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HELPP!!! Physics a level electricity question
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Hi
Ok so idk how to paste the question cos tsr won't let me so here's the link to the paper. It's question 30.
https://revisionscience.com/sites/re...n-paper-12.pdf
Can someone help me with this physics question? Been stumped for a while. I was thinking that the answer would be C because if B3 is opened more current would go through both B1 and B2 so they would both increase brightness but this is wrong.
The correct answer is B. Please explain this.
Much appreciated
Ok so idk how to paste the question cos tsr won't let me so here's the link to the paper. It's question 30.
https://revisionscience.com/sites/re...n-paper-12.pdf
Can someone help me with this physics question? Been stumped for a while. I was thinking that the answer would be C because if B3 is opened more current would go through both B1 and B2 so they would both increase brightness but this is wrong.
The correct answer is B. Please explain this.
Much appreciated
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Here's the picture for ease of access:
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Ok, so initially when the switch is closed, B2 and B3 are in parallel, so we can use the formula to calculate the resistance (1/R = 1/R1 + 1/R2 + ...), which means that the combined resistance is lower than the sum of the individual resistances of B2 and B3.
As the lamps are identical, their resistances are the same, so the total resistance of the circuit is R+1/(1/R+1/R) = R + 1/(2/R) = R + R/2 = 3R/2, 66.67% of that is B1, 33.33% of that is B2 and B3 combined.
As the resistance across B1 is initially higher than that of B2, the P.D. across B1 is high, and the P.D. across B2 is low (and the P.D. across B3 is also low, but that isn't relevant). This means that the brightness of B1 is higher than that of B3.
Once you open the switch, B3 goes out, and there is just a series circuit with B1 and B2, so the P.D. is equal across both of them (as their resistances are now the same). This means that B1's resistance has decreased to 50%, and B2's resistance has increased to 50%, so the P.D. changes accordingly, which means that the brightness of B1 decreases, and the brightness of B2 increases.
Hope this makes sense, if not, let me know and I'll explain in more detail
Ok, so initially when the switch is closed, B2 and B3 are in parallel, so we can use the formula to calculate the resistance (1/R = 1/R1 + 1/R2 + ...), which means that the combined resistance is lower than the sum of the individual resistances of B2 and B3.
As the lamps are identical, their resistances are the same, so the total resistance of the circuit is R+1/(1/R+1/R) = R + 1/(2/R) = R + R/2 = 3R/2, 66.67% of that is B1, 33.33% of that is B2 and B3 combined.
As the resistance across B1 is initially higher than that of B2, the P.D. across B1 is high, and the P.D. across B2 is low (and the P.D. across B3 is also low, but that isn't relevant). This means that the brightness of B1 is higher than that of B3.
Once you open the switch, B3 goes out, and there is just a series circuit with B1 and B2, so the P.D. is equal across both of them (as their resistances are now the same). This means that B1's resistance has decreased to 50%, and B2's resistance has increased to 50%, so the P.D. changes accordingly, which means that the brightness of B1 decreases, and the brightness of B2 increases.
Hope this makes sense, if not, let me know and I'll explain in more detail
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Thanks so much xx. Explained it really well. You made more sense than my teacher.
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(Original post by samreh)
Thanks so much xx. Explained it really well. You made more sense than my teacher.
Thanks so much xx. Explained it really well. You made more sense than my teacher.
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