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So guyz, how was the exam?? Post doubts and your expected grade! GL TO ALL
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Physics AS Paper 21 GT and Doubts
- Thread starter Ahmed Khattab
- Start date
- Messages
- 124
- Reaction score
- 343
- Points
- 73
What was the answer to that power dissipation of each bulb part?
i got 19 =|
i got 19 =|
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- 16
- Reaction score
- 15
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I got like a large number but dw your given a mark if you right the formulae so even if u got it wrong maximum 1 mark lost
- Messages
- 124
- Reaction score
- 343
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- 73
Dude i'm not even sure about my formulae. I calculated the power dissipated in the battery ( due to internal resistance) and subtracted from 48.
Is that correct?
Is that correct?
- Messages
- 16
- Reaction score
- 15
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I did p=VI since we were given both the voltage and we got the current in the second question.
- Messages
- 124
- Reaction score
- 343
- Points
- 73
Me too! & i was getting exactly 48. but then the ques. said that power dissipated in each bulb is less than 48, so i changed my ans.
inshaaAllah kher.
inshaaAllah kher.
- Messages
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- Reaction score
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it is 27 w for sure
Answer was 27w.
Let me explain why using Kirchoff's first and second law.
First, (conservation of charge):
6 Amps flowed around the circuit. This implies, because it is parallel, each lamp would get 3A. Using I^2 x R for Power, it would be 3^2 x 3 = 27
Now with 2nd law (conservation of energy)
Voltage produced by battery : 12 V. V = E - Ir E = 12 I = 6 and r = 0.5. 12-6x0.5= 9
Using the equation V^2/R, it is 9^2 / 3 = 27w.
There you go ^^
Let me explain why using Kirchoff's first and second law.
First, (conservation of charge):
6 Amps flowed around the circuit. This implies, because it is parallel, each lamp would get 3A. Using I^2 x R for Power, it would be 3^2 x 3 = 27
Now with 2nd law (conservation of energy)
Voltage produced by battery : 12 V. V = E - Ir E = 12 I = 6 and r = 0.5. 12-6x0.5= 9
Using the equation V^2/R, it is 9^2 / 3 = 27w.
There you go ^^