You need to change the equation of P=VI to
P=I^2R and since there is 3 cells in total connected in series their potential difference will be adding up and when we use this logic to the power as well, we can say the total output of 3 cells will be P=I^2Rx3 and you can get 1.68mW. Then you have the power generated by the laser which is 3.96 mW and when you divide 1.68mW to 3.96mW you get 0.424242 which is basically 0.40 when you round them up.
Original post by Amamiya
What am I doing wrong for ii)? I cant seem to get efficiency of 40%
I am also stuck on this. Please let us know if you find the answer to this question.
(Original post by shabnam.kavakeb)I am also stuck on this. Please let us know if you find the answer to this question.
Hi, I do not understand the way you think of the useful power and total power. what is the useful power?I thought it is the power from the light of the laser so it should go to the numerator and not the denominator. !!!!!
Hi, I do not understand the way you think of the useful power and total power. what is the useful power?I thought it is the power from the light of the laser so it should go to the numerator and not the denominator. !!!!!
Original post by Amamiya
What am I doing wrong for ii)? I cant seem to get efficiency of 40%
Before I explain part (ii), I want to point out that your internal resistance of the cell is incorrect. I believe is a careless mistake.
For part (ii), the efficiency of the laser point is NOT approximately 40%. MS says “analysis of data in passage to give efficiency of laser as approximately 40%”.
There is a passage before the question which is not shown by you but is shown below. It is from the passage that you deduce the efficiency of the laser to be approximately 40%.
MS mentions that “calculation of efficiency of pointer (1)”. This is where you do the actual calculation of the efficiency. Your calculations are mostly correct. As mentioned by the user lheeseung, there are 3 cells, so the total power transferred to the laser point is 3IV or 3 × 0.0518 W.
Hope it makes sense now.
Original post by lheeseung_05
You need to change the equation of P=VI to
P=I^2R and since there is 3 cells in total connected in series their potential difference will be adding up and when we use this logic to the power as well, we can say the total output of 3 cells will be P=I^2Rx3 and you can get 1.68mW. Then you have the power generated by the laser which is 3.96 mW and when you divide 1.68mW to 3.96mW you get 0.424242 which is basically 0.40 when you round them up.
You need to change the equation of P=VI to
P=I^2R and since there is 3 cells in total connected in series their potential difference will be adding up and when we use this logic to the power as well, we can say the total output of 3 cells will be P=I^2Rx3 and you can get 1.68mW. Then you have the power generated by the laser which is 3.96 mW and when you divide 1.68mW to 3.96mW you get 0.424242 which is basically 0.40 when you round them up.
I disagree with your working and explanation.
There is no need to use and cannot use P = I2R because we don’t know the resistance of the “laser pointer”.
When you use P = I2R to do the calculation, you are computing the power dissipated in the cell based on the numerical values that you are using. However, the internal resistance of each cell is indeed 28.6 Ω shown in the calculation to part (i), but not sure why you change to 2.86 Ω. This is the “wasted” power in the cells NOT power transferred to the laser pointer or power dissipated in the laser pointer.
The efficiency of the laser pointer is NOT 40%.
Original post by Eimmanuel
Before I explain part (ii), I want to point out that your internal resistance of the cell is incorrect. I believe is a careless mistake.
For part (ii), the efficiency of the laser point is NOT approximately 40%. MS says “analysis of data in passage to give efficiency of laser as approximately 40%”.
There is a passage before the question which is not shown by you but is shown below. It is from the passage that you deduce the efficiency of the laser to be approximately 40%.
MS mentions that “calculation of efficiency of pointer (1)”. This is where you do the actual calculation of the efficiency. Your calculations are mostly correct. As mentioned by the user lheeseung, there are 3 cells, so the total power transferred to the laser point is 3IV or 3 × 0.0518 W.
Hope it makes sense now.
For part (ii), the efficiency of the laser point is NOT approximately 40%. MS says “analysis of data in passage to give efficiency of laser as approximately 40%”.
There is a passage before the question which is not shown by you but is shown below. It is from the passage that you deduce the efficiency of the laser to be approximately 40%.
MS mentions that “calculation of efficiency of pointer (1)”. This is where you do the actual calculation of the efficiency. Your calculations are mostly correct. As mentioned by the user lheeseung, there are 3 cells, so the total power transferred to the laser point is 3IV or 3 × 0.0518 W.
Hope it makes sense now.
I'm still a bit confused on how you use the '50% less electricity' bit. I don't get calculation I need to do.
Original post by Amamiya
I'm still a bit confused on how you use the '50% less electricity' bit. …
MS says “analysis of data in passage to give efficiency of laser as approximately 40%”.
Original post by Amamiya
… I don't get calculation I need to do.
Your calculations seem to be good. What is it that you don’t get? Please be specific.
Your calculation shows that the efficiency of the laser is approximately 3% which is much lower than the efficiency of an industrial laser which is the last bit that the MS says “comparing data analysed to draw a conclusion that the claim is not justified.”
Original post by Eimmanuel
As mentioned by the user lheeseung, there are 3 cells, so the total power transferred to the laser point is 3IV or 3 × 0.0518 W.
As mentioned by the user lheeseung, there are 3 cells, so the total power transferred to the laser point is 3IV or 3 × 0.0518 W.
Original post by Eimmanuel
However, the internal resistance of each cell is indeed 28.6 Ω shown in the calculation to part (i), but not sure why you change to 2.86 Ω.
Just to point out, but the question gives the voltage across cells plural, not individually. And asks for the internal resistance of battery of cells plural, not of each individual cell. So there should be no multiplication by 3 for P = IV.
The start of the question was kind of misleading in saying 'A student uses a laser pointer to measure the internal resistance of a cell' singular, when the rest of the questions just gives info on and asks only about the battery of 3 cells as a whole
Original post by AGrizzlyBearo
Just to point out, but the question gives the voltage across cells plural, not individually. And asks for the internal resistance of battery of cells plural, not of each individual cell. So there should be no multiplication by 3 for P = IV.
The start of the question was kind of misleading in saying 'A student uses a laser pointer to measure the internal resistance of a cell' singular, when the rest of the questions just gives info on and asks only about the battery of 3 cells as a whole
The start of the question was kind of misleading in saying 'A student uses a laser pointer to measure the internal resistance of a cell' singular, when the rest of the questions just gives info on and asks only about the battery of 3 cells as a whole
Indeed, you got the point. Thanks.
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