg you got the answer 1/12 + 1/12 = 1/6 you flip it and you get 6. So 6 ohms is the resistance. The fixed resistor is in series with the lamps so you need to realize that voltage will be split in a series circuit. the ratio is 12 ohms (resistor) : 6 ohms ( both lamps in parallel)

so both lamps will get 6/(6+12) * 15 volts = 5 volts
therefore each lamp gets 5/2 = 2.5 volts

Please let me know if this is correct
I thought both lamps would have 5 volts each since the voltage across components in parallel are the same
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Ethanity007
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#4
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#4
The two lamps in // have a effective resistance = 6 ohms
Using potential divider principle,
the pd across the lamp = 6 / (6 + 12) x 15 = 5 V
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Arkcano
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#5
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#5
(Original post by JinTech)
I thought both lamps would have 5 volts each since the voltage across components in parallel are the same
Sorry sorry my bad its 5 volts I accidentally divided by 2. Its been 3 years since I did this topic.
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